面向对象程序设计:c++语言描述(第4版)


Robert Lafore 800 East 96th St.,Indianapolis,Indiana 46240 USA Object-Oriented Programming in C++, Fourth Edition 00 3087 FM 11/29/01 2:15 PM Page iCopyright  2002 by Sams Publishing All rights reserved. No part of this book shall be reproduced, stored in a retrieval system, or transmitted by any means, electronic, mechanical, photo- copying, recording, or otherwise, without written permission from the pub- lisher. No patent liability is assumed with respect to the use of the information contained herein. Although every precaution has been taken in the preparation of this book, the publisher and author assume no responsibility for errors or omissions. Nor is any liability assumed for damages resulting from the use of the information contained herein. International Standard Book Number: 0-672-32308-7 Library of Congress Catalog Card Number: 2001094813 Printed in the United States of America First Printing: December 2001 04 03 02 01 4321 Trademarks All terms mentioned in this book that are known to be trademarks or service marks have been appropriately capitalized. Sams Publishing cannot attest to the accuracy of this information. Use of a term in this book should not be regarded as affecting the validity of any trademark or service mark. Warning and Disclaimer Every effort has been made to make this book as complete and as accurate as possible, but no warranty or fitness is implied. The information provided is on an “as is” basis. The author and the publisher shall have neither liability nor responsibility to any person or entity with respect to any loss or damages arising from the information contained in this book. EXECUTIVE EDITOR Michael Stephens ACQUISITIONS EDITOR Michael Stephens MANAGING EDITOR Matt Purcell PROJECT EDITORS Angela Boley Christina Smith INDEXER Rebecca Salerno PROOFREADER Matt Wynalda TECHNICAL EDITOR Mark Cashman TEAM COORDINATOR Pamalee Nelson MEDIA DEVELOPER Dan Scherf INTERIOR DESIGNER Gary Adair COVER DESIGNER Alan Clements PAGE LAYOUT Ayanna Lacey 00 3087 FM 11/29/01 2:15 PM Page iiOverview Introduction 1 1The Big Picture 9 2 C++ Programming Basics 29 3 Loops and Decisions 75 4Structures 131 5 Functions 161 6 Objects and Classes 215 7Arrays and Strings 263 8 Operator Overloading 319 9 Inheritance 371 10 Pointers 429 11 Virtual Functions 503 12 Streams and Files 567 13 Multifile Programs 633 14 Templates and Exceptions 681 15 The Standard Template Library 725 16 Object-Oriented Software Development 801 A ASCII Chart 849 B C++ Precedence Table and Keywords 859 CMicrosoft Visual C++ 863 D Borland C++Builder 871 E Console Graphics Lite 881 F STL Algorithms and Member Functions 895 G Answers to Questions and Exercises 913 H Bibliography 977 Index 981 00 3087 FM 11/29/01 2:15 PM Page iiiContents Introduction 1 1 The Big Picture 9 Why Do We Need Object-Oriented Programming?..............................10 Procedural Languages ......................................................................10 The Object-Oriented Approach ........................................................13 Characteristics of Object-Oriented Languages......................................16 Objects..............................................................................................16 Classes ..............................................................................................18 Inheritance ........................................................................................18 Reusability........................................................................................21 Creating New Data Types ................................................................21 Polymorphism and Overloading ......................................................21 C++ and C..............................................................................................22 Laying the Groundwork ........................................................................23 The Unified Modeling Language (UML)..............................................23 Summary................................................................................................25 Questions................................................................................................25 2 C++ Programming Basics 29 Getting Started ......................................................................................30 Basic Program Construction..................................................................30 Functions ..........................................................................................31 Program Statements..........................................................................32 Whitespace........................................................................................33 Output Using cout ................................................................................33 String Constants................................................................................34 Directives ..............................................................................................35 Preprocessor Directives ....................................................................35 Header Files......................................................................................35 The using Directive..........................................................................36 Comments ..............................................................................................36 Comment Syntax ..............................................................................36 When to Use Comments ..................................................................37 Alternative Comment Syntax ..........................................................37 Integer Variables ....................................................................................38 Defining Integer Variables................................................................38 Declarations and Definitions ............................................................40 Variable Names ................................................................................40 Assignment Statements ....................................................................40 00 3087 FM 11/29/01 2:15 PM Page ivInteger Constants ..............................................................................41 Output Variations..............................................................................41 The endl Manipulator ......................................................................41 Other Integer Types ..........................................................................42 Character Variables................................................................................42 Character Constants..........................................................................43 Initialization......................................................................................44 Escape Sequences ............................................................................44 Input with cin ........................................................................................45 Variables Defined at Point of Use ....................................................47 Cascading << ....................................................................................47 Expressions ......................................................................................47 Precedence........................................................................................47 Floating Point Types ..............................................................................48 Type float........................................................................................48 Type double and long double ........................................................49 Floating-Point Constants ..................................................................50 The const Qualifier ..........................................................................51 The #define Directive......................................................................51 Type bool ..............................................................................................51 The setw Manipulator............................................................................52 Cascading the Insertion Operator ....................................................54 Multiple Definitions ........................................................................54 The IOMANIP Header File ..................................................................54 Variable Type Summary ........................................................................54 unsigned Data Types........................................................................55 Type Conversion ....................................................................................56 Automatic Conversions ....................................................................57 Casts..................................................................................................58 Arithmetic Operators ............................................................................60 The Remainder Operator..................................................................61 Arithmetic Assignment Operators....................................................61 Increment Operators ........................................................................63 Library Functions ..................................................................................65 Header Files......................................................................................66 Library Files ....................................................................................66 Header Files and Library Files ........................................................67 Two Ways to Use #include..............................................................67 Summary................................................................................................68 Questions................................................................................................69 Exercises ................................................................................................71 00 3087 FM 11/29/01 2:15 PM Page vOBJECT-ORIENTED PROGRAMMING IN C++, FOURTH EDITON vi 3 Loops and Decisions 75 Relational Operators ..............................................................................76 Loops......................................................................................................78 The for Loop....................................................................................78 Debugging Animation ......................................................................84 for Loop Variations..........................................................................84 The while Loop................................................................................86 Precedence: Arithmetic and Relational Operators ..........................89 The do Loop......................................................................................91 When to Use Which Loop................................................................93 Decisions................................................................................................93 The if Statement..............................................................................94 The if...else Statement ................................................................98 The else...if Construction ..........................................................106 The switch Statement ....................................................................107 The Conditional Operator ..............................................................111 Logical Operators ................................................................................114 Logical AND Operator......................................................................115 Logical OR Operator........................................................................116 Logical NOT Operator......................................................................117 Precedence Summary ..........................................................................118 Other Control Statements ....................................................................118 The break Statement ......................................................................119 The continue Statement ................................................................121 The goto Statement ........................................................................123 Summary..............................................................................................123 Questions..............................................................................................124 Exercises ..............................................................................................126 4 Structures 131 Structures ............................................................................................132 A Simple Structure ........................................................................132 Defining the Structure ....................................................................133 Defining a Structure Variable ........................................................134 Accessing Structure Members........................................................136 Other Structure Features ................................................................137 A Measurement Example ..............................................................139 Structures Within Structures ..........................................................141 A Card Game Example ..................................................................145 Structures and Classes....................................................................148 Enumerations ......................................................................................148 Days of the Week............................................................................148 One Thing or Another ....................................................................151 00 3087 FM 11/29/01 2:15 PM Page viCONTENTS vii Organizing the Cards......................................................................153 Specifying Integer Values ..............................................................155 Not Perfect......................................................................................155 Other Examples ..............................................................................155 Summary..............................................................................................156 Questions..............................................................................................156 Exercises ..............................................................................................158 5 Functions 161 Simple Functions ................................................................................162 The Function Declaration ..............................................................164 Calling the Function ......................................................................164 The Function Definition ................................................................164 Comparison with Library Functions ..............................................166 Eliminating the Declaration............................................................166 Passing Arguments to Functions..........................................................167 Passing Constants ..........................................................................167 Passing Variables ............................................................................169 Passing by Value ............................................................................170 Structures as Arguments ................................................................171 Names in the Declaration ..............................................................176 Returning Values from Functions ........................................................176 The return Statement ....................................................................177 Returning Structure Variables ........................................................180 Reference Arguments ..........................................................................182 Passing Simple Data Types by Reference......................................182 A More Complex Pass by Reference ............................................185 Passing Structures by Reference ....................................................186 Notes on Passing by Reference ......................................................188 Overloaded Functions ..........................................................................188 Different Numbers of Arguments ..................................................189 Different Kinds of Arguments........................................................191 Recursion ............................................................................................193 Inline Functions ..................................................................................195 Default Arguments ..............................................................................197 Scope and Storage Class......................................................................199 Local Variables ..............................................................................199 Global Variables..............................................................................202 Static Local Variables ....................................................................204 Storage............................................................................................205 Returning by Reference ......................................................................206 Function Calls on the Left of the Equal Sign ................................207 Don’t Worry Yet..............................................................................207 00 3087 FM 11/29/01 2:15 PM Page viiOBJECT-ORIENTED PROGRAMMING IN C++, FOURTH EDITON viii const Function Arguments ..................................................................208 Summary..............................................................................................209 Questions..............................................................................................210 Exercises ..............................................................................................212 6 Objects and Classes 215 A Simple Class ....................................................................................216 Classes and Objects........................................................................217 Defining the Class ..........................................................................218 Using the Class ..............................................................................221 Calling Member Functions ............................................................221 C++ Objects as Physical Objects ........................................................223 Widget Parts as Objects..................................................................223 Circles as Objects ..........................................................................224 C++ Objects as Data Types ................................................................226 Constructors ........................................................................................227 A Counter Example........................................................................228 A Graphics Example ......................................................................231 Destructors......................................................................................232 Objects as Function Arguments ..........................................................233 Overloaded Constructors ................................................................234 Member Functions Defined Outside the Class ..............................236 Objects as Arguments ....................................................................237 The Default Copy Constructor ............................................................238 Returning Objects from Functions ......................................................240 Arguments and Objects ..................................................................241 A Card-Game Example........................................................................243 Structures and Classes ........................................................................247 Classes, Objects, and Memory ............................................................247 Static Class Data..................................................................................249 Uses of Static Class Data ..............................................................249 An Example of Static Class Data ..................................................249 Separate Declaration and Definition ..............................................251 const and Classes ................................................................................252 const Member Functions ..............................................................252 const Objects ................................................................................255 What Does It All Mean?......................................................................256 Summary..............................................................................................257 Questions..............................................................................................257 Exercises ..............................................................................................259 00 3087 FM 11/29/01 2:15 PM Page viiiCONTENTS ix 7 Arrays and Strings 263 Array Fundamentals ............................................................................264 Defining Arrays ..............................................................................265 Array Elements ..............................................................................265 Accessing Array Elements..............................................................267 Averaging Array Elements ............................................................267 Initializing Arrays ..........................................................................268 Multidimensional Arrays................................................................270 Passing Arrays to Functions ..........................................................274 Arrays of Structures........................................................................277 Arrays as Class Member Data ............................................................279 Arrays of Objects ................................................................................283 Arrays of English Distances ..........................................................283 Arrays of Cards ..............................................................................286 C-Strings ..............................................................................................290 C-String Variables ..........................................................................290 Avoiding Buffer Overflow..............................................................292 String Constants..............................................................................292 Reading Embedded Blanks ............................................................293 Reading Multiple Lines..................................................................294 Copying a String the Hard Way ....................................................295 Copying a String the Easy Way......................................................296 Arrays of Strings ............................................................................297 Strings as Class Members ..............................................................298 A User-Defined String Type ..........................................................300 The Standard C++ string Class..........................................................302 Defining and Assigning string Objects ........................................302 Input/Output with string Objects..................................................304 Finding string Objects ..................................................................305 Modifying string Objects ............................................................306 Comparing string Objects ............................................................307 Accessing Characters in string Objects........................................309 Other string Functions..................................................................310 Summary..............................................................................................310 Questions..............................................................................................311 Exercises ..............................................................................................313 8 Operator Overloading 319 Overloading Unary Operators..............................................................320 The operator Keyword ..................................................................322 Operator Arguments ......................................................................323 00 3087 FM 11/29/01 2:15 PM Page ixOBJECT-ORIENTED PROGRAMMING IN C++, FOURTH EDITON x Operator Return Values ..................................................................323 Nameless Temporary Objects ........................................................325 Postfix Notation..............................................................................326 Overloading Binary Operators ............................................................328 Arithmetic Operators......................................................................328 Concatenating Strings ....................................................................332 Multiple Overloading ....................................................................334 Comparison Operators....................................................................334 Arithmetic Assignment Operators..................................................337 The Subscript Operator ([]) ..........................................................340 Data Conversion ..................................................................................344 Conversions Between Basic Types ................................................344 Conversions Between Objects and Basic Types ............................345 Conversions Between Objects of Different Classes ......................350 Conversions: When to Use What....................................................357 UML Class Diagrams ..........................................................................357 Associations....................................................................................357 Navigability ....................................................................................358 Pitfalls of Operator Overloading and Conversion ..............................358 Use Similar Meanings ....................................................................358 Use Similar Syntax ........................................................................359 Show Restraint................................................................................359 Avoid Ambiguity ............................................................................360 Not All Operators Can Be Overloaded ..........................................360 Keywords explicit and mutable ........................................................360 Preventing Conversions with explicit..........................................360 Changing const Object Data Using mutable ................................362 Summary..............................................................................................364 Questions..............................................................................................364 Exercises ..............................................................................................367 9Inheritance 371 Derived Class and Base Class ............................................................373 Specifying the Derived Class ........................................................375 Generalization in UML Class Diagrams........................................375 Accessing Base Class Members ....................................................376 The protected Access Specifier ....................................................377 Derived Class Constructors ................................................................380 Overriding Member Functions ............................................................382 Which Function Is Used? ....................................................................383 Scope Resolution with Overridden Functions................................384 00 3087 FM 11/29/01 2:15 PM Page xCONTENTS xi Inheritance in the English Distance Class ..........................................384 Operation of ENGLEN ......................................................................387 Constructors in DistSign ..............................................................387 Member Functions in DistSign ....................................................387 Abetting Inheritance ......................................................................388 Class Hierarchies ................................................................................388 “Abstract” Base Class ....................................................................392 Constructors and Member Functions ............................................393 Inheritance and Graphics Shapes ........................................................393 Public and Private Inheritance ............................................................396 Access Combinations ....................................................................397 Access Specifiers: When to Use What ..........................................399 Levels of Inheritance ..........................................................................399 Multiple Inheritance ............................................................................403 Member Functions in Multiple Inheritance....................................404 private Derivation in EMPMULT ..........................................................409 Constructors in Multiple Inheritance..............................................409 Ambiguity in Multiple Inheritance......................................................413 Aggregation: Classes Within Classes ..................................................414 Aggregation in the EMPCONT Program............................................416 Composition: A Stronger Aggregation ..........................................420 Inheritance and Program Development ..............................................420 Summary..............................................................................................421 Questions..............................................................................................422 Exercises ..............................................................................................424 10 Pointers 429 Addresses and Pointers........................................................................430 The Address-of Operator & ..................................................................431 Pointer Variables ............................................................................433 Syntax Quibbles..............................................................................434 Accessing the Variable Pointed To ................................................436 Pointer to void................................................................................439 Pointers and Arrays..............................................................................440 Pointer Constants and Pointer Variables ........................................442 Pointers and Functions ........................................................................443 Passing Simple Variables................................................................443 Passing Arrays ................................................................................446 Sorting Array Elements ..................................................................448 Pointers and C-Type Strings ................................................................452 Pointers to String Constants ..........................................................452 Strings as Function Arguments ......................................................453 00 3087 FM 11/29/01 2:15 PM Page xiOBJECT-ORIENTED PROGRAMMING IN C++, FOURTH EDITON xii Copying a String Using Pointers....................................................454 Library String Functions ................................................................456 The const Modifier and Pointers ..................................................456 Arrays of Pointers to Strings..........................................................456 Memory Management: new and delete ..............................................458 The new Operator............................................................................459 The delete Operator ......................................................................461 A String Class Using new ..............................................................462 Pointers to Objects ..............................................................................464 Referring to Members ....................................................................465 Another Approach to new ..............................................................465 An Array of Pointers to Objects ....................................................467 A Linked List Example........................................................................469 A Chain of Pointers........................................................................469 Adding an Item to the List ............................................................471 Displaying the List Contents..........................................................472 Self-Containing Classes..................................................................473 Augmenting LINKLIST......................................................................473 Pointers to Pointers..............................................................................474 Sorting Pointers ..............................................................................476 The person** Data Type ................................................................476 Comparing Strings..........................................................................478 A Parsing Example ..............................................................................479 Parsing Arithmetic Expressions......................................................479 The PARSE Program ........................................................................481 Simulation: A Horse Race ..................................................................484 Designing the Horse Race..............................................................485 Multiplicity in the UML ................................................................489 UML State Diagrams ..........................................................................490 States ..............................................................................................491 Transitions ......................................................................................491 Racing from State to State..............................................................492 Debugging Pointers..............................................................................492 Summary..............................................................................................493 Questions..............................................................................................494 Exercises ..............................................................................................497 11 Virtual Functions 503 Virtual Functions..................................................................................504 Normal Member Functions Accessed with Pointers ......................505 Virtual Member Functions Accessed with Pointers ......................507 Late Binding ..................................................................................509 00 3087 FM 11/29/01 2:15 PM Page xiiCONTENTS xiii Abstract Classes and Pure Virtual Functions ................................510 Virtual Functions and the person Class ........................................511 Virtual Functions in a Graphics Example ......................................514 Virtual Destructors..........................................................................517 Virtual Base Classes ......................................................................518 Friend Functions ..................................................................................520 Friends as Bridges ..........................................................................520 Breaching the Walls........................................................................522 English Distance Example..............................................................522 friends for Functional Notation ....................................................526 friend Classes................................................................................528 Static Functions....................................................................................529 Accessing static Functions ..........................................................531 Numbering the Objects ..................................................................532 Investigating Destructors................................................................532 Assignment and Copy Initialization ....................................................532 Overloading the Assignment Operator ..........................................533 The Copy Constructor ....................................................................536 UML Object Diagrams ..................................................................539 A Memory-Efficient String Class ................................................540 The this Pointer ..................................................................................547 Accessing Member Data with this................................................547 Using this for Returning Values....................................................548 Revised STRIMEM Program..............................................................550 Dynamic Type Information..................................................................553 Checking the Type of a Class with dynamic_cast ........................553 Changing Pointer Types with dynamic_cast..................................554 The typeid Operator ......................................................................556 Summary..............................................................................................557 Questions..............................................................................................558 Exercises ..............................................................................................561 12 Streams and Files 567 Stream Classes ....................................................................................568 Advantages of Streams ..................................................................568 The Stream Class Hierarchy ..........................................................568 The ios Class ................................................................................570 The istream Class..........................................................................574 The ostream Class..........................................................................575 The iostream and the _withassign Classes ..................................576 Stream Errors ......................................................................................577 Error-Status Bits ............................................................................577 Inputting Numbers..........................................................................578 00 3087 FM 11/29/01 2:15 PM Page xiiiOBJECT-ORIENTED PROGRAMMING IN C++, FOURTH EDITON Too Many Characters ....................................................................579 No-Input Input................................................................................579 Inputting Strings and Characters....................................................580 Error-Free Distances ......................................................................580 Disk File I/O with Streams..................................................................583 Formatted File I/O ..........................................................................583 Strings with Embedded Blanks ......................................................586 Character I/O ..................................................................................588 Binary I/O ......................................................................................589 The reinterpret_cast Operator....................................................591 Closing Files ..................................................................................591 Object I/O ......................................................................................591 I/O with Multiple Objects ..............................................................594 File Pointers ........................................................................................597 Specifying the Position ..................................................................598 Specifying the Offset......................................................................598 The tellg() Function ....................................................................601 Error Handling in File I/O ..................................................................601 Reacting to Errors ..........................................................................601 Analyzing Errors ............................................................................602 File I/O with Member Functions ........................................................604 Objects That Read and Write Themselves ....................................604 Classes That Read and Write Themselves ....................................607 Overloading the Extraction and Insertion Operators ..........................616 Overloading for cout and cin ........................................................616 Overloading for Files......................................................................618 Memory as a Stream Object ................................................................620 Command-Line Arguments..................................................................622 Printer Output ......................................................................................624 Summary..............................................................................................626 Questions..............................................................................................627 Exercises ..............................................................................................628 13 Multifile Programs 633 Reasons for Multifile Programs ..........................................................634 Class Libraries................................................................................634 Organization and Conceptualization ..............................................635 Creating a Multifile Program ..............................................................637 Header Files....................................................................................637 Directory ........................................................................................637 Projects ..........................................................................................637 xiv 00 3087 FM 11/29/01 2:15 PM Page xivCONTENTS Inter-File Communication....................................................................638 Communication Among Source Files ............................................638 Header Files....................................................................................643 Namespaces ....................................................................................647 A Very Long Number Class ................................................................651 Numbers as Strings ........................................................................652 The Class Specifier ........................................................................652 The Member Functions ..................................................................654 The Application Program ..............................................................657 A High-Rise Elevator Simulation........................................................658 Running the ELEV Program ............................................................658 Designing the System ....................................................................660 Listings for ELEV ............................................................................662 Elevator Strategy ............................................................................674 State Diagram for the ELEV Program..............................................675 Summary..............................................................................................676 Questions..............................................................................................677 Projects ................................................................................................679 14 Templates and Exceptions 681 Function Templates..............................................................................682 A Simple Function Template..........................................................684 Function Templates with Multiple Arguments ..............................686 Class Templates ..................................................................................690 Class Name Depends on Context ..................................................694 A Linked List Class Using Templates............................................696 Storing User-Defined Data Types ..................................................698 The UML and Templates................................................................702 Exceptions............................................................................................703 Why Do We Need Exceptions?......................................................703 Exception Syntax............................................................................704 A Simple Exception Example ........................................................706 Multiple Exceptions........................................................................710 Exceptions with the Distance Class ..............................................712 Exceptions with Arguments............................................................714 The bad_alloc Class ......................................................................717 Exception Notes..............................................................................718 Summary..............................................................................................720 Questions..............................................................................................720 Exercises ..............................................................................................722 xv 00 3087 FM 11/29/01 2:15 PM Page xvOBJECT-ORIENTED PROGRAMMING IN C++, FOURTH EDITON 15 The Standard Template Library 725 Introduction to the STL ......................................................................726 Containers ......................................................................................727 Algorithms......................................................................................732 Iterators ..........................................................................................733 Potential Problems with the STL ..................................................734 Algorithms ..........................................................................................735 The find() Algorithm ....................................................................735 The count() Algorithm ..................................................................736 The sort() Algorithm ....................................................................737 The search() Algorithm ................................................................737 The merge() Algorithm ..................................................................738 Function Objects ............................................................................739 The for_each() Algorithm ............................................................742 The transform() Algorithm ..........................................................742 Sequence Containers............................................................................743 Vectors ............................................................................................743 Lists ................................................................................................747 Deques ............................................................................................750 Iterators ................................................................................................751 Iterators as Smart Pointers..............................................................752 Iterators as an Interface ..................................................................753 Matching Algorithms with Containers ..........................................755 Iterators at Work ............................................................................759 Specialized Iterators ............................................................................763 Iterator Adapters ............................................................................763 Stream Iterators ..............................................................................767 Associative Containers ........................................................................771 Sets and Multisets ..........................................................................771 Maps and Multimaps......................................................................775 Storing User-Defined Objects..............................................................778 A Set of person Objects ................................................................778 A List of person Objects................................................................782 Function Objects..................................................................................786 Predefined Function Objects ..........................................................786 Writing Your Own Function Objects..............................................789 Function Objects Used to Modify Container Behavior ................794 Summary..............................................................................................794 Questions..............................................................................................795 Exercises ..............................................................................................797 xvi 00 3087 FM 11/29/01 2:15 PM Page xviCONTENTS 16 Object-Oriented Software Development 801 Evolution of the Software Development Processes ............................802 The Seat-of-the-Pants Process........................................................802 The Waterfall Process ....................................................................802 Object-Oriented Programming ......................................................803 Modern Processes ..........................................................................803 Use Case Modeling..............................................................................805 Actors..............................................................................................805 Use Cases........................................................................................806 Scenarios ........................................................................................806 Use Case Diagrams ........................................................................806 Use Case Descriptions....................................................................807 From Use Cases to Classes ............................................................808 The Programming Problem..................................................................809 Hand-Written Forms ......................................................................809 Assumptions ..................................................................................811 The Elaboration Phase for the LANDLORD Program ............................812 Actors..............................................................................................812 Use Cases........................................................................................812 Use Case Descriptions....................................................................813 Scenarios ........................................................................................815 UML Activity Diagrams ................................................................815 From Use Cases to Classes..................................................................816 Listing the Nouns ..........................................................................816 Refining the List ............................................................................817 Discovering Attributes....................................................................818 From Verbs to Messages ................................................................818 Class Diagram ................................................................................820 Sequence Diagrams ........................................................................820 Writing the Code..................................................................................824 The Header File..............................................................................825 The .CPP Files..................................................................................831 More Simplifications ......................................................................841 Interacting with the Program ..............................................................841 Final Thoughts ....................................................................................843 Summary..............................................................................................844 Questions..............................................................................................844 Projects ................................................................................................846 A ASCII Chart 849 B C++ Precedence Table and Keywords 859 Precedence Table ................................................................................860 Keywords ............................................................................................860 xvii 00 3087 FM 11/29/01 2:15 PM Page xviiOBJECT-ORIENTED PROGRAMMING IN C++, FOURTH EDITON CMicrosoft Visual C++ 863 Screen Elements ..................................................................................864 Single-File Programs ..........................................................................864 Building an Existing File................................................................864 Writing a New File ........................................................................865 Errors ..............................................................................................865 Run-Time Type Information (RTTI) ..............................................866 Multifile Programs ..............................................................................866 Projects and Workspaces ................................................................866 Developing the Project ..................................................................867 Saving, Closing, and Opening Projects..........................................868 Compiling and Linking ..................................................................868 Building Console Graphics Lite Programs..........................................868 Debugging............................................................................................868 Single-Stepping ..............................................................................869 Watching Variables ........................................................................869 Stepping Into Functions..................................................................869 Breakpoints ....................................................................................870 D Borland C++Builder 871 Running the Example Programs in C++Builder..................................872 Cleaning Up the Screen ......................................................................873 Creating a New Project........................................................................873 Naming and Saving a Project ..............................................................874 Starting with Existing Files ................................................................875 Compiling, Linking, and Executing ....................................................875 Executing from C++Builder ..........................................................875 Executing from MS-DOS ..............................................................875 Precompiled Header Files ..............................................................876 Closing and Opening Projects........................................................876 Adding a Header File to Your Project ................................................876 Creating a New Header File ..........................................................876 Editing an Existing Header File ....................................................876 Telling C++Builder the Header File’s Location ............................877 Projects with Multiple Source Files ....................................................877 Creating Additional Source Files ..................................................877 Adding Existing Source Files to Your Project ..............................877 The Project Manager ......................................................................878 Console Graphics Lite Programs ........................................................878 Debugging............................................................................................878 Single-Stepping ..............................................................................879 Watching Variables ........................................................................879 Tracing into Functions....................................................................879 Breakpoints ....................................................................................879 xviii 00 3087 FM 11/29/01 2:15 PM Page xviiiCONTENTS E Console Graphics Lite 881 Using the Console Graphics Lite Routines ........................................882 The Console Graphics Lite Functions ................................................883 Implementations of the Console Graphics Lite Functions ..................884 Microsoft Compilers ......................................................................885 Borland Compilers..........................................................................885 Source Code Listings ..........................................................................885 Listing for MSOFTCON.H ..................................................................886 Listing for MSOFTCON.CPP ..............................................................886 Listing for BORLACON.H ..................................................................890 Listing for BORLACON.CPP ..............................................................891 F STL Algorithms and Member Functions 895 Algorithms ..........................................................................................896 Member Functions ..............................................................................907 Iterators ................................................................................................909 G Answers to Questions and Exercises 913 Chapter 1..............................................................................................914 Answers to Questions ....................................................................914 Chapter 2..............................................................................................914 Answers to Questions ....................................................................914 Solutions to Exercises ....................................................................916 Chapter 3..............................................................................................917 Answers to Questions ....................................................................917 Solutions to Exercises ....................................................................918 Chapter 4..............................................................................................921 Answers to Questions ....................................................................921 Solutions to Exercises ....................................................................922 Chapter 5..............................................................................................924 Answers to Questions ....................................................................924 Solutions to Exercises ....................................................................925 Chapter 6..............................................................................................928 Answers to Questions ....................................................................928 Solutions to Exercises ....................................................................929 Chapter 7..............................................................................................932 Answers to Questions ....................................................................932 Solutions to Exercises ....................................................................933 Chapter 8..............................................................................................937 Answers to Questions ....................................................................937 Solutions to Exercises ....................................................................938 Chapter 9..............................................................................................943 Answers to Questions ....................................................................943 Solutions to Exercises ....................................................................944 xix 00 3087 FM 11/29/01 2:15 PM Page xixChapter 10............................................................................................949 Answers to Questions ....................................................................949 Solutions to Exercises ....................................................................950 Chapter 11............................................................................................954 Answers to Questions ....................................................................954 Solutions to Exercises ....................................................................956 Chapter 12............................................................................................960 Answers to Questions ....................................................................960 Solutions to Exercises ....................................................................961 Chapter 13............................................................................................963 Answers to Questions ....................................................................963 Chapter 14............................................................................................964 Answers to Questions ....................................................................964 Solutions to Exercises ....................................................................965 Chapter 15............................................................................................969 Answers to Questions ....................................................................969 Solutions to Exercises ....................................................................970 Chapter 16............................................................................................974 Answers to Questions ....................................................................974 HBibliography 977 Advanced C++ ....................................................................................978 Defining Documents ............................................................................978 The Unified Modeling Language ........................................................978 The History of C++ ............................................................................979 Other Topics ........................................................................................979 Index 981 00 3087 FM 11/29/01 2:15 PM Page xxPreface The major changes to this Fourth Edition include an earlier introduction to UML, a new section on inter-file communication in Chapter 13, and a revised approach to software develop- ment in Chapter 16. Introducing the UML at the beginning allows the use of UML diagrams where they fit naturally with topics in the text, so there are many new UML diagrams throughout the book. The section on inter-file communication gathers together many concepts that were previously scattered throughout the book. The industry’s approach to object-oriented analysis and design has evolved since the last edition, and accordingly we’ve modified the chapter on this topic to reflect recent developments. C++ itself has changed very little since the last edition. However, besides the revisions just mentioned, we’ve made many smaller changes to clarify existing topics and correct typos and inaccuracies in the text. 00 3087 FM 11/29/01 2:15 PM Page xxiAbout the Author Robert Lafore has been writing books about computer programming since 1982. His best- selling titles include Assembly Language Programming for the IBM PC, C Programming Using Turbo C++, C++ Interactive Course, and Data Structures and Algorithms in Java. Mr. Lafore holds degrees in mathematics and electrical engineering, and has been active in programming since the days of the PDP-5, when 4K of main memory was considered luxurious. His interests include hiking, windsurfing, and recreational mathematics. 00 3087 FM 11/29/01 2:15 PM Page xxiiDedication This book is dedicated to GGL and her indomitable spirit. Acknowledgments to the Fourth Edition My thanks to many readers who e-mailed comments and corrections. I am also indebted to the following professors of computer science who offered their suggestions and corrections: Bill Blomberg of Regis University in Denver; Richard Daehler-Wilking of the College of Charleston in South Carolina; Frank Hoffmann of the Royal Institute of Technology in Sweden, and David Blockus of San Jose State University in California. My special thanks to David Topham of Ohlone College in Fremont, California, for his many detailed ideas and his sharp eye for problems. At Sams Publishing, Michael Stephens provided an expert and friendly liaison with the details of publishing. Reviewer Robin Rowe and Technical Editor Mark Cashman attempted with great care to save me from myself; any lack of success is entirely my fault. Project Manager Christina Smith made sure that everything came together in an amazingly short time, Angela Boley helped keep everything moving smoothly, and Matt Wynalda provided expert proofread- ing. I’m grateful to you all. Acknowledgments to the Third Edition I’d like to thank the entire team at MacMillan Computer Publishing. In particular, Tracy Dunkelberger ably spearheaded the entire project and exhibited great patience with what turned out to be a lengthy schedule. Jeff Durham handled the myriad details involved in inter- facing between me and the editors with skill and good humor. Andrei Kossorouko lent his expertise in C++ to ensure that I didn’t make this edition worse instead of better. Acknowledgments to the Second Edition My thanks to the following professors—users of this book as a text at their respective colleges and universities—for their help in planning the second edition: Dave Bridges, Frank Cioch, Jack Davidson, Terrence Fries, Jimmie Hattemer, Jack Van Luik, Kieran Mathieson, Bill McCarty, Anita Millspaugh, Ian Moraes, Jorge Prendes, Steve Silva, and Edward Wright. I would like to thank the many readers of the first edition who wrote in with corrections and suggestions, many of which were invaluable. At Waite Group Press, Joanne Miller has ably ridden herd on my errant scheduling and filled in as academic liaison, and Scott Calamar, as always, has made sure that everyone knew what they were doing. Deirdre Greene provided an uncannily sharp eye as copy editor. 00 3087 FM 11/29/01 2:15 PM Page xxiiiThanks, too, to Mike Radtke and Harry Henderson for their expert technical reviews. Special thanks to Edward Wright, of Western Oregon State College, for reviewing and experi- menting with the new exercises. Acknowledgments to the First Edition My primary thanks go to Mitch Waite, who poured over every inch of the manuscript with painstaking attention to detail and made a semi-infinite number of helpful suggestions. Bill McCarty of Azusa Pacific University reviewed the content of the manuscript and its suit- ability for classroom use, suggested many excellent improvements, and attempted to correct my dyslexic spelling. George Leach ran all the programs, and, to our horror, found several that didn’t perform cor- rectly in certain circumstances. I trust these problems have all been fixed; if not, the fault is entirely mine. Scott Calamar of the Waite Group dealt with the myriad organizational aspects of writing and producing this book. His competence and unfailing good humor were an important ingredient in its completion. I would also like to thank Nan Borreson of Borland for supplying the latest releases of the software (among other useful tidbits), Harry Henderson for reviewing the exercises, Louise Orlando of the Waite Group for ably shepherding the book through production, Merrill Peterson of Matrix Productions for coordinating the most trouble-free production run I’ve ever been involved with, Juan Vargas for the innovative design, and Frances Hasegawa for her uncanny ability to decipher my sketches and produce beautiful and effective art. 00 3087 FM 11/29/01 2:15 PM Page xxivTell Us What You Think! As the reader of this book, you are our most important critic and commentator. We value your opinion and want to know what we’re doing right, what we could do better, what areas you’d like to see us publish in, and any other words of wisdom you’re willing to pass our way. As an executive editor for Sams Publishing,I welcome your comments. You cane-mail or write me directly to let me know what you did or didn’t like about this book—as well as what we can do to make our books stronger. Please note that I cannot help you with technical problems related to the topic of this book, and that due to the high volume of mail I receive, I might not be able to reply to every mes- sage. When you write, please be sure to include this book’s title and author’s name as well as your name and phone or fax number. I will carefully review your comments and share them with the author and editors who worked on the book. E-mail: feedback@samspublishing.com Mail: Sams Publishing 201 West 103rd Street Indianapolis, IN 46290 USA 00 3087 FM 11/29/01 2:15 PM Page xxvIntroduction This book teaches you how to write programs in a the C++ programming language. However, it does more than that. In the past few years, several major innovations in software develop- ment have appeared on the scene. This book teaches C++ in the context of these new develop- ments. Let’s see what they are. Programming Innovations In the old days, 20 or so years ago, programmers starting a project would sit down almost immediately and start writing code. However, as programming projects became large and more complicated, it was found that this approach did not work very well. The problem was com- plexity. Large programs are probably the most complicated entities ever created by humans. Because of this complexity, programs are prone to error, and software errors can be expensive and even life threatening (in air traffic control, for example). Three major innovations in programming have been devised to cope with the problem of complexity. They are • Object-oriented programming (OOP) • The Unified Modeling Language (UML) •Improved software development processes This book teaches the C++ language with these developments in mind. You will not only learn a computer language, but new ways of conceptualizing software development. Object-Oriented Programming Why has object-oriented programming become the preferred approach for most software pro- jects? OOP offers a new and powerful way to cope with complexity. Instead of viewing a pro- gram as a series of steps to be carried out, it views it as a group of objects that have certain properties and can take certain actions. This may sound obscure until you learn more about it, but it results in programs that are clearer, more reliable, and more easily maintained. A major goal of this book is to teach object-oriented programming. We introduce it as early as possible, and cover all its major features. The majority of our example programs are object- oriented. The Unified Modeling Language The Unified Modeling Language (UML) is a graphical language consisting of many kinds of diagrams. It helps program analysts figure out what a program should do, and helps program- mers design and understand how a program works. The UML is a powerful tool that can make programming easier and more effective. 01 3087 Intro 11/29/01 2:23 PM Page 1OBJECT-ORIENTED PROGRAMMING IN C++, FOURTH EDITION We give an overview of the UML in Chapter 1, and then discuss specific features of the UML throughout the book. We introduce each UML feature where it will help to clarify the OOP topic being discussed. In this way you learn the UML painlessly at the same time the UML helps you to learn C++. Languages and Development Platforms Of the object-oriented programming languages, C++ is by far the most widely used. Java, a recent addition to the field of OO languages, lacks certain features—such as pointers, tem- plates, and multiple inheritance—that make it less powerful and versatile than C++. (If you ever do want to learn Java, its syntax is very similar to that of C++, so learning C++ gives you a head start in Java.) Several other OO languages have been introduced recently, such as C#, but they have not yet attained the wide acceptance of C++. Until recently the standards for C++ were in a constant state of evolution. This meant that each compiler vendor handled certain details differently. However, in November 1997, the ANSI/ISO C++ standards committee approved the final draft of what is now known as Standard C++. (ANSI stands for American National Standards Institute, and ISO stands for International Standards Institute.) Standard C++ adds many new features to the language, such as the Standard Template Library (STL). In this book we follow Standard C++ (in all but a few places, which we’ll note as we go along). The most popular development environments for C++ are manufactured by Microsoft and Borland (Inprise) and run on the various flavors of Microsoft Windows. In this book we’ve attempted to ensure that all sample programs run on the current versions of both Borland and Microsoft compilers. (See Appendix C, “Microsoft Visual C++,” and Appendix D, “Borland C++Builder,” for more on these compilers.) What This Book Does This book teaches object-oriented programming with the C++ programming language, using either Microsoft or Borland compilers. It also introduces the UML and software development processes. It is suitable for professional programmers, students, and kitchen-table enthusiasts. New Concepts OOP involves concepts that are new to programmers of traditional languages such as Pascal, Basic, and C. These ideas, such as classes, inheritance, and polymorphism, lie at the heart of object-oriented programming. But it’s easy to lose sight of these concepts when discussing the specifics of an object-oriented language. Many books overwhelm the reader with the details of language features, while ignoring the reason these features exist. This book attempts to keep an eye on the big picture and relate the details to the larger concepts. 2 01 3087 Intro 11/29/01 2:23 PM Page 2INTRODUCTION The Gradual Approach We take a gradual approach in this book, starting with very simple programming examples and working up to full-fledged object-oriented applications. We introduce new concepts slowly so that you will have time to digest one idea before going on to the next. We use illustrations whenever possible to help clarify new ideas. There are questions and programming exercises at the end of most chapters to enhance the book’s usefulness in the classroom. Answers to the questions and to the first few (starred) exercises can be found in Appendix G. The exercises vary in difficulty to pose a variety of challenges for the student. What You Need to Know to Use This Book You can use this book even if you have no previous programming experience. However, such experience, in Visual Basic for example, certainly won’t hurt. You do not need to know the C language to use this book. Many books on C++ assume that you already know C, but this one does not. It teaches C++ from the ground up. If you do know C, it won’t hurt, but you may be surprised at how little overlap there is between C and C++. You should be familiar with the basic operations of Microsoft Windows, such as starting appli- cations and copying files. Software and Hardware You will need a C++ compiler. The programs in this book have been tested with Microsoft Visual C++ and Borland C++Builder. Both compilers come in low-priced “Learning Editions” suitable for students. Appendix C provides detailed information on operating the Microsoft compiler, while Appendix D does the same for the Inprise (Borland) product. Other compilers, if they adhere to Standard C++, will probably handle most of the programs in this book as written. Your computer should have enough processor speed, memory, and hard disk space to run the compiler you’ve chosen. You can check the manufacturer’s specifications to determine these requirements. Console-Mode Programs There are numerous example programs throughout the book. They are console-mode programs, which run in a character-mode window within the compiler environment, or directly within an MS-DOS box. This avoids the complexity of full-scale graphics-oriented Windows programs. 3 01 3087 Intro 11/29/01 2:23 PM Page 3OBJECT-ORIENTED PROGRAMMING IN C++, FOURTH EDITION Example Program Source Code You can obtain the source code for the example programs from the Sams Publishing Web site at http://www.samspublishing.com Type the ISBN (found at the front of the book) or the book’s title and click Search to find the data on this book. Then click Source Code to download the program examples. Console Graphics Lite A few example programs draw pictures using a graphics library we call Console Graphics Lite. The graphics rely on console characters, so they are not very sophisticated, but they allow some interesting programs. The files for this library are provided on the publisher’s Web site, along with the source files for the example programs. To compile and run these graphics examples, you’ll need to include a header file in your pro- gram, either MSOFTCON.H or BORLACON.H, depending on your compiler. You’ll also need to add either MSOFTCON.CPP or BORLACON.CPP to the project for the graphics example. Appendix E, “Console Graphics Lite,” provides listings of these files and tells how to use them. Appendixes C and D explain how to work with files and projects in a specific compiler’s environment. Programming Exercises Each chapter contains roughly 12 exercises, each requiring the creation of a complete C++ program. Solutions for the first three or four exercises in each chapter are provided in Appendix G. For the remainder of the exercises, readers are on their own. (However, if you are teaching a C++ course, see the “Note to Teachers” at the end of this Introduction.) Easier Than You Think You may have heard that C++ is difficult to learn, but it’s really quite similar to other lan- guages, with two or three “grand ideas” thrown in. These new ideas are fascinating in them- selves, and we think you’ll have fun learning about them. They are also becoming part of the programming culture; they’re something everyone should know a little bit about, like evolution and psychoanalysis. We hope this book will help you enjoy learning about these new ideas, at the same time that it teaches you the details of programming in C++. 4 01 3087 Intro 11/29/01 2:23 PM Page 4INTRODUCTION A Note to Teachers Teachers, and others who already know something about C++ or C, may be interested in some details of the approach we use in this book and how it’s organized. Standard C++ All the programs in this book are compatible with Standard C++, with a few minor exceptions that are needed to accommodate compiler quirks. We devote a chapter to the STL (Standard Template Library), which is included in Standard C++. The Unified Modeling Language (UML) In the previous edition, we introduced the UML in the final chapter. In this edition we have integrated the UML into the body of the book, introducing UML topics in appropriate places. For example, UML class diagrams are introduced where we first show different classes com- municating, and generalization is covered in the chapter on inheritance. Chapter 1, “The Big Picture,” includes a list showing where the various UML topics are intro- duced. Software Development Processes Formal software development processes are becoming an increasingly important aspect of pro- gramming. Also, students are frequently mystified by the process of designing an object- oriented program. For these reasons we include a chapter on software development processes, with an emphasis on object-oriented programming. In the last edition we focused on CRC cards, but the emphasis in software development has shifted more in the direction of use case analysis, so we use that to analyze our programming projects. C++ Is Not the Same as C A few institutions still want their students to learn C before learning C++. In our view this is a mistake. C and C++ are entirely separate languages. It’s true that their syntax is similar, and C is actually a subset of C++. But the similarity is largely a historical accident. In fact, the basic approach in a C++ program is radically different from that in a C program. C++ has overtaken C as the preferred language for serious software development. Thus we don’t believe it is necessary or advantageous to teach C before teaching C++. Students who don’t know C are saved the time and trouble of learning C and then learning C++, an ineffi- cient approach. Students who already know C may be able to skim parts of some chapters, but they will find that a remarkable percentage of the material is new. 5 01 3087 Intro 11/29/01 2:23 PM Page 5OBJECT-ORIENTED PROGRAMMING IN C++, FOURTH EDITION Optimize Organization for OOP We could have begun the book by teaching the procedural concepts common to C and C++, and moved on to the new OOP concepts once the procedural approach had been digested. That seemed counterproductive, however, because one of our goals is to begin true object-oriented programming as quickly as possible. Accordingly, we provide a minimum of procedural groundwork before getting to classes in Chapter 6. Even the initial chapters are heavily steeped in C++, as opposed to C, usage. We introduce some concepts earlier than is traditional in books on C. For example, structures are a key feature for understanding C++ because classes are syntactically an extension of struc- tures. For this reason, we introduce structures in Chapter 5 so that they will be familiar when we discuss classes. Some concepts, such as pointers, are introduced later than in traditional C books. It’s not nec- essary to understand pointers to follow the essentials of OOP, and pointers are usually a stum- bling block for C and C++ students. Therefore, we defer a discussion of pointers until the main concepts of OOP have been thoroughly digested. Substitute Superior C++ Features Some features of C have been superseded by new approaches in C++. For instance, the printf() and scanf() functions, input/output workhorses in C, are seldom used in C++ because cout and cin do a better job. Consequently, we leave out descriptions of these func- tions. Similarly, #define constants and macros in C have been largely superseded by the const qualifier and inline functions in C++, and need be mentioned only briefly. Minimize Irrelevant Capabilities Because the focus in this book is on object-oriented programming, we can leave out some fea- tures of C that are seldom used and are not particularly relevant to OOP. For instance, it isn’t necessary to understand the C bit-wise operators (used to operate on individual bits) to learn object-oriented programming. These and a few other features can be dropped from our discus- sion, or mentioned only briefly, with no loss in understanding of the major features of C++. The result is a book that focuses on the fundamentals of OOP, moving the reader gently but briskly toward an understanding of new concepts and their application to real programming problems. 6 01 3087 Intro 11/29/01 2:23 PM Page 6INTRODUCTION Programming Exercises No answers to the unstarred exercises are provided in this book. However, qualified instructors can obtain suggested solutions from the Sams Publishing Web site. Type the ISBN or title and click Search to move to this book’s page, then click Downloads. The exercises vary considerably in their degree of difficulty. In each chapter the early exercises are fairly easy, while later ones are more challenging. Instructors will probably want to assign only those exercises suited to the level of a particular class. 7 01 3087 Intro 11/29/01 2:23 PM Page 701 3087 Intro 11/29/01 2:23 PM Page 8CHAPTER 1 The Big Picture IN THIS CHAPTER • Why Do We Need Object-Oriented Programming? 10 • Characteristics of Object-Oriented Languages 16 • C++ and C 22 • Laying the Groundwork 23 • The Unified Modeling Language (UML) 23 02 3087 CH01 11/29/01 2:15 PM Page 9Chapter 110 This book teaches you how to program in C++, a computer language that supports object- oriented programming (OOP). Why do we need OOP? What does it do that traditional lan- guages such as C, Pascal, and BASIC don’t? What are the principles behind OOP? Two key concepts in OOP are objects and classes. What do these terms mean? What is the relationship between C++ and the older C language? This chapter explores these questions and provides an overview of the features to be discussed in the balance of the book. What we say here will necessarily be rather general (although mer- cifully brief). If you find the discussion somewhat abstract, don’t worry. The concepts we men- tion here will come into focus as we demonstrate them in detail in subsequent chapters. Why Do We Need Object-Oriented Programming? Object-oriented programming was developed because limitations were discovered in earlier approaches to programming. To appreciate what OOP does, we need to under- stand what these limitations are and how they arose from traditional programming languages. Procedural Languages C, Pascal, FORTRAN, and similar languages are procedural languages. That is, each statement in the language tells the computer to do something: Get some input, add these numbers, divide by six, display that output. A program in a procedural language is a list of instructions. For very small programs, no other organizing principle (often called a paradigm) is needed. The programmer creates the list of instructions, and the computer carries them out. Division into Functions When programs become larger, a single list of instructions becomes unwieldy. Few programmers can comprehend a program of more than a few hundred statements unless it is broken down into smaller units. For this reason the function was adopted as a way to make programs more comprehensible to their human creators. (The term function is used in C++ and C. In other languages the same concept may be referred to as a subroutine, a subprogram, or a procedure.) A procedural program is divided into functions, and (ideally, at least) each function has a clearly defined purpose and a clearly defined interface to the other functions in the program. 02 3087 CH01 11/29/01 2:15 PM Page 10The idea of breaking a program into functions can be further extended by grouping a number of functions together into a larger entity called a module (which is often a file), but the princi- ple is similar: a grouping of components that execute lists of instructions. Dividing a program into functions and modules is one of the cornerstones of structured pro- gramming,the somewhat loosely defined discipline that influenced programming organization for several decades before the advent of object-oriented programming. Problems with Structured Programming As programs grow ever larger and more complex, even the structured programming approach begins to show signs of strain. You may have heard about, or been involved in, horror stories of program development. The project is too complex, the schedule slips, more programmers are added, complexity increases, costs skyrocket, the sched- ule slips further, and disaster ensues. (See The Mythical Man-Month by Frederick P. Brooks, Jr. [Addison Wesley, 1982] for a vivid description of this process.) Analyzing the reasons for these failures reveals that there are weaknesses in the procedural paradigm itself. No matter how well the structured programming approach is implemented, large programs become excessively complex. What are the reasons for these problems with procedural languages? There are two related problems. First, functions have unrestricted access to global data. Second, unrelated functions and data, the basis of the procedural paradigm, provide a poor model of the real world. Let’s examine these problems in the context of an inventory program. One important global data item in such a program is the collection of items in the inventory. Various functions access this data to input a new item, display an item, modify an item, and so on. Unrestricted Access In a procedural program, one written in C for example, there are two kinds of data. Local data is hidden inside a function, and is used exclusively by the function. In the inventory program a display function might use local data to remember which item it was displaying. Local data is closely related to its function and is safe from modifica- tion by other functions. However, when two or more functions must access the same data—and this is true of the most important data in a program—then the data must be made global,as our collection of inven- tory items is. Global data can be accessed by any function in the program. (We ignore the issue of grouping functions into modules, which doesn’t materially affect our argument.) The arrangement of local and global variables in a procedural program is shown in Figure 1.1. The Big Picture 1 T HE B IG P ICTURE 11 02 3087 CH01 11/29/01 2:15 PM Page 11FIGURE 1.1 Global and local variables. In a large program, there are many functions and many global data items. The problem with the procedural paradigm is that this leads to an even larger number of potential connections between functions and data, as shown in Figure 1.2. Chapter 112 FIGURE 1.2 The procedural paradigm. This large number of connections causes problems in several ways. First, it makes a program’s structure difficult to conceptualize. Second, it makes the program difficult to modify. A change made in a global data item may necessitate rewriting all the functions that access that item. 02 3087 CH01 11/29/01 2:15 PM Page 12For example, in our inventory program, someone may decide that the product codes for the inventory items should be changed from 5 digits to 12 digits. This may necessitate a change from a short to a long data type. Now all the functions that operate on the data must be modified to deal with a long instead of a short. It’s similar to what happens when your local supermarket moves the bread from aisle 4 to aisle 7. Everyone who patronizes the supermarket must then figure out where the bread has gone, and adjust their shopping habits accordingly. When data items are modified in a large program it may not be easy to tell which functions access the data, and even when you figure this out, modifications to the functions may cause them to work incorrectly with other global data items. Everything is related to everything else, so a modification anywhere has far-reaching, and often unintended, consequences. Real-World Modeling The second—and more important—problem with the procedural paradigm is that its arrangement of separate data and functions does a poor job of modeling things in the real world. In the physical world we deal with objects such as people and cars. Such objects aren’t like data and they aren’t like functions. Complex real-world objects have both attributes and behavior. Attributes Examples of attributes (sometimes called characteristics) are, for people, eye color and job title; and, for cars, horsepower and number of doors. As it turns out, attributes in the real world are equivalent to data in a program: they have a certain specific val- ues, such as blue (for eye color) or four (for the number of doors). Behavior Behavior is something a real-world object does in response to some stimulus. If you ask your boss for a raise, she will generally say yes or no. If you apply the brakes in a car, it will generally stop. Saying something and stopping are examples of behavior. Behavior is like a function: you call a function to do something (display the inventory, for example) and it does it. So neither data nor functions, by themselves, model real-world objects effectively. The Object-Oriented Approach The fundamental idea behind object-oriented languages is to combine into a single unit both data and the functions that operate on that data. Such a unit is called an object. The Big Picture 1 T HE B IG P ICTURE 13 02 3087 CH01 11/29/01 2:15 PM Page 13An object’s functions, called member functions in C++, typically provide the only way to access its data. If you want to read a data item in an object, you call a member function in the object. It will access the data and return the value to you. You can’t access the data directly. The data is hidden,so it is safe from accidental alteration. Data and its functions are said to be encapsulated into a single entity. Data encapsulation and data hiding are key terms in the description of object-oriented languages. If you want to modify the data in an object, you know exactly what functions interact with it: the member functions in the object. No other functions can access the data. This simplifies writing, debugging, and maintaining the program. A C++ program typically consists of a number of objects, which communicate with each other by calling one another’s member functions. The organization of a C++ program is shown in Figure 1.3. Chapter 114 FIGURE 1.3 The object-oriented paradigm. 02 3087 CH01 11/29/01 2:15 PM Page 14We should mention that what are called member functions in C++ are called methods in some other object-oriented (OO) languages (such as Smalltalk, one of the first OO languages). Also, data items are referred to as attributes or instance variables. Calling an object’s member func- tion is referred to as sending a message to the object. These terms are not official C++ termi- nology, but they are used with increasing frequency, especially in object-oriented design. An Analogy You might want to think of objects as departments—such as sales, accounting, per- sonnel, and so on—in a company. Departments provide an important approach to cor- porate organization. In most companies (except very small ones), people don’t work on personnel problems one day, the payroll the next, and then go out in the field as salespeople the week after. Each department has its own personnel, with clearly assigned duties. It also has its own data: the accounting department has payroll fig- ures, the sales department has sales figures, the personnel department keeps records of each employee, and so on. The people in each department control and operate on that department’s data. Dividing the company into departments makes it easier to comprehend and control the company’s activities, and helps maintain the integrity of the information used by the company. The accounting department, for instance, is responsible for the payroll data. If you’re a sales manager, and you need to know the total of all the salaries paid in the southern region in July, you don’t just walk into the accounting department and start rummaging through file cabinets. You send a memo to the appropriate person in the department, then wait for that person to access the data and send you a reply with the information you want. This ensures that the data is accessed accurately and that it is not corrupted by inept outsiders. This view of corporate organization is shown in Figure 1.4. In the same way, objects provide an approach to program organization while help- ing to maintain the integrity of the program’s data. OOP: An Approach to Organization Keep in mind that object-oriented programming is not primarily concerned with the details of program operation. Instead, it deals with the overall organization of the pro- gram. Most individual program statements in C++ are similar to statements in proce- dural languages, and many are identical to statements in C. Indeed, an entire member function in a C++ program may be very similar to a procedural function in C. It is only when you look at the larger context that you can determine whether a statement or a function is part of a procedural C program or an object-oriented C++ program. The Big Picture 1 T HE B IG P ICTURE 15 02 3087 CH01 11/29/01 2:15 PM Page 15FIGURE 1.4 The corporate paradigm. Characteristics of Object-Oriented Languages Let’s briefly examine a few of the major elements of object-oriented languages in general, and C++ in particular. Objects When you approach a programming problem in an object-oriented language, you no longer ask how the problem will be divided into functions, but how it will be divided into objects. Thinking in terms of objects, rather than functions, has a surprisingly helpful effect on how easily programs can be designed. This results from the close match between objects in the programming sense and objects in the real world. This process is described in detail in Chapter 16, “Object-Oriented Software Development.” Chapter 116 Sales Department Sales Manager Sales data Secretary Personnel Department Personnel Manager Personnel data Personnel Staff Finance Department Chief Financial Officer Financial data Financial Assistant 02 3087 CH01 11/29/01 2:15 PM Page 16What kinds of things become objects in object-oriented programs? The answer to this is lim- ited only by your imagination, but here are some typical categories to start you thinking: • Physical objects Automobiles in a traffic-flow simulation Electrical components in a circuit-design program Countries in an economics model Aircraft in an air traffic control system • Elements of the computer-user environment Windows Menus Graphics objects (lines, rectangles, circles) The mouse, keyboard, disk drives, printer • Data-storage constructs Customized arrays Stacks Linked lists Binary trees • Human entities Employees Students Customers Salespeople • Collections of data An inventory A personnel file A dictionary A table of the latitudes and longitudes of world cities • User-defined data types Time Angles Complex numbers Points on the plane The Big Picture 1 T HE B IG P ICTURE 17 02 3087 CH01 11/29/01 2:15 PM Page 17• Components in computer games Cars in an auto race Positions in a board game (chess, checkers) Animals in an ecological simulation Opponents and friends in adventure games The match between programming objects and real-world objects is the happy result of combin- ing data and functions: The resulting objects offer a revolution in program design. No such close match between programming constructs and the items being modeled exists in a procedural language. Classes In OOP we say that objects are members of classes. What does this mean? Let’s look at an analogy. Almost all computer languages have built-in data types. For instance, a data type int, meaning integer, is predefined in C++ (as we’ll see in Chapter 3, “Loops and Decisions”). You can declare as many variables of type int as you need in your program: int day; int count; int divisor; int answer; In a similar way, you can define many objects of the same class, as shown in Figure 1.5. A class serves as a plan, or blueprint. It specifies what data and what functions will be included in objects of that class. Defining the class doesn’t create any objects, just as the mere existence of data type int doesn’t create any variables. A class is thus a description of a number of similar objects. This fits our non-technical under- standing of the word class. Prince, Sting, and Madonna are members of the rock musician class. There is no one person called “rock musician,” but specific people with specific names are members of this class if they possess certain characteristics. An object is often called an “instance” of a class. Inheritance The idea of classes leads to the idea of inheritance. In our daily lives, we use the con- cept of classes divided into subclasses. We know that the animal class is divided into mammals, amphibians, insects, birds, and so on. The vehicle class is divided into cars, trucks, buses, motorcycles, and so on. Chapter 118 02 3087 CH01 11/29/01 2:15 PM Page 18FIGURE 1.5 A class and its objects. The principle in this sort of division is that each subclass shares common characteristics with the class from which it’s derived. Cars, trucks, buses, and motorcycles all have wheels and a motor; these are the defining characteristics of vehicles. In addition to the characteristics shared with other members of the class, each subclass also has its own particular characteris- tics: Buses, for instance, have seats for many people, while trucks have space for hauling heavy loads. This idea is shown in Figure 1.6. Notice in the figure that features A and B, which are part of the base class, are common to all the derived classes, but that each derived class also has fea- tures of its own. The Big Picture 1 T HE B IG P ICTURE 19 02 3087 CH01 11/29/01 2:15 PM Page 19FIGURE 1.6 Inheritance. In a similar way, an OOP class can become a parent of several subclasses. In C++ the original class is called the base class; other classes can be defined that share its characteristics, but add their own as well. These are called derived classes. Don’t confuse the relation of objects to classes, on the one hand, with the relation of a base class to derived classes, on the other. Objects, which exist in the computer’s memory, each embody the exact characteristics of their class, which serves as a template. Derived classes inherit some characteristics from their base class, but add new ones of their own. Inheritance is somewhat analogous to using functions to simplify a traditional procedural pro- gram. If we find that three different sections of a procedural program do almost exactly the same thing, we recognize an opportunity to extract the common elements of these three sec- tions and put them into a single function. The three sections of the program can call the func- tion to execute the common actions, and they can perform their own individual processing as well. Similarly, a base class contains elements common to a group of derived classes. As func- tions do in a procedural program, inheritance shortens an object-oriented program and clarifies the relationship among program elements. Chapter 120 02 3087 CH01 11/29/01 2:15 PM Page 20Reusability Once a class has been written, created, and debugged, it can be distributed to other programmers for use in their own programs. This is called reusability. It is similar to the way a library of functions in a procedural language can be incorporated into dif- ferent programs. However, in OOP, the concept of inheritance provides an important extension to the idea of reusability. A programmer can take an existing class and, without modifying it, add additional features and capabilities to it. This is done by deriving a new class from the existing one. The new class will inherit the capabilities of the old one, but is free to add new features of its own. For example, you might have written (or purchased from someone else) a class that creates a menu system, such as that used in Windows or other Graphic User Interfaces (GUIs). This class works fine, and you don’t want to change it, but you want to add the capability to make some menu entries flash on and off. To do this, you simply create a new class that inherits all the capabilities of the existing one but adds flashing menu entries. The ease with which existing software can be reused is an important benefit of OOP. Many companies find that being able to reuse classes on a second project provides an increased return on their original programming investment. We’ll have more to say about this in later chapters. Creating New Data Types One of the benefits of objects is that they give the programmer a convenient way to construct new data types. Suppose you work with two-dimensional positions (such as x and y coordinates, or latitude and longitude) in your program. You would like to express operations on these positional values with normal arithmetic operations, such as position1 = position2 + origin where the variables position1, position2, and origin each represent a pair of inde- pendent numerical quantities. By creating a class that incorporates these two values, and declaring position1, position2, and origin to be objects of this class, we can, in effect, create a new data type. Many features of C++ are intended to facilitate the creation of new data types in this manner. Polymorphism and Overloading Note that the = (equal) and + (plus) operators, used in the position arithmetic shown above, don’t act the same way they do in operations on built-in types such as int. The objects position1 and so on are not predefined in C++, but are programmer-defined The Big Picture 1 T HE B IG P ICTURE 21 02 3087 CH01 11/29/01 2:15 PM Page 21objects of class Position. How do the = and + operators know how to operate on objects? The answer is that we can define new behaviors for these operators. These operations will be member functions of the Position class. Using operators or functions in different ways, depending on what they are operating on, is called polymorphism (one thing with several distinct forms). When an existing operator, such as + or =, is given the capability to operate on a new data type, it is said to be overloaded. Overloading is a kind of polymorphism; it is also an important feature of OOP. C++ and C C++ is derived from the C language. Strictly speaking, it is a superset of C: Almost every correct statement in C is also a correct statement in C++, although the reverse is not true. The most important elements added to C to create C++ concern classes, objects, and object-oriented programming. (C++ was originally called “C with classes.”) However, C++ has many other new features as well, including an improved approach to input/output (I/O) and a new way to write comments. Figure 1.7 shows the relationship of C and C++. Chapter 122 FIGURE 1.7 The relationship between C and C++. 02 3087 CH01 11/29/01 2:15 PM Page 22In fact, the practical differences between C and C++ are larger than you might think. Although you can write a program in C++ that looks like a program in C, hardly anyone does. C++ pro- grammers not only make use of the new features of C++, they also emphasize the traditional C features in different proportions than do C programmers. If you already know C, you will have a head start in learning C++ (although you may also have some bad habits to unlearn), but much of the material will be new. Laying the Groundwork Our goal is to help you begin writing OOP programs as soon as possible. However, as we noted, much of C++ is inherited from C, so while the overall structure of a C++ program may be OOP, down in the trenches you need to know some old-fashioned procedural fundamentals. Chapters 2–5 therefore deal with the “traditional” aspects of C++, many of which are also found in C. You will learn about variables and I/O, about control structures such as loops and decisions, and about functions themselves. You will also learn about structures, since the same syntax that’s used for structures is used for classes. If you already know C, you might be tempted to skip these chapters. However, you will find that there are many differences, some obvious and some rather subtle, between C and C++. Our advice is to read these chapters, skimming what you know, and concentrating on the ways C++ differs from C. The specific discussion of OOP starts in Chapter 6, “Objects and Classes.” From then on the examples will be object oriented. The Unified Modeling Language (UML) The UML is a graphical “language” for modeling computer programs. “Modeling” means to create a simplified representation of something, as a blueprint models a house. The UML pro- vides a way to visualize the higher-level organization of programs without getting mired down in the details of actual code. The UML began as three separate modeling languages, one created by Grady Booch at Rational Software, one by James Rumbaugh at General Electric, and one by Ivar Jacobson at Ericson. Eventually Rumbaugh and Jacobson joined Booch at Rational, where they became known as the three amigos. During the late 1990s they unified (hence the name) their modeling languages into the Unified Modeling Language. The result was adopted by the Object Management Group (OMG), a consortium of companies devoted to industry standards. The Big Picture 1 T HE B IG P ICTURE 23 02 3087 CH01 11/29/01 2:15 PM Page 23Why do we need the UML? One reason is that in a large computer program it’s often hard to understand, simply by looking at the code, how the parts of the program relate to each other. As we’ve seen, object-oriented programming is a vast improvement over procedural programs. Nevertheless, figuring out what a program is supposed to do requires, at best, considerable study of the program listings. The trouble with code is that it’s very detailed. It would be nice if there were a way to see a bigger picture, one that depicts the major parts of the program and how they work together. The UML answers this need. The most important part of the UML is a set of different kinds of diagrams. Class diagrams show the relationships among classes, object diagrams show how specific objects relate, sequence diagrams show the communication among objects over time, use case diagrams show how a program’s users interact with the program, and so on. These diagrams provide a variety of ways to look at a program and its operation. The UML plays many roles besides helping us to understand how a program works. As we’ll see in Chapter 16, it can help in the initial design of a program. In fact, the UML is useful throughout all phases of software development, from initial specification to documentation, testing, and maintenance. The UML is not a software development process. Many such processes exist for specifying the stages of the development process. The UML is simply a way to look at the software being developed. Although it can be applied to any kind of programming language, the UML is espe- cially attuned to OOP. As we noted in the Introduction, we introduce specific features of the UML in stages through- out the book. • Chapter 1: (this section) introduction to the UML • Chapter 8: class diagrams, associations, and navigability • Chapter 9: generalization, aggregation, and composition • Chapter 10: state diagrams and multiplicity • Chapter 11: object diagrams • Chapter 13: more complex state diagrams • Chapter 14: templates, dependencies, and stereotypes • Chapter 16: use cases, use case diagrams, activity diagrams, and sequence diagrams Chapter 124 02 3087 CH01 11/29/01 2:15 PM Page 24Summary OOP is a way of organizing programs. The emphasis is on the way programs are designed, not on coding details. In particular, OOP programs are organized around objects, which contain both data and functions that act on that data. A class is a tem- plate for a number of objects. Inheritance allows a class to be derived from an existing class without modifying it. The derived class has all the data and functions of the parent class, but adds new ones of its own. Inheritance makes possible reusability, or using a class over and over in different programs. C++ is a superset of C. It adds to the C language the capability to implement OOP. It also adds a variety of other features. In addition, the emphasis is changed in C++ so that some features common to C, although still available in C++, are seldom used, while others are used far more frequently. The result is a surprisingly different language. The Unified Modeling Language (UML) is a standardized way to visualize a program’s struc- ture and operation using diagrams. The general concepts discussed in this chapter will become more concrete as you learn more about the details of C++. You may want to refer back to this chapter as you progress further into this book. Questions Answers to these questions can be found in Appendix G. Note that throughout this book, multiple-choice questions can have more than one correct answer. 1. Pascal, BASIC, and C are p_____ languages, while C++ is an o_____ language. 2. A widget is to the blueprint for a widget as an object is to a. a member function. b. a class. c. an operator. d. a data item. 3. The two major components of an object are _____ and functions that _____. 4. In C++, a function contained within a class is called a. a member function. b. an operator. c. a class function. d. a method. The Big Picture 1 T HE B IG P ICTURE 25 02 3087 CH01 11/29/01 2:15 PM Page 255. Protecting data from access by unauthorized functions is called _____. 6. Which of the following are good reasons to use an object-oriented language? a. You can define your own data types. b. Program statements are simpler than in procedural languages. c. An OO program can be taught to correct its own errors. d. It’s easier to conceptualize an OO program. 7. _____ model entities in the real world more closely than do functions. 8. True or false: A C++ program is similar to a C program except for the details of coding. 9. Bundling data and functions together is called _____. 10. When a language has the capability to produce new data types, it is said to be a. reprehensible. b. encapsulated. c. overloaded. d. extensible. 11. True or false: You can easily tell, from any two lines of code, whether a pro- gram is written in C or C++. 12. The ability of a function or operator to act in different ways on different data types is called __________. 13. A normal C++ operator that acts in special ways on newly defined data types is said to be a. glorified. b. encapsulated. c. classified. d. overloaded. 14. Memorizing the new terms used in C++ is a. critically important. b. something you can return to later. c. the key to wealth and success. d. completely irrelevant. Chapter 126 02 3087 CH01 11/29/01 2:15 PM Page 2615. The Unified Modeling Language is a. a program that builds physical models. b. a way to look at the organization of a program. c. the combination of C++ and FORTRAN. d. helpful in developing software systems. The Big Picture 1 T HE B IG P ICTURE 27 02 3087 CH01 11/29/01 2:15 PM Page 2702 3087 CH01 11/29/01 2:15 PM Page 28CHAPTER 2 C++ Programming Basics IN THIS CHAPTER • Getting Started 30 • Basic Program Construction 30 • Output Using cout 33 •Directives 35 • Comments 36 •Integer Variables 38 • Character Variables 42 •Input with cin 45 •Floating Point Types 48 •Type bool 51 • The setw Manipulator 52 •Variable Type Summary 54 •Type Conversion 56 •Arithmetic Operators 60 •Library Functions 65 03 3087 CH02 11/29/01 2:19 PM Page 29Chapter 230 In any language there are some fundamentals you need to know before you can write even the most elementary programs. This chapter introduces three such fundamentals: basic program construction, variables, and input/output (I/O). It also touches on a variety of other language features, including comments, arithmetic operators, the increment operator, data conversion, and library functions. These topics are not conceptually difficult, but you may find that the style in C++ is a little austere compared with, say, BASIC or Pascal. Before you learn what it’s all about, a C++ pro- gram may remind you more of a mathematics formula than a computer program. Don’t worry about this. You’ll find that as you gain familiarity with C++, it starts to look less forbidding, while other languages begin to seem unnecessarily fancy and verbose. Getting Started As we noted in the Introduction, you can use either a Microsoft or a Borland compiler with this book. Appendixes C and D provide details about their operation. (Other compilers may work as well.) Compilers take source code and transform it into executable files, which your computer can run as it does other programs. Source files are text files (extension .CPP) that cor- respond with the listings printed in this book. Executable files have the .EXE extension, and can be executed either from within your compiler, or, if you’re familiar with MS-DOS, directly from a DOS window. The programs run without modification on the Microsoft compiler or in an MS-DOS window. If you’re using the Borland compiler, you’ll need to modify the programs slightly before run- ning them; otherwise the output won’t remain on the screen long enough to see. Make sure to read Appendix D, “Borland C++Builder,” to see how this is done. Basic Program Construction Let’s look at a very simple C++ program. This program is called FIRST,so its source file is FIRST.CPP. It simply prints a sentence on the screen. Here it is: #include using namespace std; int main() { cout << “Every age has a language of its own\n”; return 0; } Despite its small size, this program demonstrates a great deal about the construction of C++ programs. Let’s examine it in detail. 03 3087 CH02 11/29/01 2:19 PM Page 30Functions Functions are one of the fundamental building blocks of C++. The FIRST program consists almost entirely of a single function called main(). The only parts of this program that are not part of the function are the first two lines—the ones that start with #include and using. (We’ll see what these lines do in a moment.) We noted in Chapter 1, “The Big Picture,” that a function can be part of a class, in which case it is called a member function. However, functions can also exist independently of classes. We are not yet ready to talk about classes, so we will show functions that are separate standalone entities, as main() is here. Function Name The parentheses following the word main are the distinguishing feature of a function. Without the parentheses the compiler would think that main refers to a variable or to some other pro- gram element. When we discuss functions in the text, we’ll follow the same convention that C++ uses: We’ll put parentheses following the function name. Later on we’ll see that the parentheses aren’t always empty. They’re used to hold function arguments:values passed from the calling program to the function. The word int preceding the function name indicates that this particular function has a return value of type int. Don’t worry about this now; we’ll learn about data types later in this chapter and return values in Chapter 5, “Functions.” Braces and the Function Body The body of a function is surrounded by braces (sometimes called curly brackets). These braces play the same role as the BEGIN and END keywords in some other languages: They sur- round or delimit a block of program statements. Every function must use this pair of braces around the function body. In this example there are only two statements in the function body: the line starting with cout, and the line starting with return. However, a function body can consist of many statements. Always Start with main() When you run a C++ program, the first statement executed will be at the beginning of a func- tion called main(). (At least that’s true of the console mode programs in this book.) The pro- gram may consist of many functions, classes, and other program elements, but on startup, control always goes to main(). If there is no function called main() in your program, an error will be reported when you run the program. In most C++ programs, as we’ll see later, main() calls member functions in various objects to carry out the program’s real work. The main() function may also contain calls to other stand- alone functions. This is shown in Figure 2.1. C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 31 03 3087 CH02 11/29/01 2:19 PM Page 31FIGURE 2.1 Objects, functions, and main(). Program Statements The program statement is the fundamental unit of C++ programming. There are two statements in the FIRST program: the line cout << “Every age has a language of its own\n”; and the return statement return 0; The first statement tells the computer to display the quoted phrase. Most statements tell the computer to do something. In this respect, statements in C++ are similar to statements in other languages. In fact, as we’ve noted, the majority of statements in C++ are identical to state- ments in C. A semicolon signals the end of the statement. This is a crucial part of the syntax but easy to forget. In some languages (like BASIC), the end of a statement is signaled by the end of the line, but that’s not true in C++. If you leave out the semicolon, the compiler will often (although not always) signal an error. Chapter 232 03 3087 CH02 11/29/01 2:19 PM Page 32The last statement in the function body is return 0;. This tells main() to return the value 0 to whoever called it, in this case the operating system or compiler. In older versions of C++ you could give main() the return type of void and dispense with the return statement, but this is not considered correct in Standard C++. We’ll learn more about return in Chapter 5. Whitespace We mentioned that the end of a line isn’t important to a C++ compiler. Actually, the compiler ignores whitespace almost completely. Whitespace is defined as spaces, carriage returns, line- feeds, tabs, vertical tabs, and formfeeds. These characters are invisible to the compiler. You can put several statements on one line, separated by any number of spaces or tabs, or you can run a statement over two or more lines. It’s all the same to the compiler. Thus the FIRST program could be written this way: #include using namespace std; int main () { cout << “Every age has a language of its own\n” ; return 0;} We don’t recommend this syntax—it’s nonstandard and hard to read—but it does compile cor- rectly. There are several exceptions to the rule that whitespace is invisible to the compiler. The first line of the program, starting with #include, is a preprocessor directive, which must be written on one line. Also, string constants, such as “Every age has a language of its own”, can- not be broken into separate lines. (If you need a long string constant, you can insert a back- slash (\) at the line break or divide the string into two separate strings, each surrounded by quotes.) Output Using cout As you have seen, the statement cout << “Every age has a language of its own\n”; causes the phrase in quotation marks to be displayed on the screen. How does this work? A complete description of this statement requires an understanding of objects, operator overload- ing, and other topics we won’t discuss until later in the book, but here’s a brief preview. C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 33 03 3087 CH02 11/29/01 2:19 PM Page 33The identifier cout (pronounced “C out”) is actually an object. It is predefined in C++ to corre- spond to the standard output stream. A stream is an abstraction that refers to a flow of data. The standard output stream normally flows to the screen display—although it can be redirected to other output devices. We’ll discuss streams (and redirection) in Chapter 12, “Streams and Files.” The operator << is called the insertion or put to operator. It directs the contents of the variable on its right to the object on its left. In FIRST it directs the string constant “Every age has a language of its own\n” to cout, which sends it to the display. (If you know C, you’ll recognize << as the left-shift bit-wise operator and wonder how it can also be used to direct output. In C++, operators can be overloaded. That is, they can perform different activities, depending on the context. We’ll learn about overloading in Chapter 8, “Operator Overloading.”) Although the concepts behind the use of cout and << may be obscure at this point, using them is easy. They’ll appear in almost every example program. Figure 2.2 shows the result of using cout and the insertion operator <<. Chapter 234 FIGURE 2.2 Output with cout. String Constants The phrase in quotation marks, “Every age has a language of its own\n”, is an example of a string constant. As you probably know, a constant, unlike a variable, cannot be given a new value as the program runs. Its value is set when the program is written, and it retains this value throughout the program’s existence. As we’ll see later, the situation regarding strings is rather complicated in C++. Two ways of handling strings are commonly used. A string can be represented by an array of characters, or it can be represented as an object of a class. We’ll learn more about both kinds of strings in Chapter 7, “Arrays and Strings.” 03 3087 CH02 11/29/01 2:19 PM Page 34The ‘\n’ character at the end of the string constant is an example of an escape sequence. It causes the next text output to be displayed on a new line. We use it here so that the phrases such as “Press any key to continue,” inserted by some compilers for display after the program terminates, will appear on a new line. We’ll discuss escape sequences later in this chapter. Directives The two lines that begin the FIRST program are directives. The first is a preprocessor directive, and the second is a using directive. They occupy a sort of gray area: They’re not part of the basic C++ language, but they’re necessary anyway Preprocessor Directives The first line of the FIRST program #include might look like a program statement, but it’s not. It isn’t part of a function body and doesn’t end with a semicolon, as program statements must. Instead, it starts with a number sign (#). It’s called a preprocessor directive. Recall that program statements are instructions to the com- puter to do something, such as adding two numbers or printing a sentence. A preprocessor directive, on the other hand, is an instruction to the compiler. A part of the compiler called the preprocessor deals with these directives before it begins the real compilation process. The preprocessor directive #include tells the compiler to insert another file into your source file. In effect, the #include directive is replaced by the contents of the file indicated. Using an #include directive to insert another file into your source file is similar to pasting a block of text into a document with your word processor. #include is only one of many preprocessor directives, all of which can be identified by the ini- tial # sign. The use of preprocessor directives is not as common in C++ as it is in C, but we’ll look at a few additional examples as we go along. The type file usually included by #include is called a header file. Header Files In the FIRST example, the preprocessor directive #include tells the compiler to add the source file IOSTREAM to the FIRST.CPP source file before compiling. Why do this? IOSTREAM is an exam- ple of a header file (sometimes called an include file). It’s concerned with basic input/output operations, and contains declarations that are needed by the cout identifier and the << operator. Without these declarations, the compiler won’t recognize cout and will think << is being used incorrectly. There are many such include files. The newer Standard C++ header files don’t have a file extension, but some older header files, left over from the days of the C language, have the extension .H. C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 35 03 3087 CH02 11/29/01 2:19 PM Page 35If you want to see what’s in IOSTREAM, you can find the include directory for your compiler and display it as a source file in the Edit window. (See the appropriate appendix for hints on how to do this.) Or you can look at it with the WordPad or Notepad utilities. The contents won’t make much sense at this point, but you will at least prove to yourself that IOSTREAM is a source file, written in normal ASCII characters. We’ll return to the topic of header files at the end of this chapter, when we introduce library functions. The using Directive A C++ program can be divided into different namespaces. A namespace is a part of the pro- gram in which certain names are recognized; outside of the namespace they’re unknown. The directive using namespace std; says that all the program statements that follow are within the std namespace. Various program components such as cout are declared within this namespace. If we didn’t use the using direc- tive, we would need to add the std name to many program elements. For example, in the FIRST program we’d need to say std::cout << “Every age has a language of its own.”; To avoid adding std:: dozens of times in programs we use the using directive instead. We’ll discuss namespaces further in Chapter 13, “Multifile Programs.” Comments Comments are an important part of any program. They help the person writing a program, and anyone else who must read the source file, understand what’s going on. The compiler ignores comments, so they do not add to the file size or execution time of the executable program. Comment Syntax Let’s rewrite our FIRST program, incorporating comments into our source file. We’ll call the new program COMMENTS: // comments.cpp // demonstrates comments #include //preprocessor directive using namespace std; //”using” directive Chapter 236 03 3087 CH02 11/29/01 2:19 PM Page 36int main() //function name “main” { //start function body cout << “Every age has a language of its own\n”; //statement return 0; //statement } //end function body Comments start with a double slash symbol (//) and terminate at the end of the line. (This is one of the exceptions to the rule that the compiler ignores whitespace.) A comment can start at the beginning of the line or on the same line following a program statement. Both possibilities are shown in the COMMENTS example. When to Use Comments Comments are almost always a good thing. Most programmers don’t use enough of them. If you’re tempted to leave out comments, remember that not everyone is as smart as you; they may need more explanation than you do about what your program is doing. Also, you may not be as smart next month, when you’ve forgotten key details of your program’s operation, as you are today. Use comments to explain to the person looking at the listing what you’re trying to do. The details are in the program statements themselves, so the comments should concentrate on the big picture, clarifying your reasons for using a certain statement or group of statements. Alternative Comment Syntax There’s a second comment style available in C++: /* this is an old-style comment */ This type of comment (the only comment originally available in C) begins with the /* charac- ter pair and ends with */ (not with the end of the line). These symbols are harder to type (since / is lowercase while * is uppercase) and take up more space on the line, so this style is not generally used in C++. However, it has advantages in special situations. You can write a multi- line comment with only two comment symbols: /* this is a potentially very long multiline comment */ This is a good approach to making a comment out of a large text passage, since it saves insert- ing the // symbol on every line. C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 37 03 3087 CH02 11/29/01 2:19 PM Page 37You can also insert a /* */ comment anywhere within the text of a program line: func1() { /* empty function body */ } If you attempt to use the // style comment in this case, the closing brace won’t be visible to the compiler—since a // style comment runs to the end of the line—and the code won’t com- pile correctly. Integer Variables Variables are the most fundamental part of any language. A variable has a symbolic name and can be given a variety of values. Variables are located in particular places in the computer’s memory. When a variable is given a value, that value is actually placed in the memory space assigned to the variable. Most popular languages use the same general variable types, such as integers, floating-point numbers, and characters, so you are probably already familiar with the ideas behind them. Integer variables represent integer numbers like 1, 30,000, and –27. Such numbers are used for counting discrete numbers of objects, like 11 pencils or 99 bottles of beer. Unlike floating- point numbers, integers have no fractional part; you can express the idea of four using integers, but not four and one-half. Defining Integer Variables Integer variables exist in several sizes, but the most commonly used is type int. The amount of memory occupied by the integer types is system dependent. On a 32-bit system such as Windows, an int occupies 4 bytes (which is 32 bits) of memory. This allows an int to hold numbers in the range from –2,147,483,648 to 2,147,483,647. Figure 2.3 shows an integer vari- able in memory. While type int occupies 4 bytes on current Windows computers, it occupied only 2 bytes in MS-DOS and earlier versions of Windows. The ranges occupied by the various types are listed in the header file LIMITS; you can also look them up using your compiler’s help system. Here’s a program that defines and uses several variables of type int: // intvars.cpp // demonstrates integer variables #include using namespace std; int main() { int var1; //define var1 int var2; //define var2 Chapter 238 03 3087 CH02 11/29/01 2:19 PM Page 38var1 = 20; //assign value to var1 var2 = var1 + 10; //assign value to var2 cout << “var1+10 is “; //output text cout << var2 << endl; //output value of var2 return 0; } C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 39 FIGURE 2.3 Variable of type int in memory. Type this program into your compiler’s edit screen (or load it from the Web site), compile and link it, and then run it. Examine the output window. The statements int var1; int var2; define two integer variables, var1 and var2. The keyword int signals the type of variable. These statements, which are called declarations,must terminate with a semicolon, like other program statements. You must declare a variable before using it. However, you can place variable declarations any- where in a program. It’s not necessary to declare variables before the first executable statement (as was necessary in C). However, it’s probably more readable if commonly-used variables are located at the beginning of the program. 03 3087 CH02 11/29/01 2:19 PM Page 39Declarations and Definitions Let’s digress for a moment to note a subtle distinction between the terms definition and decla- ration as applied to variables. A declaration introduces a variable’s name (such as var1) into a program and specifies its type (such as int). However, if a declaration also sets aside memory for the variable, it is also called a definition. The statements int var1; int var2; in the INTVARS program are definitions, as well as declarations, because they set aside memory for var1 and var2. We’ll be concerned mostly with declarations that are also definitions, but later on we’ll see various kinds of declarations that are not definitions. Variable Names The program INTVARS uses variables named var1 and var2. The names given to variables (and other program features) are called identifiers. What are the rules for writing identifiers? You can use upper- and lowercase letters, and the digits from 1 to 9. You can also use the under- score (_). The first character must be a letter or underscore. Identifiers can be as long as you like, but most compilers will only recognize the first few hundred characters. The compiler dis- tinguishes between upper- and lowercase letters, so Var is not the same as var or VAR. You can’t use a C++ keyword as a variable name. A keyword is a predefined word with a spe- cial meaning. int, class, if, and while are examples of keywords. A complete list of key- words can be found in Appendix B, “C++ Precedence Table and Keywords,” and in your compiler’s documentation. Many C++ programmers follow the convention of using all lowercase letters for variable names. Other programmers use a mixture of upper- and lowercase, as in IntVar or dataCount. Still others make liberal use of underscores. Whichever approach you use, it’s good to be con- sistent throughout a program. Names in all uppercase are sometimes reserved for constants (see the discussion of const that follows). These same conventions apply to naming other pro- gram elements such as classes and functions. A variable’s name should make clear to anyone reading the listing the variable’s purpose and how it is used. Thus boilerTemperature is better than something cryptic like bT or t. Assignment Statements The statements var1 = 20; var2 = var1 + 10; Chapter 240 03 3087 CH02 11/29/01 2:19 PM Page 40assign values to the two variables. The equal sign (=), as you might guess, causes the value on the right to be assigned to the variable on the left. The = in C++ is equivalent to the := in Pascal or the = in BASIC. In the first line shown here, var1, which previously had no value, is given the value 20. Integer Constants The number 20 is an integer constant. Constants don’t change during the course of the pro- gram. An integer constant consists of numerical digits. There must be no decimal point in an integer constant, and it must lie within the range of integers. In the second program line shown here, the plus sign (+) adds the value of var1 and 10, in which 10 is another constant. The result of this addition is then assigned to var2. Output Variations The statement cout << “var1+10 is “; displays a string constant, as we’ve seen before. The next statement cout << var2 << endl; displays the value of the variable var2. As you can see in your console output window, the out- put of the program is var1+10 is 30 Note that cout and the << operator know how to treat an integer and a string differently. If we send them a string, they print it as text. If we send them an integer, they print it as a number. This may seem obvious, but it is another example of operator overloading, a key feature of C++. (C programmers will remember that such functions as printf() need to be told not only the variable to be displayed, but the type of the variable as well, which makes the syntax far less intuitive.) As you can see, the output of the two cout statements appears on the same line on the output screen. No linefeed is inserted automatically. If you want to start on a new line, you must insert a linefeed yourself. We’ve seen how to do this with the ‘\n’ escape sequence. Now we’ll see another way: using something called a manipulator. The endl Manipulator The last cout statement in the INTVARS program ends with an unfamiliar word: endl. This causes a linefeed to be inserted into the stream, so that subsequent text is displayed on the next line. It has the same effect as sending the ‘\n’ character, but is somewhat clearer. It’s an C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 41 03 3087 CH02 11/29/01 2:19 PM Page 41example of a manipulator. Manipulators are instructions to the output stream that modify the output in various ways; we’ll see more of them as we go along. Strictly speaking, endl (unlike ‘\n’) also causes the output buffer to be flushed, but this happens invisibly so for most pur- poses the two are equivalent. Other Integer Types There are several numerical integer types besides type int. The two most common types are long and short. (Strictly speaking type char is an integer type as well, but we’ll cover it sepa- rately.) We noted that the size of type int is system dependent. In contrast, types long and short have fixed sizes no matter what system is used. Type long always occupies four bytes, which is the same as type int on 32-bit Windows sys- tems. Thus it has the same range, from –2,147,483,648 to 2,147,483,647. It can also be written as long int; this means the same as long. There’s little point in using type long on 32-bit sys- tems, since it’s the same as int. However, if your program may need to run on a 16-bit system such as MS-DOS, or on older versions of Windows, specifying type long will guarantee a four-bit integer type. In 16-bit systems, type int has the same range as type short. On all systems type short occupies two bytes, giving it a range of –32,768 to 32,767. There’s probably not much point using type short on modern Windows systems unless it’s important to save memory. Type int,although twice as large, is accessed faster than type short. If you want to create a constant of type long,use the letter L following the numerical value, as in longvar = 7678L; // assigns long constant 7678 to longvar Many compilers offer integer types that explicitly specify the number of bits used. (Remember there are 8 bits to a byte.) These type names are preceded by two underscores. They are __int8, __int16, __int32, and __int64. The __int8 type corresponds to char, and (at least in 32-bit systems) the type name __int16 corresponds to short and __int32 corresponds to both int and long. The __int64 type holds huge integers with up to 19 decimal digits. Using these type names has the advantage that the number of bytes used for a variable is not implementa- tion dependent. However, this is not usually an issue, and these types are seldom used. Character Variables Type char stores integers that range in value from –128 to 127. Variables of this type occupy only 1 byte (eight bits) of memory. Character variables are sometimes used to store numbers that confine themselves to this limited range, but they are much more commonly used to store ASCII characters. Chapter 242 03 3087 CH02 11/29/01 2:19 PM Page 42As you may already know, the ASCII character set is a way of representing characters such as ‘a’, ‘B’, ‘$’, ‘3’, and so on, as numbers. These numbers range from 0 to 127. Most Windows systems extend this range to 255 to accommodate various foreign-language and graphics char- acters. Appendix A, “ASCII Table,” shows the ASCII character set. Complexities arise when foreign languages are used, and even when programs are transferred between computer systems in the same language. This is because the characters in the range 128 to 255 aren’t standardized and because the one-byte size of type char is too small to accommodate the number of characters in many languages, such as Japanese. Standard C++ provides a larger character type called wchar_t to handle foreign languages. This is important if you’re writing programs for international distribution. However, in this book we’ll ignore type wchar_t and assume that we’re dealing with the ASCII character set found in current ver- sions of Windows. Character Constants Character constants use single quotation marks around a character, like ‘a’ and ‘b’. (Note that this differs from string constants, which use double quotation marks.) When the C++ compiler encounters such a character constant, it translates it into the corresponding ASCII code. The constant ‘a’ appearing in a program, for example, will be translated into 97, as shown in Figure 2.4. C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 43 FIGURE 2.4 Variable of type char in memory. 03 3087 CH02 11/29/01 2:19 PM Page 43Character variables can be assigned character constants as values. The following program shows some examples of character constants and variables. // charvars.cpp // demonstrates character variables #include //for cout, etc. using namespace std; int main() { char charvar1 = ‘A’; //define char variable as character char charvar2 = ‘\t’; //define char variable as tab cout << charvar1; //display character cout << charvar2; //display character charvar1 = ‘B’; //set char variable to char constant cout << charvar1; //display character cout << ‘\n’; //display newline character return 0; } Initialization Variables can be initialized at the same time they are defined. In this program two variables of type char—charvar1 and charvar2—are initialized to the character constants ‘A’ and ‘\t’. Escape Sequences This second character constant, ‘\t’, is an odd one. Like ‘\n’, which we encountered earlier, it’s an example of an escape sequence. The name reflects the fact that the backslash causes an “escape” from the normal way characters are interpreted. In this case the t is interpreted not as the character ‘t’ but as the tab character. A tab causes printing to continue at the next tab stop. In console-mode programs, tab stops are positioned every eight spaces. Another character con- stant, ‘\n’, is sent directly to cout in the last line of the program. Escape sequences can be used as separate characters or embedded in string constants. Table 2.1 shows a list of common escape sequences. TABLE 2.1 Common Escape Sequences Escape Sequence Character \ a Bell (beep) \ b Backspace \ f Formfeed Chapter 244 03 3087 CH02 11/29/01 2:19 PM Page 44\ n Newline \ r Return \ t Tab \ \ Backslash \ ‘ Single quotation mark \ “ Double quotation marks \ xdd Hexadecimal notation Since the backslash, the single quotation marks, and the double quotation marks all have spe- cialized meanings when used in constants, they must be represented by escape sequences when we want to display them as characters. Here’s an example of a quoted phrase in a string con- stant: cout << “\”Run, Spot, run,\” she said.”; This translates to “Run, Spot, run,” she said. Sometimes you need to represent a character constant that doesn’t appear on the keyboard, such as the graphics characters above ASCII code 127. To do this, you can use the ‘\xdd’ rep- resentation, where each d stands for a hexadecimal digit. If you want to print a solid rectangle, for example, you’ll find such a character listed as decimal number 178, which is hexadecimal number B2 in the ASCII table. This character would be represented by the character constant ‘\xB2’. We’ll see some examples of this later. The CHARVARS program prints the value of charvar1 (‘A’) and the value of charvar2 (a tab). It then sets charvar1 to a new value (‘B’), prints that, and finally prints the newline. The output looks like this: A B Input with cin Now that we’ve seen some variable types in use, let’s see how a program accomplishes input. The next example program asks the user for a temperature in degrees Fahrenheit, converts it to Celsius, and displays the result. It uses integer variables. C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 45 TABLE 2.1 Continued Escape Sequence Character 03 3087 CH02 11/29/01 2:19 PM Page 45// fahren.cpp // demonstrates cin, newline #include using namespace std; int main() { int ftemp; //for temperature in fahrenheit cout << “Enter temperature in fahrenheit: “; cin >> ftemp; int ctemp = (ftemp-32) * 5 / 9; cout << “Equivalent in Celsius is: “ << ctemp << ‘\n’; return 0; } The statement cin >> ftemp; causes the program to wait for the user to type in a number. The resulting number is placed in the variable ftemp. The keyword cin (pronounced “C in”) is an object, predefined in C++ to correspond to the standard input stream. This stream represents data coming from the keyboard (unless it has been redirected). The >> is the extraction or get from operator. It takes the value from the stream object on its left and places it in the variable on its right. Here’s some sample interaction with the program: Enter temperature in fahrenheit: 212 Equivalent in Celsius is: 100 Figure 2.5 shows input using cin and the extraction operator >>. Chapter 246 FIGURE 2.5 Input with cin. 03 3087 CH02 11/29/01 2:19 PM Page 46Variables Defined at Point of Use The FAHREN program has several new wrinkles besides its input capability. Look closely at the listing. Where is the variable ctemp defined? Not at the beginning of the program, but in the next-to-the-last line, where it’s used to store the result of the arithmetic operation. As we noted earlier, you can define variables throughout a program, not just at the beginning. (Many lan- guages, including C, require all variables to be defined before the first executable statement.) Defining variables where they are used can make the listing easier to understand, since you don’t need to refer repeatedly to the start of the listing to find the variable definitions. However, the practice should be used with discretion. Variables that are used in many places in a function are better defined at the start of the function. Cascading << The insertion operator << is used repeatedly in the second cout statement in FAHREN. This is perfectly legal. The program first sends the phrase Equivalent in Celsius is: to cout,then it sends the value of ctemp, and finally the newline character ‘\n’. The extraction operator >> can be cascaded with cin in the same way, allowing the user to enter a series of values. However, this capability is not used so often, since it eliminates the opportunity to prompt the user between inputs. Expressions Any arrangement of variables, constants, and operators that specifies a computation is called an expression. Thus, alpha+12 and (alpha-37)*beta/2 are expressions. When the computa- tions specified in the expression are performed, the result is usually a value. Thus if alpha is 7, the first expression shown has the value 19. Parts of expressions may also be expressions. In the second example, alpha-37 and beta/2 are expressions. Even single variables and constants, like alpha and 37,are considered to be expressions. Note that expressions aren’t the same as statements. Statements tell the compiler to do some- thing and terminate with a semicolon, while expressions specify a computation. There can be several expressions in a statement. Precedence Note the parentheses in the expression (ftemp-32) * 5 / 9 C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 47 03 3087 CH02 11/29/01 2:19 PM Page 47Without the parentheses, the multiplication would be carried out first, since * has higher prior- ity than -. With the parentheses, the subtraction is done first, then the multiplication, since all operations inside parentheses are carried out first. What about the precedence of the * and / signs? When two arithmetic operators have the same precedence, the one on the left is exe- cuted first, so in this case the multiplication will be carried out next, then the division. Precedence and parentheses are normally applied this same way in algebra and in other com- puter languages, so their use probably seems quite natural. However, precedence is an impor- tant topic in C++. We’ll return to it later when we introduce different kinds of operators. Floating Point Types We’ve talked about type int and type char, both of which represent numbers as integers—that is, numbers without a fractional part. Now let’s examine a different way of storing numbers— as floating-point variables. Floating-point variables represent numbers with a decimal place—like 3.1415927, 0.0000625, and –10.2. They have both an integer part, to the left of the decimal point, and a fractional part, to the right. Floating-point variables represent what mathematicians call real numbers, which are used for measurable quantities such as distance, area, and temperature. They typically have a fractional part. There are three kinds of floating-point variables in C++: type float,type double, and type long double. Let’s start with the smallest of these, type float. Type float Type float stores numbers in the range of about 3.4x10–38 to 3.4x1038,with a precision of seven digits. It occupies 4 bytes (32 bits) in memory, as shown in Figure 2.6. The following example program prompts the user to type in a floating-point number represent- ing the radius of a circle. It then calculates and displays the circle’s area. // circarea.cpp // demonstrates floating point variables #include //for cout, etc. using namespace std; int main() { float rad; //variable of type float const float PI = 3.14159F; //type const float cout << “Enter radius of circle: “; //prompt cin >> rad; //get radius Chapter 248 03 3087 CH02 11/29/01 2:19 PM Page 48float area = PI * rad * rad; //find area cout << “Area is “ << area << endl; //display answer return 0; } C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 49 FIGURE 2.6 Variable of type float in memory. Here’s a sample interaction with the program: Enter radius of circle: 0.5 Area is 0.785398 This is the area in square feet of a 12-inch LP record (which has a radius of 0.5 feet). At one time this was an important quantity for manufacturers of vinyl. Type double and long double The larger floating point types, double and long double,are similar to float except that they require more memory space and provide a wider range of values and more precision. Type double requires 8 bytes of storage and handles numbers in the range from 1.7x10–308 to 1.7x10308 with a precision of 15 digits. Type long double is compiler-dependent but is often the same as double. Type double is shown in Figure 2.7. 03 3087 CH02 11/29/01 2:19 PM Page 49FIGURE 2.7 Variable of type double. Floating-Point Constants The number 3.14159F in CIRCAREA is an example of a floating-point constant. The decimal point signals that it is a floating-point constant, and not an integer, and the F specifies that it’s type float,rather than double or long double. The number is written in normal decimal notation. You don’t need a suffix letter with constants of type double; it’s the default. With type long double,use the letter L. You can also write floating-point constants using exponential notation. Exponential notation is a way of writing large numbers without having to write out a lot of zeros. For example, 1,000,000,000 can be written as 1.0E9 in exponential notation. Similarly, 1234.56 would be written 1.23456E3. (This is the same as 1.23456 times 103.) The number following the E is called the exponent. It indicates how many places the decimal point must be moved to change the number to ordinary decimal notation. The exponent can be positive or negative. The exponential number 6.35239E–5 is equivalent to 0.0000635239 in decimal notation. This is the same as 6.35239 times 10–5. Chapter 250 03 3087 CH02 11/29/01 2:19 PM Page 50The const Qualifier Besides demonstrating variables of type float,the CIRCAREA example also introduces the qual- ifier const. It’s used in the statement const float PI = 3.14159F; //type const float The keyword const (for constant) precedes the data type of a variable. It specifies that the value of a variable will not change throughout the program. Any attempt to alter the value of a variable defined with this qualifier will elicit an error message from the compiler. The qualifier const ensures that your program does not inadvertently alter a variable that you intended to be a constant, such as the value of PI in CIRCAREA. It also reminds anyone reading the listing that the variable is not intended to change. The const modifier can apply to other entities besides simple variables. We’ll learn more about this as we go along. The #define Directive Although the construction is not recommended in C++, constants can also be specified using the preprocessor directive #define. This directive sets up an equivalence between an identifier and a text phrase. For example, the line #define PI 3.14159 appearing at the beginning of your program specifies that the identifier PI will be replaced by the text 3.14159 throughout the program. This construction has long been popular in C. However, you can’t specify the data type of the constant using #define, which can lead to pro- gram bugs; so even in C #define has been superseded by const used with normal variables. However, you may encounter this construction in older programs. Type bool For completeness we should mention type bool here, although it won’t be important until we discuss relational operators in the next chapter. We’ve seen that variables of type int can have billions of possible values, and those of type char can have 256. Variables of type bool can have only two possible values: true and false. In theory a bool type requires only one bit (not byte) of storage, but in practice compilers often store them as bytes because a byte can be quickly accessed, while an individual bit must be extracted from a byte, which requires additional time. As we’ll see, type bool is most commonly used to hold the results of comparisons. Is alpha less than beta? If so, a bool value is given the value true; if not, it’s given the value false. C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 51 03 3087 CH02 11/29/01 2:19 PM Page 51Type bool gets its name from George Boole, a 19th century English mathematician who invented the concept of using logical operators with true-or-false values. Thus such true/false values are often called Boolean values. The setw Manipulator We’ve mentioned that manipulators are operators used with the insertion operator (<<) to mod- ify—or manipulate—the way data is displayed. We’ve already seen the endl manipulator; now we’ll look at another one: setw, which changes the field width of output. You can think of each value displayed by cout as occupying a field: an imaginary box with a certain width. The default field is just wide enough to hold the value. That is, the integer 567 will occupy a field three characters wide, and the string “pajamas” will occupy a field seven characters wide. However, in certain situations this may not lead to optimal results. Here’s an example. The WIDTH1program prints the names of three cities in one column, and their popula- tions in another. // width1.cpp // demonstrates need for setw manipulator #include using namespace std; int main() { long pop1=2425785, pop2=47, pop3=9761; cout << “LOCATION “ << “POP.” << endl << “Portcity “ << pop1 << endl << “Hightown “ << pop2 << endl << “Lowville “ << pop3 << endl; return 0; } Here’s the output from this program: LOCATION POP. Portcity 2425785 Hightown 47 Lowville 9761 Unfortunately, this format makes it hard to compare the numbers; it would be better if they lined up to the right. Also, we had to insert spaces into the names of the cities to separate them from the numbers. This is an inconvenience. Chapter 252 03 3087 CH02 11/29/01 2:19 PM Page 52Here’s a variation of this program, WIDTH2, that uses the setw manipulator to eliminate these problems by specifying field widths for the names and the numbers: // width2.cpp // demonstrates setw manipulator #include #include // for setw using namespace std; int main() { long pop1=2425785, pop2=47, pop3=9761; cout << setw(8) << “LOCATION” << setw(12) << “POPULATION” << endl << setw(8) << “Portcity” << setw(12) << pop1 << endl << setw(8) << “Hightown” << setw(12) << pop2 << endl << setw(8) << “Lowville” << setw(12) << pop3 << endl; return 0; } The setw manipulator causes the number (or string) that follows it in the stream to be printed within a field n characters wide, where n is the argument to setw(n). The value is right- justified within the field. Figure 2.8 shows how this looks. Type long is used for the population figures, which prevents a potential overflow problem on systems that use 2-byte integer types, in which the largest integer value is 32,767. C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 53 FIGURE 2.8 Field widths and setw. 03 3087 CH02 11/29/01 2:19 PM Page 53Here’s the output of WIDTH2: LOCATION POPULATION Portcity 2425785 Hightown 47 Lowville 9761 Cascading the Insertion Operator Note that there’s only one cout statement in WIDTH1 and WIDTH2, although it’s written on mul- tiple lines. In doing this, we take advantage of the fact that the compiler ignores whitespace, and that the insertion operator can be cascaded. The effect is the same as using four separate statements, each beginning with cout. Multiple Definitions We initialized the variables pop1, pop2, and pop3 to specific values at the same time we defined them. This is similar to the way we initialized char variables in the CHARVARS example. Here, however, we’ve defined and initialized all three variables on one line, using the same long keyword and separating the variable names with commas. This saves space where a num- ber of variables are all the same type. The IOMANIP Header File The declarations for the manipulators (except endl) are not in the usual IOSTREAM header file, but in a separate header file called IOMANIP. When you use these manipulators you must #include this header file in your program, as we do in the WIDTH2example. Variable Type Summary Our program examples so far have used four data types—int, char, float,and long. In addition we’ve mentioned types bool, short, double, and long double. Let’s pause now to summarize these data types. Table 2.2 shows the keyword used to define the type, the numerical range the type can accommodate, the digits of precision (in the case of floating- point numbers), and the bytes of memory occupied in a 32-bit environment. TABLE 2.2 Basic C++ Variable Types Numerical Range Digits of Bytes of Keyword Low High Precision Memory bool false true n/a 1 char –128 127 n/a 1 short –32,768 32,767 n/a 2 Chapter 254 03 3087 CH02 11/29/01 2:19 PM Page 54int –2,147,483,648 2,147,483,647 n/a 4 long –2,147,483,648 2,147,483,647 n/a 4 float 3.4 x 10–38 3.4 x 1038 74 double 1.7 x 10–308 1.7 x 10308 15 8 unsigned Data Types By eliminating the sign of the character and integer types, you can change their range to start at 0 and include only positive numbers. This allows them to represent numbers twice as big as the signed type. Table 2.3 shows the unsigned versions. TABLE 2.3 Unsigned Integer Types Numerical Range Bytes of Keyword Low High Memory unsigned char 0 255 1 unsigned short 0 65,535 2 unsigned int 04,294,967,295 4 unsigned long 04,294,967,295 4 The unsigned types are used when the quantities represented are always positive—such as when representing a count of something—or when the positive range of the signed types is not quite large enough. To change an integer type to an unsigned type, precede the data type keyword with the key- word unsigned. For example, an unsigned variable of type char would be defined as unsigned char ucharvar; Exceeding the range of signed types can lead to obscure program bugs. In certain (probably rare) situations such bugs can be eliminated by using unsigned types. For example, the follow- ing program stores the constant 1,500,000,000 (1.5 billion) both as an int in signedVar and as an unsigned int in unsignVar. // signtest.cpp // tests signed and unsigned integers #include C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 55 TABLE 2.2 Continued Numerical Range Digits of Bytes of Keyword Low High Precision Memory 03 3087 CH02 11/29/01 2:19 PM Page 55using namespace std; int main() { int signedVar = 1500000000; //signed unsigned int unsignVar = 1500000000; //unsigned signedVar = (signedVar * 2) / 3; //calculation exceeds range unsignVar = (unsignVar * 2) / 3; //calculation within range cout << “signedVar = “ << signedVar << endl; //wrong cout << “unsignVar = “ << unsignVar << endl; //OK return 0; } The program multiplies both variables by 2, then divides them by 3. Although the result is smaller than the original number, the intermediate calculation is larger than the original num- ber. This is a common situation, but it can lead to trouble. In SIGNTEST we expect that two- thirds the original value, or 1,000,000,000, will be restored to both variables. Unfortunately, in signedVar the multiplication created a result—3,000,000,000—that exceeded the range of the int variable (–2,147,483,648 to 2,147,483,647). Here’s the output: signedVar = -431,655,765 unsignVar = 1,000,000,000 The signed variable now displays an incorrect answer, while the unsigned variable, which is large enough to hold the intermediate result of the multiplication, records the result correctly. The moral is this: Be careful that all values generated in your program are within the range of the variables that hold them. (The results will be different on 16-bit or 64-bit computers, which use different numbers of bytes for type int.) Type Conversion C++, like C, is more forgiving than some languages in the way it treats expressions involving several different data types. As an example, consider the MIXED program: // mixed.cpp // shows mixed expressions #include using namespace std; int main() { int count = 7; float avgWeight = 155.5F; Chapter 256 03 3087 CH02 11/29/01 2:19 PM Page 56double totalWeight = count * avgWeight; cout << “totalWeight=” << totalWeight << endl; return 0; } Here a variable of type int is multiplied by a variable of type float to yield a result of type double. This program compiles without error; the compiler considers it normal that you want to multiply (or perform any other arithmetic operation on) numbers of different types. Not all languages are this relaxed. Some don’t permit mixed expressions, and would flag the line that performs the arithmetic in MIXED as an error. Such languages assume that when you mix types you’re making a mistake, and they try to save you from yourself. C++ and C, how- ever, assume that you must have a good reason for doing what you’re doing, and they help carry out your intentions. This is one reason for the popularity of C++ and C. They give you more freedom. Of course, with more freedom, there’s also more opportunity for you to make a mistake. Automatic Conversions Let’s consider what happens when the compiler confronts such mixed-type expressions as the one in MIXED. Types are considered “higher” or “lower,” based roughly on the order shown in Table 2.4. TABLE 2.4 Order of Data Types Data Type Order long double Highest double float long int short char Lowest The arithmetic operators such as + and * like to operate on two operands of the same type. When two operands of different types are encountered in the same expression, the lower-type variable is converted to the type of the higher-type variable. Thus in MIXED, the int value of count is converted to type float and stored in a temporary variable before being multiplied by the float variable avgWeight. The result (still of type float) is then converted to double so that it can be assigned to the double variable totalWeight. This process is shown in Figure 2.9. C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 57 03 3087 CH02 11/29/01 2:19 PM Page 57FIGURE 2.9 Data conversion. These conversions take place invisibly, and ordinarily you don’t need to think too much about them; C++ automatically does what you want. However, sometimes the compiler isn’t so happy about conversions, as we’ll see in a moment. Also, when we start to use objects, we will in effect be defining our own data types. We may want to use these new data types in mixed expressions, just as we use normal variables in mixed expressions. When this is the case, we must be careful to create our own conversion routines to change objects of one type into objects of another. The compiler won’t do it for us, as it does here with the built-in data types. Casts Casts sounds like something to do with social classes in India, but in C++ the term applies to data conversions specified by the programmer, as opposed to the automatic data conversions we just described. Casts are also called type casts. What are casts for? Sometimes a program- mer needs to convert a value from one type to another in a situation where the compiler will not do it automatically or without complaining. There are several kinds of casts in Standard C++: static casts, dynamic casts, reinterpret casts, and const casts. Here we’ll be concerned only with static casts; we’ll learn about the others, which are used in more specialized situations, in later chapters. Chapter 258 03 3087 CH02 11/29/01 2:19 PM Page 58C++ casts have a rather forbidding appearance. Here’s a statement that uses a C++ cast to change a variable of type int into a variable of type char: aCharVar = static_cast(anIntVar); Here the variable to be cast (anIntVar) is placed in parentheses and the type it’s to be changed to (char) is placed in angle brackets. The result is that anIntVar is changed to type char before it’s assigned to aCharVar. In this case the assignment statement would have carried out the cast itself, but there are situations where the cast is essential. Recall that in the SIGNTEST example an intermediate result exceeded the capacity of the vari- able type, resulting in an erroneous result. We fixed the problem by using unsigned int instead of int. This worked because the intermediate result—3,000,000,000—would fit in the range of the unsigned variable. But suppose an intermediate result won’t fit the unsigned type either. In such a case we might be able to solve the problem by using a cast. Here’s an example: // cast.cpp // tests signed and unsigned integers #include using namespace std; int main() { int intVar = 1500000000; //1,500,000,000 intVar = (intVar * 10) / 10; //result too large cout << “intVar = “ << intVar << endl; //wrong answer intVar = 1500000000; //cast to double intVar = (static_cast(intVar) * 10) / 10; cout << “intVar = “ << intVar << endl; //right answer return 0; } When we multiply the variable intVar by 10, the result—15,000,000,000—is far too large to fit in a variable of type int or unsigned int. This leads to the wrong answer, as shown by the output of the first part of the program. We could redefine the data type of the variables to be double; this provides plenty of room, since this type holds numbers with up to 15 digits. But suppose that for some reason, such as keeping the program small, we don’t want to change the variables to type double. In this case there’s another solution: We can cast intVar to type double before multiplying. This is some- times called coercion; the data is coerced into becoming another type. The expression static_cast(intVar) C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 59 03 3087 CH02 11/29/01 2:19 PM Page 59casts intVar to type double. It generates a temporary variable of type double with the same value as intVar. It is this temporary variable that is multiplied by 10. Since it is type double, the result fits. This result is then divided by 10 and assigned to the normal int variable intVar. Here’s the program’s output: intVar = 211509811 intVar = 1500000000 The first answer, without the cast, is wrong; but in the second answer, the cast produces the correct result. Before Standard C++, casts were handled using quite a different format. Instead of aCharVar = static_cast(anIntVar); you could say aCharVar = (char)anIntVar; or alternatively aCharVar = char(anIntVar); One problem with these approaches is that they are hard to see; the syntax blends into the rest of the listing. They are also hard to search for using a Find operation with your source code editor. The new format solves this problem: static_cast is easy to see and easy to search for. These old casts still work, but their use is discouraged (or deprecated,to use the technical term). Casts should be used only when absolutely necessary. They are a controlled way of evading type safety (which means making sure that variables don’t change types by mistake) and can lead to trouble because they make it impossible for the compiler to spot potential problems. However, sometimes casts can’t be avoided. We’ll see some examples of situations where casts are necessary as we go along. Arithmetic Operators As you have probably gathered by this time, C++ uses the four normal arithmetic operators +, -, *, and / for addition, subtraction, multiplication, and division. These operators work on all the data types, both integer and floating-point. They are used in much the same way that they are used in other languages, and are closely analogous to their use in algebra. However, there are some other arithmetic operators whose use is not so obvious. Chapter 260 03 3087 CH02 11/29/01 2:19 PM Page 60The Remainder Operator There is a fifth arithmetic operator that works only with integer variables (types char, short, int, and long). It’s called the remainder operator, and is represented by the percent symbol (%). This operator (also called the modulus operator) finds the remainder when one number is divided by another. The REMAIND program demonstrates the effect. // remaind.cpp // demonstrates remainder operator #include using namespace std; int main() { cout << 6 % 8 << endl // 6 << 7 % 8 << endl // 7 << 8 % 8 << endl // 0 << 9 % 8 << endl // 1 << 10 % 8 << endl; // 2 return 0; } Here the numbers 6–10 are divided by 8, using the remainder operator. The answers are 6, 7, 0, 1, and 2—the remainders of these divisions. The remainder operator is used in a wide variety of situations. We’ll show examples as we go along. A note about precedence: In the expression cout << 6 % 8 the remainder operator is evaluated first because it has higher precedence than the << operator. If it did not, we would need to put parentheses around 6%8to ensure it was evaluated before being acted on by <<. Arithmetic Assignment Operators C++ offers several ways to shorten and clarify your code. One of these is the arithmetic assignment operator. This operator helps to give C++ listings their distinctive appearance. The following kind of statement is common in most languages. total = total + item; // adds “item” to “total” In this situation you add something to an existing value (or you perform some other arithmetic operation on it). But the syntax of this statement offends those for whom brevity is important, because the name total appears twice. So C++ offers a condensed approach: the arithmetic assignment operator, which combines an arithmetic operator and an assignment operator and C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 61 03 3087 CH02 11/29/01 2:19 PM Page 61eliminates the repeated operand. Here’s a statement that has exactly the same effect as the pre- ceding one. total += item; // adds “item” to “total” Figure 2.10 emphasizes the equivalence of the two forms. Chapter 262 FIGURE 2.10 Arithmetic assignment operator. There are arithmetic assignment operators corresponding to all the arithmetic operations: +=, -=, *=, /=, and %= (and some other operators as well). The following example shows the arith- metic assignment operators in use: // assign.cpp // demonstrates arithmetic assignment operators #include using namespace std; int main() { int ans = 27; ans += 10; //same as: ans = ans + 10; cout << ans << “, “; ans -= 7; //same as: ans = ans - 7; cout << ans << “, “; ans *= 2; //same as: ans = ans * 2; cout << ans << “, “; 03 3087 CH02 11/29/01 2:19 PM Page 62ans /= 3; //same as: ans = ans / 3; cout << ans << “, “; ans %= 3; //same as: ans = ans % 3; cout << ans << endl; return 0; } Here’s the output from this program: 37, 30, 60, 20, 2 You don’t need to use arithmetic assignment operators in your code, but they are a common feature of the language; they’ll appear in numerous examples in this book. Increment Operators Here’s an even more specialized operator. You often need to add 1 to the value of an existing variable. You can do this the “normal” way: count = count + 1; // adds 1 to “count” Or you can use an arithmetic assignment operator: count += 1; // adds 1 to “count” But there’s an even more condensed approach: ++count; // adds 1 to “count” The ++ operator increments (adds 1 to) its argument. Prefix and Postfix As if this weren’t weird enough, the increment operator can be used in two ways: as a prefix, meaning that the operator precedes the variable; and as a postfix, meaning that the operator fol- lows the variable. What’s the difference? Often a variable is incremented within a statement that performs some other operation on it. For example totalWeight = avgWeight * ++count; The question here is this: Is the multiplication performed before or after count is incremented? In this case count is incremented first. How do we know that? Because prefix notation is used: ++count. If we had used postfix notation, count++,the multiplication would have been per- formed first, then count would have been incremented. This is shown in Figure 2.11. C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 63 03 3087 CH02 11/29/01 2:19 PM Page 63FIGURE 2.11 The increment operator. Here’s an example that shows both the prefix and postfix versions of the increment operator: // increm.cpp // demonstrates the increment operator #include using namespace std; int main() { int count = 10; cout << “count=” << count << endl; //displays 10 cout << “count=” << ++count << endl; //displays 11 (prefix) cout << “count=” << count << endl; //displays 11 cout << “count=” << count++ << endl; //displays 11 (postfix) cout << “count=” << count << endl; //displays 12 return 0; } Here’s the program’s output: count=10 count=11 Chapter 264 03 3087 CH02 11/29/01 2:19 PM Page 64count=11 count=11 count=12 The first time count is incremented, the prefix ++ operator is used. This causes the increment to happen at the beginning of the statement evaluation, before the output operation has been carried out. When the value of the expression ++count is displayed, it has already been incre- mented, and << sees the value 11. The second time count is incremented, the postfix ++ opera- tor is used. When the expression count++ is displayed, it retains its unincremented value of 11. Following the completion of this statement, the increment takes effect, so that in the last state- ment of the program we see that count has acquired the value 12. The Decrement (--) Operator The decrement operator, --, behaves very much like the increment operator, except that it sub- tracts 1 from its operand. It too can be used in both prefix and postfix forms. Library Functions Many activities in C++ are carried out by library functions. These functions perform file access, mathematical computations, and data conversion, among other things. We don’t want to dig too deeply into library functions before we explain how functions work (see Chapter 5), but you can use simple library functions without a thorough understanding of their operation. The next example, SQRT,uses the library function sqrt() to calculate the square root of a num- ber entered by the user. // sqrt.cpp // demonstrates sqrt() library function #include //for cout, etc. #include //for sqrt() using namespace std; int main() { double number, answer; //sqrt() requires type double cout << “Enter a number: “; cin >> number; //get the number answer = sqrt(number); //find square root cout << “Square root is “ << answer << endl; //display it return 0; } C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 65 03 3087 CH02 11/29/01 2:19 PM Page 65The program first obtains a number from the user. This number is then used as an argument to the sqrt() function, in the statement answer = sqrt(number); An argument is the input to the function; it is placed inside the parentheses following the func- tion name. The function then processes the argument and returns a value; this is the output from the function. In this case the return value is the square root of the original number. Returning a value means that the function expression takes on this value, which can then be assigned to another variable—in this case answer. The program then displays this value. Here’s some output from the program: Enter a number: 1000 Square root is 31.622777 Multiplying 31.622777 by itself on your pocket calculator will verify that this answer is pretty close. The arguments to a function, and their return values, must be the correct data type. You can find what these data types are by looking at the description of the library function in your com- piler’s help file, which describes each of the hundreds of library functions. For sqrt(),the description specifies both an argument and a return value of type double,so we use variables of this type in the program. Header Files As with cout and other such objects, you must #include a header file that contains the decla- ration of any library functions you use. In the documentation for the sqrt() function, you’ll see that the specified header file is CMATH. In SQRT the preprocessor directive #include takes care of incorporating this header file into our source file. If you don’t include the appropriate header file when you use a library function, you’ll get an error message like this from the compiler: ‘sqrt’ unidentified identifier. Library Files We mentioned earlier that various files containing library functions and objects will be linked to your program to create an executable file. These files contain the actual machine-executable code for the functions. Such library files often have the extension .LIB. The sqrt() function is found in such a file. It is automatically extracted from the file by the linker, and the proper connections are made so that it can be called (that is, invoked or accessed) from the SQRT pro- gram. Your compiler takes care of all these details for you, so ordinarily you don’t need to worry about the process. However, you should understand what these files are for. Chapter 266 03 3087 CH02 11/29/01 2:19 PM Page 66Header Files and Library Files The relationship between library files and header files can be confusing, so let’s review it. To use a library function like sqrt(), you must link the library file that contains it to your pro- gram. The appropriate functions from the library file are then connected to your program by the linker. However, that’s not the end of the story. The functions in your source file need to know the names and types of the functions and other elements in the library file. They are given this information in a header file. Each header file contains information for a particular group of functions. The functions themselves are grouped together in a library file, but the information about them is scattered throughout a number of header files. The IOSTREAM header file contains information for various I/O functions and objects, including cout, while the CMATH header file contains information for mathematics functions such as sqrt(). If you were using string func- tions such as strcpy(), you would include STRING.H, and so on. Figure 2.12 shows the relationship of header files and library files to the other files used in program development. The use of header files is common in C++. Whenever you use a library function or a prede- fined object or operator, you will need to use a header file that contains appropriate declara- tions. Two Ways to Use #include You can use #include in two ways. The angle brackets < and > surrounding the filenames IOSTREAM and CMATH in the SQRT example indicate that the compiler should begin searching for these files in the standard INCLUDE directory. This directory, which is traditionally called INCLUDE, holds the header files supplied by the compiler manufacturer for the system. Instead of angle brackets around the filename, you can also use quotation marks, as in #include “myheader.h” Quotation marks instruct the compiler to begin its search for the header file in the current directory; this is usually the directory that contains the source file. You normally use quotation marks for header files you write yourself (a situation we’ll explore in Chapter 13, “Multifile Programs”). Quotation marks or angle brackets work in any case, but making the appropriate choice speeds up the compilation process slightly by giving the compiler a hint about where to find the file. Appendix C, “Microsoft Visual C++,” and Appendix D, “Borland C++Builder,” explain how to handle header files with specific compilers. C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 67 03 3087 CH02 11/29/01 2:19 PM Page 67Chapter 268 Compiler Linker Source file MYPROG1.CPP Object file Library file MYPROG1.OBJ SOMELIB.LIB MYPROG1.EXE Library header file SOMELIB.HUser header file #include #include "myprog.h" MYPROG.H FIGURE 2.12 Header and library files. Summary In this chapter we’ve learned that a major building block of C++ programs is the function. A function named main() is always the first one executed when a program is executed. A function is composed of statements, which tell the computer to do something. Each state- ment ends with a semicolon. A statement may contain one or more expressions, which are sequences of variables and operators that usually evaluate to a specific value. Output is most commonly handled in C++ with the cout object and << insertion operator, which together cause variables or constants to be sent to the standard output device—usually the screen. Input is handled with cin and the extraction operator >>, which cause values to be received from the standard input device—usually the keyboard. 03 3087 CH02 11/29/01 2:19 PM Page 68Various data types are built into C++: char, int, long, and short are the integer types and float, double, and long double are the floating-point types. All of these types are signed. Unsigned versions of the integer types, signaled by the keyword unsigned, don’t hold negative numbers but hold positive ones twice as large. Type bool is used for Boolean variables and can hold only the constants true or false. The const keyword stipulates that a variable’s value will not change in the course of a pro- gram. Strictly speaking, the variable is no longer a variable but a constant. A variable is automatically converted from one type to another in mixed expressions (those involving different data types) and by casting, which allows the programmer to specify a con- version. C++ employs the usual arithmetic operators +, -, *, and /. In addition, the remainder operator, %,returns the remainder of integer division. The arithmetic assignment operators +=, +-, and so on perform an arithmetic operation and an assignment simultaneously. The increment and decrement operators ++ and -- increase or decrease a variable by 1. Preprocessor directives consist of instructions to the compiler, rather than to the computer. The #include directive tells the compiler to insert another file into the present source file, and the #define directive tells it to substitute one thing for another. The using directive tells the com- piler to recognize names that are in a certain namespace. If you use a library function in your program, the code for the function is in a library file, which is automatically linked to your program. A header file containing the function’s declara- tion must be inserted into your source file with an #include statement. Questions Answers to these questions can be found in Appendix G. 1. Dividing a program into functions a. is the key to object-oriented programming. b. makes the program easier to conceptualize. c. may reduce the size of the program. d. makes the program run faster. 2. A function name must be followed by ________. 3. A function body is delimited by ________. 4. Why is the main() function special? 5. A C++ instruction that tells the computer to do something is called a ________. C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 69 03 3087 CH02 11/29/01 2:19 PM Page 696. Write an example of a normal C++ comment and an example of an old-fashioned /* comment. 7. An expression a. usually evaluates to a numerical value. b. indicates the emotional state of the program. c. always occurs outside a function. d. may be part of a statement. 8. Specify how many bytes are occupied by the following data types in a 32-bit system: a. Type int b. Type long double c. Type float d. Type long 9. True or false: A variable of type char can hold the value 301. 10. What kind of program elements are the following? a. 12 b. ‘a’ c. 4.28915 d. JungleJim e. JungleJim() 11. Write statements that display on the screen a. the character ‘x’ b. the name Jim c. the number 509 12. True or false: In an assignment statement, the value on the left of the equal sign is always equal to the value on the right. 13. Write a statement that displays the variable george in a field 10 characters wide. 14. What header file must you #include with your source file to use cout and cin? 15. Write a statement that gets a numerical value from the keyboard and places it in the vari- able temp. 16. What header file must you #include with your program to use setw? 17. Two exceptions to the rule that the compiler ignores whitespace are ________ and ________. Chapter 270 03 3087 CH02 11/29/01 2:19 PM Page 7018. True or false: It’s perfectly all right to use variables of different data types in the same arithmetic expression. 19. The expression 11%3 evaluates to ________. 20. An arithmetic assignment operator combines the effect of what two operators? 21. Write a statement that uses an arithmetic assignment operator to increase the value of the variable temp by 23. Write the same statement without the arithmetic assignment operator. 22. The increment operator increases the value of a variable by how much? 23. Assuming var1 starts with the value 20, what will the following code fragment print out? cout << var1--; cout << ++var1; 24. In the examples we’ve seen so far, header files have been used for what purpose? 25. The actual code for library functions is contained in a ________ file. Exercises Answers to the starred exercises can be found in Appendix G. *1. Assuming there are 7.481 gallons in a cubic foot, write a program that asks the user to enter a number of gallons, and then displays the equivalent in cubic feet. *2. Write a program that generates the following table: 1990 135 1991 7290 1992 11300 1993 16200 Use a single cout statement for all output. *3. Write a program that generates the following output: 10 20 19 Use an integer constant for the 10, an arithmetic assignment operator to generate the 20, and a decrement operator to generate the 19. 4. Write a program that displays your favorite poem. Use an appropriate escape sequence for the line breaks. If you don’t have a favorite poem, you can borrow this one by Ogden Nash: Candy is dandy, But liquor is quicker. C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 71 03 3087 CH02 11/29/01 2:19 PM Page 715. A library function, islower(),takes a single character (a letter) as an argument and returns a nonzero integer if the letter is lowercase, or zero if it is uppercase. This func- tion requires the header file CTYPE.H. Write a program that allows the user to enter a let- ter, and then displays either zero or nonzero, depending on whether a lowercase or uppercase letter was entered. (See the SQRT program for clues.) 6. On a certain day the British pound was equivalent to $1.487 U.S., the French franc was $0.172, the German deutschemark was $0.584, and the Japanese yen was $0.00955. Write a program that allows the user to enter an amount in dollars, and then displays this value converted to these four other monetary units. 7. You can convert temperature from degrees Celsius to degrees Fahrenheit by multiplying by 9/5 and adding 32. Write a program that allows the user to enter a floating-point num- ber representing degrees Celsius, and then displays the corresponding degrees Fahrenheit. 8. When a value is smaller than a field specified with setw(),the unused locations are, by default, filled in with spaces. The manipulator setfill() takes a single character as an argument and causes this character to be substituted for spaces in the empty parts of a field. Rewrite the WIDTH program so that the characters on each line between the location name and the population number are filled in with periods instead of spaces, as in Portcity.....2425785 9. If you have two fractions, a/b and c/d, their sum can be obtained from the formula a c a*d + b*c --- + --- = ----------- b d b*d For example, 1/4 plus 2/3 is 1 2 1*3 + 4*2 3 + 8 11 --- + --- = ----------- = ------- = ---- 4 3 4*3 12 12 Write a program that encourages the user to enter two fractions, and then displays their sum in fractional form. (You don’t need to reduce it to lowest terms.) The interaction with the user might look like this: Enter first fraction: 1/2 Enter second fraction: 2/5 Sum = 9/10 You can take advantage of the fact that the extraction operator (>>) can be chained to read in more than one quantity at once: cin >> a >> dummychar >> b; Chapter 272 03 3087 CH02 11/29/01 2:19 PM Page 7210. In the heyday of the British empire, Great Britain used a monetary system based on pounds, shillings, and pence. There were 20 shillings to a pound, and 12 pence to a shilling. The notation for this old system used the pound sign, £, and two decimal points, so that, for example, £5.2.8 meant 5 pounds, 2 shillings, and 8 pence. (Pence is the plural of penny.) The new monetary system, introduced in the 1950s, consists of only pounds and pence, with 100 pence to a pound (like U.S. dollars and cents). We’ll call this new system decimal pounds. Thus £5.2.8 in the old notation is £5.13 in decimal pounds (actu- ally £5.1333333). Write a program to convert the old pounds-shillings-pence format to decimal pounds. An example of the user’s interaction with the program would be Enter pounds: 7 Enter shillings: 17 Enter pence: 9 Decimal pounds = £7.89 In most compilers you can use the decimal number 156 (hex character constant ‘\x9c’) to represent the pound sign (£). In some compilers, you can put the pound sign into your program directly by pasting it from the Windows Character Map accessory. 11. By default, output is right-justified in its field. You can left-justify text output using the manipulator setiosflags(ios::left). (For now, don’t worry about what this new notation means.) Use this manipulator, along with setw(), to help generate the following output: Last name First name Street address Town State ------------------------------------------------------------ Jones Bernard 109 Pine Lane Littletown MI O’Brian Coleen 42 E. 99th Ave. Bigcity NY Wong Harry 121-A Alabama St. Lakeville IL 12. Write the inverse of Exercise 10, so that the user enters an amount in Great Britain’s new decimal-pounds notation (pounds and pence), and the program converts it to the old pounds-shillings-pence notation. An example of interaction with the program might be Enter decimal pounds: 3.51 Equivalent in old notation = £3.10.2. Make use of the fact that if you assign a floating-point value (say 12.34) to an integer variable, the decimal fraction (0.34) is lost; the integer value is simply 12. Use a cast to avoid a compiler warning. You can use statements like float decpounds; // input from user (new-style pounds) int pounds; // old-style (integer) pounds float decfrac; // decimal fraction (smaller than 1.0) pounds = static_cast(decpounds); // remove decimal fraction decfrac = decpounds - pounds; // regain decimal fraction You can then multiply decfrac by 20 to find shillings. A similar operation obtains pence. C++ Programming Basics 2 C++ P ROGRAMMING B ASICS 73 03 3087 CH02 11/29/01 2:19 PM Page 7303 3087 CH02 11/29/01 2:19 PM Page 74CHAPTER 3 Loops and Decisions IN THIS CHAPTER • Relational Operators 76 • Loops 78 • Decisions 93 • Logical Operators 114 •Precedence Summary 118 •Other Control Statements 118 04 3087 CH03 11/29/01 2:17 PM Page 75Chapter 376 Not many programs execute all their statements in strict order from beginning to end. Most programs (like many humans) decide what to do in response to changing circumstances. The flow of control jumps from one part of the program to another, depending on calculations per- formed in the program. Program statements that cause such jumps are called control statements. There are two major categories: loops and decisions. How many times a loop is executed, or whether a decision results in the execution of a section of code, depends on whether certain expressions are true or false. These expressions typically involve a kind of operator called a relational operator, which compares two values. Since the operation of loops and decisions is so closely involved with these operators, we’ll examine them first. Relational Operators A relational operator compares two values. The values can be any built-in C++ data type, such as char, int, and float,or—as we’ll see later—they can be user-defined classes. The compar- ison involves such relationships as equal to, less than, and greater than. The result of the com- parison is true or false; for example, either two values are equal (true), or they’re not (false). Our first program, RELAT, demonstrates relational operators in a comparison of integer vari- ables and constants. // relat.cpp // demonstrates relational operators #include using namespace std; int main() { int numb; cout << “Enter a number: “; cin >> numb; cout << “numb<10 is “ << (numb < 10) << endl; cout << “numb>10 is “ << (numb > 10) << endl; cout << “numb==10 is “ << (numb == 10) << endl; return 0; } This program performs three kinds of comparisons between 10 and a number entered by the user. Here’s the output when the user enters 20: Enter a number: 20 numb<10 is 0 numb>10 is 1 numb==10 is 0 04 3087 CH03 11/29/01 2:17 PM Page 76The first expression is true if numb is less than 10. The second expression is true if numb is greater than 10, and the third is true if numb is equal to 10. As you can see from the output, the C++ compiler considers that a true expression has the value 1, while a false expression has the value 0. As we mentioned in the last chapter, Standard C++ includes a type bool, which can hold one of two constant values, true or false. You might think that results of relational expressions like numb<10 would be of type bool, and that the program would print false instead of 0 and true instead of 1. In fact, C++ is rather schizophrenic on this point. Displaying the results of relational operations, or even the values of type bool variables, with cout<< yields 0 or 1, not false or true. Historically this is because C++ started out with no bool type. Before the advent of Standard C++, the only way to express false and true was with 0 and 1. Now false can be represented by either a bool value of false,or by an integer value of 0; and true can be represented by either a bool value of true or an integer value of 1. In most simple situations the difference isn’t apparent because we don’t need to display true/false values; we just use them in loops and decisions to influence what the program will do next. Here’s the complete list of C++ relational operators: Operator Meaning > Greater than (greater than) < Less than == Equal to != Not equal to >= Greater than or equal to <= Less than or equal to Now let’s look at some expressions that use relational operators, and also look at the value of each expression. The first two lines are assignment statements that set the values of the variables harry and jane. You might want to hide the comments with your old Jose Canseco baseball card and see whether you can predict which expressions evaluate to true and which to false. jane = 44; //assignment statement harry = 12; //assignment statement (jane == harry) //false (harry <= 12) //true (jane > harry) //true (jane >= 44) //true (harry != 12) // false (7 < harry) //true (0) //false (by definition) (44) //true (since it’s not 0) Loops and Decisions 3 L OOPS AND D ECISIONS 77 04 3087 CH03 11/29/01 2:17 PM Page 77Note that the equal operator, ==,uses two equal signs. A common mistake is to use a single equal sign—the assignment operator—as a relational operator. This is a nasty bug, since the compiler may not notice anything wrong. However, your program won’t do what you want (unless you’re very lucky). Although C++ generates a 1 to indicate true, it assumes that any value other than 0 (such as –7 or 44) is true; only 0 is false. Thus, the last expression in the list is true. Now let’s see how these operators are used in typical situations. We’ll examine loops first, then decisions. Loops Loops cause a section of your program to be repeated a certain number of times. The repetition continues while a condition is true. When the condition becomes false, the loop ends and con- trol passes to the statements following the loop. There are three kinds of loops in C++: the for loop, the while loop, and the do loop. The for Loop The for loop is (for many people, anyway) the easiest C++ loop to understand. All its loop- control elements are gathered in one place, while in the other loop constructions they are scat- tered about the program, which can make it harder to unravel how these loops work. The for loop executes a section of code a fixed number of times. It’s usually (although not always) used when you know, before entering the loop, how many times you want to execute the code. Here’s an example, FORDEMO,that displays the squares of the numbers from 0 to 14: // fordemo.cpp // demonstrates simple FOR loop #include using namespace std; int main() { int j; //define a loop variable for(j=0; j<15; j++) //loop from 0 to 14, cout << j * j << “ “; //displaying the square of j cout << endl; return 0; } Chapter 378 04 3087 CH03 11/29/01 2:17 PM Page 78Here’s the output: 0 1 4 9 16 25 36 49 64 81 100 121 144 169 196 How does this work? The for statement controls the loop. It consists of the keyword for,fol- lowed by parentheses that contain three expressions separated by semicolons: for(j=0; j<15; j++) These three expressions are the initialization expression,the test expression, and the increment expression,as shown in Figure 3.1. Loops and Decisions 3 L OOPS AND D ECISIONS 79 FIGURE 3.1 Syntax of the for loop. These three expressions usually (but not always) involve the same variable, which we call the loop variable. In the FORDEMO example the loop variable is j. It’s defined before the statements within the loop body start to execute. The body of the loop is the code to be executed each time through the loop. Repeating this code is the raison d’être for the loop. In this example the loop body consists of a single state- ment: cout << j * j << “ “; 04 3087 CH03 11/29/01 2:17 PM Page 79This statement prints out the square of j,followed by two spaces. The square is found by mul- tiplying j by itself. As the loop executes, j goes through the sequence 0, 1, 2, 3, and so on up to 14; so the squares of these numbers are displayed—0, 1, 4, 9, up to 196. Note that the for statement is not followed by a semicolon. That’s because the for statement and the loop body are together considered to be a program statement. This is an important detail. If you put a semicolon after the for statement, the compiler will think there is no loop body, and the program will do things you probably don’t expect. Let’s see how the three expressions in the for statement control the loop. The Initialization Expression The initialization expression is executed only once, when the loop first starts. It gives the loop variable an initial value. In the FORDEMO example it sets j to 0. The Test Expression The test expression usually involves a relational operator. It is evaluated each time through the loop, just before the body of the loop is executed. It determines whether the loop will be exe- cuted again. If the test expression is true, the loop is executed one more time. If it’s false, the loop ends, and control passes to the statements following the loop. In the FORDEMO example the statement cout << endl; is executed following the completion of the loop. The Increment Expression The increment expression changes the value of the loop variable, often by incrementing it. It is always executed at the end of the loop, after the loop body has been executed. Here the incre- ment operator ++ adds 1 to j each time through the loop. Figure 3.2 shows a flowchart of a for loop’s operation. How Many Times? The loop in the FORDEMO example executes exactly 15 times. The first time, j is 0. This is ensured in the initialization expression. The last time through the loop, j is 14. This is deter- mined by the test expression j<15. When j becomes 15, the loop terminates; the loop body is not executed when j has this value. The arrangement shown is commonly used to do some- thing a fixed number of times: start at 0, use a test expression with the less-than operator and a value equal to the desired number of iterations, and increment the loop variable after each iter- ation. Chapter 380 04 3087 CH03 11/29/01 2:17 PM Page 80FIGURE 3.2 Operation of the for loop. Here’s another for loop example: for(count=0; count<100; count++) // loop body How many times will the loop body be repeated here? Exactly 100 times, with count going from 0 to 99. Multiple Statements in the Loop Body Of course you may want to execute more than one statement in the loop body. Multiple state- ments are delimited by braces, just as functions are. Note that there is no semicolon following the final brace of the loop body, although there are semicolons following the individual state- ments in the loop body. The next example, CUBELIST,uses three statements in the loop body. It prints out the cubes of the numbers from 1 to 10, using a two-column format. // cubelist.cpp // lists cubes from 1 to 10 #include Loops and Decisions 3 L OOPS AND D ECISIONS 81 04 3087 CH03 11/29/01 2:17 PM Page 81#include //for setw using namespace std; int main() { int numb; //define loop variable for(numb=1; numb<=10; numb++) //loop from 1 to 10 { cout << setw(4) << numb; //display 1st column int cube = numb*numb*numb; //calculate cube cout << setw(6) << cube << endl; //display 2nd column } return 0; } Here’s the output from the program: 1 1 2 8 3 27 4 64 5 125 6 216 7 343 8 512 9 729 10 1000 We’ve made another change in the program to show there’s nothing immutable about the for- mat used in the last example. The loop variable is initialized to 1, not to 0, and it ends at 10, not at 9, by virtue of <=,the less-than-or-equal-to operator. The effect is that the loop body is executed 10 times, with the loop variable running from 1 to 10 (not from 0 to 9). We should note that you can also put braces around the single statement loop body shown pre- viously. They’re not necessary, but many programmers feel it improves clarity to use them whether the loop body consists of a single statement or not. Blocks and Variable Visibility The loop body, which consists of braces delimiting several statements, is called a block of code. One important aspect of a block is that a variable defined inside the block is not visible outside it. Visible means that program statements can access or “see” the variable. (We’ll dis- cuss visibility further in Chapter 5, “Functions.”) In CUBELIST we define the variable cube inside the block, in the statement int cube = numb*numb*numb; Chapter 382 04 3087 CH03 11/29/01 2:17 PM Page 82You can’t access this variable outside the block; it’s only visible within the braces. Thus if you placed the statement cube = 10; after the loop body, the compiler would signal an error because the variable cube would be undefined outside the loop. One advantage of restricting the visibility of variables is that the same variable name can be used within different blocks in the same program. (Defining variables inside a block, as we did in CUBELIST, is common in C++ but is not popular in C.) Indentation and Loop Style Good programming style dictates that the loop body be indented—that is, shifted right, relative to the loop statement (and to the rest of the program). In the FORDEMO example one line is indented, and in CUBELIST the entire block, including the braces, is indented. This indentation is an important visual aid to the programmer: It makes it easy to see where the loop body begins and ends. The compiler doesn’t care whether you indent or not (at least there’s no way to tell if it cares). There is a common variation on the style we use for loops in this book. We show the braces aligned vertically, but some programmers prefer to place the opening brace just after the loop statement, like this: for(numb=1; numb<=10; numb++) { cout << setw(4) << numb; int cube = numb*numb*numb; cout << setw(6) << cube << endl; } This saves a line in the listing but makes it more difficult to read, since the opening brace is harder to see and harder to match with the corresponding closing brace. Another style is to indent the body but not the braces: for(numb=1; numb<=10; numb++) { cout << setw(4) << numb; int cube = numb*numb*numb; cout << setw(6) << cube << endl; } This is a common approach, but at least for some people it makes it harder for the eye to con- nect the braces to the loop body. However, you can get used to almost anything. Whatever style you choose, use it consistently. Loops and Decisions 3 L OOPS AND D ECISIONS 83 04 3087 CH03 11/29/01 2:17 PM Page 83Debugging Animation You can use the debugging features built into your compiler to create a dramatic animated dis- play of loop operation. The key feature is single-stepping. Your compiler makes this easy. Start by opening a project for the program to be debugged, and a window containing the source file. The exact instructions necessary to launch the debugger vary with different compilers, so con- sult Appendix C, “Microsoft Visual C++,” or Appendix D, “Borland C++Builder,” as appropri- ate. By pressing a certain function key you can cause one line of your program to be executed at a time. This will show you the sequence of statements executed as the program proceeds. In a loop you’ll see the statements within the loop executed; then control will jump back to the start of the loop and the cycle will be repeated. You can also use the debugger to watch what happens to the values of different variables as you single-step through the program. This is a powerful tool when you’re debugging your pro- gram. You can experiment with this technique with the CUBELIST program by putting the numb and cube variables in a Watch window in your debugger and seeing how they change as the program proceeds. Again, consult the appropriate appendix for instructions on how to use Watch windows. Single-stepping and the Watch window are powerful debugging tools. If your program doesn’t behave as you think it should, you can use these features to monitor the values of key variables as you step through the program. Usually the source of the problem will become clear. for Loop Variations The increment expression doesn’t need to increment the loop variable; it can perform any oper- ation it likes. In the next example it decrements the loop variable. This program, FACTOR, asks the user to type in a number, and then calculates the factorial of this number. (The factorial is calculated by multiplying the original number by all the positive integers smaller than itself. Thus the factorial of 5 is 5*4*3*2*1, or 120.) // factor.cpp // calculates factorials, demonstrates FOR loop #include using namespace std; int main() { unsigned int numb; unsigned long fact=1; //long for larger numbers cout << “Enter a number: “; cin >> numb; //get number Chapter 384 04 3087 CH03 11/29/01 2:17 PM Page 84for(int j=numb; j>0; j--) //multiply 1 by fact *= j; //numb, numb-1, ..., 2, 1 cout << “Factorial is “ << fact << endl; return 0; } In this example the initialization expression sets j to the value entered by the user. The test expression causes the loop to execute as long as j is greater than 0. The increment expression decrements j after each iteration. We’ve used type unsigned long for the factorial, since the factorials of even small numbers are very large. On 32-bit systems such as Windows int is the same as long,but long gives added capacity on 16-bit systems. The following output shows how large factorials can be, even for small input numbers: Enter a number: 10 Factorial is 3628800 The largest number you can use for input is 12. You won’t get an error message for larger inputs, but the results will be wrong, as the capacity of type long will be exceeded. Variables Defined in for Statements There’s another wrinkle in this program: The loop variable j is defined inside the for state- ment: for(int j=numb; j>0; j--) This is a common construction in C++, and in most cases it’s the best approach to loop vari- ables. It defines the variable as closely as possible to its point of use in the listing. Variables defined in the loop statement this way are visible in the loop body only. (The Microsoft com- piler makes them visible from the point of definition onward to the end of the file, but this is not Standard C++.) Multiple Initialization and Test Expressions You can put more than one expression in the initialization part of the for statement, separating the different expressions by commas. You can also have more than one increment expression. You can have only one test expression. Here’s an example: for( j=0, alpha=100; j<50; j++, beta-- ) { // body of loop } This example has a normal loop variable j,but it also initializes another variable, alpha,and decrements a third, beta. The variables alpha and beta don’t need to have anything to do with each other, or with j. Multiple initialization expressions and multiple increment expressions are separated by commas. Loops and Decisions 3 L OOPS AND D ECISIONS 85 04 3087 CH03 11/29/01 2:17 PM Page 85Actually, you can leave out some or all of the expressions if you want to. The expression for(;;) is the same as a while loop with a test expression of true. We’ll look at while loops next. We’ll avoid using such multiple or missing expressions. While these approaches can make the listing more concise, they also tend to decrease its readability. It’s always possible to use stand- alone statements or a different form of loop to achieve the same effect. The while Loop The for loop does something a fixed number of times. What happens if you don’t know how many times you want to do something before you start the loop? In this case a different kind of loop may be used: the while loop. The next example, ENDON0, asks the user to enter a series of numbers. When the number entered is 0, the loop terminates. Notice that there’s no way for the program to know in advance how many numbers will be typed before the 0 appears; that’s up to the user. // endon0.cpp // demonstrates WHILE loop #include using namespace std; int main() { int n = 99; // make sure n isn’t initialized to 0 while( n != 0 ) // loop until n is 0 cin >> n; // read a number into n cout << endl; return 0; } Here’s some sample output. The user enters numbers, and the loop continues until 0 is entered, at which point the loop and the program terminate. 1 27 33 144 9 0 The while loop looks like a simplified version of the for loop. It contains a test expression but no initialization or increment expressions. Figure 3.3 shows the syntax of the while loop. Chapter 386 04 3087 CH03 11/29/01 2:17 PM Page 86FIGURE 3.3 Syntax of the while loop. As long as the test expression is true, the loop continues to be executed. In ENDON0, the text expression n != 0 (n not equal to 0) is true until the user enters 0. Figure 3.4 shows the operation of a while loop. The simplicity of the while loop is a bit illu- sory. Although there is no initialization expression, the loop variable (n in ENDON0) must be initialized before the loop begins. The loop body must also contain some statement that changes the value of the loop variable; otherwise the loop would never end. In ENDON0 it’s cin>>n;. Multiple Statements in a while Loop The next example, WHILE4, uses multiple statements in a while loop. It’s a variation of the CUBELIST program shown earlier with a for loop, but it calculates the fourth power, instead of the cube, of a series of integers. Let’s assume that in this program it’s important to put the results in a column four digits wide. To ensure that the results fit this column width, we must stop the loop before the results become larger than 9999. Without prior calculation we don’t know what number will generate a result of this size, so we let the program figure it out. The Loops and Decisions 3 L OOPS AND D ECISIONS 87 04 3087 CH03 11/29/01 2:17 PM Page 87test expression in the while statement terminates the program before the powers become too large. Chapter 388 FIGURE 3.4 Operation of the while loop. // while4.cpp // prints numbers raised to fourth power #include #include //for setw using namespace std; int main() { int pow=1; //power initially 1 int numb=1; //numb goes from 1 to ??? while( pow<10000 ) //loop while power <= 4 digits { cout << setw(2) << numb; //display number cout << setw(5) << pow << endl; //display fourth power ++numb; //get ready for next power pow = numb*numb*numb*numb; //calculate fourth power } cout << endl; return 0; } 04 3087 CH03 11/29/01 2:17 PM Page 88To find the fourth power of numb, we simply multiply it by itself four times. Each time through the loop we increment numb. But we don’t use numb in the test expression in while; instead, the resulting value of pow determines when to terminate the loop. Here’s the output: 1 1 2 16 3 81 4 256 5 625 6 1296 7 2401 8 4096 9 6561 The next number would be 10,000—too wide for our four-digit column; but by this time the loop has terminated. Precedence: Arithmetic and Relational Operators The next program touches on the question of operator precedence. It generates the famous sequence of numbers called the Fibonacci series. Here are the first few terms of the series: 1 1 2 3 5 8 13 21 34 55 Each term is found by adding the two previous ones: 1+1 is 2, 1+2 is 3, 2+3 is 5, 3+5 is 8, and so on. The Fibonacci series has applications in amazingly diverse fields, from sorting methods in computer science to the number of spirals in sunflowers. One of the most interesting aspects of the Fibonacci series is its relation to the golden ratio. The golden ratio is supposed to be the ideal proportion in architecture and art, and was used in the design of ancient Greek temples. As the Fibonacci series is carried out further and further, the ratio of the last two terms approaches closer and closer to the golden ratio. Here’s the list- ing for FIBO.CPP: // fibo.cpp // demonstrates WHILE loops using fibonacci series #include using namespace std; int main() { //largest unsigned long const unsigned long limit = 4294967295; unsigned long next=0; //next-to-last term unsigned long last=1; //last term Loops and Decisions 3 L OOPS AND D ECISIONS 89 04 3087 CH03 11/29/01 2:17 PM Page 89while( next < limit / 2 ) //don’t let results get too big { cout << last << “ “; //display last term long sum = next + last; //add last two terms next = last; //variables move forward last = sum; // in the series } cout << endl; return 0; } Here’s the output: 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1346269 2178309 3524578 5702887 9227465 14930352 24157817 39088169 63245986 102334155 165580141 267914296 433494437 701408733 1134903170 1836311903 2971215073 For you temple builders, the ratio of the last two terms gives an approximation of the golden ratio as 0.618033988—close enough for government work. The FIBO program uses type unsigned long,the type that holds the largest positive integers. The test expression in the while statement terminates the loop before the numbers exceed the limit of this type. We define this limit as a const type, since it doesn’t change. We must stop when next becomes larger than half the limit; otherwise, sum would exceed the limit. The test expression uses two operators: (next < limit / 2) Our intention is to compare next with the result of limit/2. That is, we want the division to be performed before the comparison. We could put parentheses around the division, to ensure that it’s performed first. (next < (limit/2) ) But we don’t need the parentheses. Why not? Because arithmetic operators have a higher precedence than relational operators. This guarantees that limit/2 will be evaluated before the comparison is made, even without the parentheses. We’ll summarize the precedence situation later in this chapter, when we look at logical operators. Chapter 390 04 3087 CH03 11/29/01 2:17 PM Page 90The do Loop In a while loop, the test expression is evaluated at the beginning of the loop. If the test expres- sion is false when the loop is entered, the loop body won’t be executed at all. In some situa- tions this is what you want. But sometimes you want to guarantee that the loop body is executed at least once, no matter what the initial state of the test expression. When this is the case you should use the do loop, which places the test expression at the end of the loop. Our example, DIVDO,invites the user to enter two numbers: a dividend (the top number in a division) and a divisor (the bottom number). It then calculates the quotient (the answer) and the remainder, using the / and % operators, and prints out the result. // divdo.cpp // demonstrates DO loop #include using namespace std; int main() { long dividend, divisor; char ch; do //start of do loop { //do some processing cout << “Enter dividend: “; cin >> dividend; cout << “Enter divisor: “; cin >> divisor; cout << “Quotient is “ << dividend / divisor; cout << “, remainder is “ << dividend % divisor; cout << “\nDo another? (y/n): “; //do it again? cin >> ch; } while( ch != ‘n’ ); //loop condition return 0; } Most of this program resides within the do loop. First, the keyword do marks the beginning of the loop. Then, as with the other loops, braces delimit the body of the loop. Finally, a while statement provides the test expression and terminates the loop. This while statement looks much like the one in a while loop, except for its position at the end of the loop and the fact that it ends with a semicolon (which is easy to forget!). The syntax of the do loop is shown in Figure 3.5. Loops and Decisions 3 L OOPS AND D ECISIONS 91 04 3087 CH03 11/29/01 2:17 PM Page 91FIGURE 3.5 Syntax of the do loop. Following each computation, DIVDO asks if the user wants to do another. If so, the user enters a ‘y’ character, and the test expression ch != ‘n’ remains true. If the user enters ‘n’, the test expression becomes false and the loop terminates. Figure 3.6 charts the operation of the do loop. Here’s an example of DIVDO’s output: Enter dividend: 11 Enter divisor: 3 Quotient is 3, remainder is 2 Do another? (y/n): y Enter dividend: 222 Enter divisor: 17 Quotient is 13, remainder is 1 Do another? (y/n): n Chapter 392 04 3087 CH03 11/29/01 2:17 PM Page 92FIGURE 3.6 Operation of the do loop. When to Use Which Loop We’ve made some general statements about how loops are used. The for loop is appropriate when you know in advance how many times the loop will be executed. The while and do loops are used when you don’t know in advance when the loop will terminate (the while loop when you may not want to execute the loop body even once, and the do loop when you’re sure you want to execute the loop body at least once). These criteria are somewhat arbitrary. Which loop type to use is more a matter of style than of hard-and-fast rules. You can actually make any of the loop types work in almost any situation. You should choose the type that makes your program the clearest and easiest to follow. Decisions The decisions in a loop always relate to the same question: Should we do this (the loop body) again? As humans we would find it boring to be so limited in our decision-making processes. We need to decide not only whether to go to work again today (continuing the loop), but also whether to buy a red shirt or a green one (or no shirt at all), whether to take a vacation, and if so, in the mountains or by the sea. Programs also need to make these one-time decisions. In a program a decision causes a one- time jump to a different part of the program, depending on the value of an expression. Loops and Decisions 3 L OOPS AND D ECISIONS 93 04 3087 CH03 11/29/01 2:17 PM Page 93Decisions can be made in C++ in several ways. The most important is with the if...else statement, which chooses between two alternatives. This statement can be used without the else,as a simple if statement. Another decision statement, switch,creates branches for multi- ple alternative sections of code, depending on the value of a single variable. Finally, the condi- tional operator is used in specialized situations. We’ll examine each of these constructions. The if Statement The if statement is the simplest of the decision statements. Our next program, IFDEMO,pro- vides an example. // ifdemo.cpp // demonstrates IF statement #include using namespace std; int main() { int x; cout << “Enter a number: “; cin >> x; if( x > 100 ) cout << “That number is greater than 100\n”; return 0; } The if keyword is followed by a test expression in parentheses. The syntax of the if statement is shown in Figure 3.7. As you can see, the syntax of if is very much like that of while. The difference is that the statements following the if are executed only once if the test expression is true; the statements following while are executed repeatedly until the test expression becomes false. Figure 3.8 shows the operation of the if statement. Here’s an example of the IFDEMO program’s output when the number entered by the user is greater than 100: Enter a number: 2000 That number is greater than 100 If the number entered is not greater than 100, the program will terminate without printing the second line. Chapter 394 04 3087 CH03 11/29/01 2:17 PM Page 94FIGURE 3.8 Operation of the if statement. Loops and Decisions 3 L OOPS AND D ECISIONS 95 FIGURE 3.7 Syntax of the if statement. 04 3087 CH03 11/29/01 2:17 PM Page 95Multiple Statements in the if Body As in loops, the code in an if body can consist of a single statement—as shown in the IFDEMO example—or a block of statements delimited by braces. This variation on IFDEMO,called IF2, shows how that looks. // if2.cpp // demonstrates IF with multiline body #include using namespace std; int main() { int x; cout << “Enter a number: “; cin >> x; if( x > 100 ) { cout << “The number “ << x; cout << “ is greater than 100\n”; } return 0; } Here’s some output from IF2: Enter a number: 12345 The number 12345 is greater than 100 Nesting ifs Inside Loops The loop and decision structures we’ve seen so far can be nested inside one another. You can nest ifs inside loops, loops inside ifs, ifs inside ifs, and so on. Here’s an example, PRIME, that nests an if within a for loop. This example tells you whether a number you enter is a prime number. (Prime numbers are integers divisible only by themselves and 1. The first few primes are 2, 3, 5, 7, 11, 13, 17.) // prime.cpp // demonstrates IF statement with prime numbers #include using namespace std; #include //for exit() int main() { unsigned long n, j; Chapter 396 04 3087 CH03 11/29/01 2:17 PM Page 96cout << “Enter a number: “; cin >> n; //get number to test for(j=2; j <= n/2; j++) //divide by every integer from if(n%j == 0) //2 on up; if remainder is 0, { //it’s divisible by j cout << “It’s not prime; divisible by “ << j << endl; exit(0); //exit from the program } cout << “It’s prime\n”; return 0; } In this example the user enters a number that is assigned to n. The program then uses a for loop to divide n by all the numbers from 2 up to n/2. The divisor is j,the loop variable. If any value of j divides evenly into n,then n is not prime. When a number divides evenly into another, the remainder is 0; we use the remainder operator % in the if statement to test for this condition with each value of j. If the number is not prime, we tell the user and we exit from the program. Here’s output from three separate invocations of the program: Enter a number: 13 It’s prime Enter a number: 22229 It’s prime Enter a number: 22231 It’s not prime; divisible by 11 Notice that there are no braces around the loop body. This is because the if statement, and the statements in its body, are considered to be a single statement. If you like you can insert braces for readability, even though the compiler doesn’t need them. Library Function exit() When PRIME discovers that a number is not prime, it exits immediately, since there’s no use proving more than once that a number isn’t prime. This is accomplished with the library func- tion exit(). This function causes the program to terminate, no matter where it is in the listing. It has no return value. Its single argument, 0 in our example, is returned to the operating sys- tem when the program exits. (This value is useful in batch files, where you can use the ERRORLEVEL value to query the return value provided by exit(). The value 0 is normally used for a successful termination; other numbers indicate errors.) Loops and Decisions 3 L OOPS AND D ECISIONS 97 04 3087 CH03 11/29/01 2:17 PM Page 97The if...else Statement The if statement lets you do something if a condition is true. If it isn’t true, nothing happens. But suppose we want to do one thing if a condition is true, and do something else if it’s false. That’s where the if...else statement comes in. It consists of an if statement, followed by a statement or block of statements, followed by the keyword else,followed by another state- ment or block of statements. The syntax is shown in Figure 3.9. Chapter 398 FIGURE 3.9 Syntax of the if...else statement. Here’s a variation of our IF example, with an else added to the if: // ifelse.cpp // demonstrates IF...ELSE statememt #include using namespace std; 04 3087 CH03 11/29/01 2:17 PM Page 98int main() { int x; cout << “\nEnter a number: “; cin >> x; if( x > 100 ) cout << “That number is greater than 100\n”; else cout << “That number is not greater than 100\n”; return 0; } If the test expression in the if statement is true, the program prints one message; if it isn’t, it prints the other. Here’s output from two different invocations of the program: Enter a number: 300 That number is greater than 100 Enter a number: 3 That number is not greater than 100 The operation of the if...else statement is shown in Figure 3.10. Loops and Decisions 3 L OOPS AND D ECISIONS 99 FIGURE 3.10 Operation of the if...else statement. 04 3087 CH03 11/29/01 2:17 PM Page 99The getche() Library Function Our next example shows an if...else statement embedded in a while loop. It also introduces a new library function: getche(). This program, CHCOUNT, counts the number of words and the number of characters in a phrase typed in by the user. // chcount.cpp // counts characters and words typed in #include using namespace std; #include //for getche() int main() { int chcount=0; //counts non-space characters int wdcount=1; //counts spaces between words char ch = ‘a’; //ensure it isn’t ‘\r’ cout << “Enter a phrase: “; while( ch != ‘\r’ ) //loop until Enter typed { ch = getche(); //read one character if( ch==’ ‘ ) //if it’s a space wdcount++; //count a word else //otherwise, chcount++; //count a character } //display results cout << “\nWords=” << wdcount << endl << “Letters=” << (chcount-1) << endl; return 0; } So far we’ve used only cin and >> for input. That approach requires that the user always press the Enter key to inform the program that the input is complete. This is true even for single characters: The user must type the character, then press Enter. However, as in the present example, a program often needs to process each character typed by the user without waiting for an Enter. The getche() library function performs this service. It returns each character as soon as it’s typed. It takes no arguments, and requires the CONIO.H header file. In CHCOUNT the value of the character returned from getche() is assigned to ch. (The getche() function echoes the character to the screen. That’s why there’s an e at the end of getche. Another function, getch(), is similar to getche() but doesn’t echo the character to the screen.) The if...else statement causes the word count wdcount to be incremented if the character is a space, and the character count chcount to be incremented if the character is anything but a space. Thus anything that isn’t a space is assumed to count as a character. (Note that this pro- gram is fairly naïve; it will be fooled by multiple spaces between words.) Chapter 3100 04 3087 CH03 11/29/01 2:17 PM Page 100Here’s some sample interaction with CHCOUNT: For while and do Words=4 Letters=13 The test expression in the while statement checks to see if ch is the ‘\r’ character, which is the character received from the keyboard when the Enter key is pressed. If so, the loop and the program terminate. Assignment Expressions The CHCOUNT program can be rewritten to save a line of code and demonstrate some important points about assignment expressions and precedence. The result is a construction that looks rather peculiar but is commonly used in C++ (and in C). Here’s the rewritten version, called CHCNT2: // chcnt2.cpp // counts characters and words typed in #include using namespace std; #include // for getche() int main() { int chcount=0; int wdcount=1; // space between two words char ch; while( (ch=getche()) != ‘\r’ ) // loop until Enter typed { if( ch==’ ‘ ) // if it’s a space wdcount++; // count a word else // otherwise, chcount++; // count a character } // display results cout << “\nWords=” << wdcount << endl << “Letters=” << chcount << endl; return 0; } The value returned by getche() is assigned to ch as before, but this entire assignment expres- sion has been moved inside the test expression for while. The assignment expression is com- pared with ‘\r’ to see whether the loop should terminate. This works because the entire assignment expression takes on the value used in the assignment. That is, if getche() returns ‘a’,then not only does ch take on the value ‘a’,but the expression Loops and Decisions 3 L OOPS AND D ECISIONS 101 04 3087 CH03 11/29/01 2:17 PM Page 101(ch=getche()) also takes on the value ‘a’. This is then compared with ‘\r’. The fact that assignment expressions have a value is also used in statements such as x = y = z = 0; This is perfectly legal in C++. First, z takes on the value 0, then z=0takes on the value 0, which is assigned to y. Then the expression y=z=0likewise takes on the value 0, which is assigned to x. The parentheses around the assignment expression in (ch=getche()) are necessary because the assignment operator = has a lower precedence than the relational operator !=. Without the parentheses the expression would be evaluated as while( ch = (getche() != ‘\r’) ) // not what we want which would assign a true or false value to ch (not what we want). The while statement in CHCNT2provides a lot of power in a small space. It is not only a test expression (checking ch to see whether it’s ‘\r’); it also gets a character from the keyboard and assigns it to ch. It’s also not easy to unravel the first time you see it. Nested if...else Statements You’re probably too young to remember adventure games on early character-mode MS-DOS systems, but let’s resurrect the concept here. You moved your “character” around an imaginary landscape and discovered castles, sorcerers, treasure, and so on, using text—not pictures—for input and output. The next program, ADIFELSE, models a small part of such an adventure game. // adifelse.cpp // demonstrates IF...ELSE with adventure program #include using namespace std; #include //for getche() int main() { char dir=’a’; int x=10, y=10; cout << “Type Enter to quit\n”; while( dir != ‘\r’ ) //until Enter is typed { cout << “\nYour location is “ << x << “, “ << y; cout << “\nPress direction key (n, s, e, w): “; Chapter 3102 04 3087 CH03 11/29/01 2:17 PM Page 102dir = getche(); //get character if( dir==’n’) //go north y--; else if( dir==’s’ ) //go south y++; else if( dir==’e’ ) //go east x++; else if( dir==’w’ ) //go west x--; } //end while return 0; } //end main When the game starts, you find yourself on a barren moor. You can go one “unit” north, south, east, or west, while the program keeps track of where you are and reports your position, which starts at coordinates 10,10. Unfortunately, nothing exciting happens to your character, no matter where you go; the moor stretches almost limitlessly in all directions, as shown in Figure 3.11. We’ll try to provide a little more excitement to this game later on. Here’s some sample interaction with ADIFELSE: Your location is 10, 10 Press direction key (n, s, e, w): n Your location is 10, 9 Press direction key (n, s, e, w): e Your location is 11, 9 Press direction key (n, s, e, w): You can press the Enter key to exit the program. This program may not cause a sensation in the video arcades, but it does demonstrate one way to handle multiple branches. It uses an if statement nested inside an if...else statement, which is nested inside another if...else statement, which is nested inside yet another if...else statement. If the first test condition is false, the second one is examined, and so on until all four have been checked. If any one proves true, the appropriate action is taken— changing the x or y coordinate—and the program exits from all the nested decisions. Such a nested group of if...else statements is called a decision tree. Loops and Decisions 3 L OOPS AND D ECISIONS 103 04 3087 CH03 11/29/01 2:17 PM Page 103FIGURE 3.11 The barren moor. Matching the else There’s a potential problem in nested if...else statements: You can inadvertently match an else with the wrong if. BADELSE provides an example: // badelse.cpp // demonstrates ELSE matched with wrong IF #include using namespace std; int main() { int a, b, c; cout << “Enter three numbers, a, b, and c:\n”; cin >> a >> b >> c; Chapter 3104 04 3087 CH03 11/29/01 2:17 PM Page 104if( a==b ) if( b==c ) cout << “a, b, and c are the same\n”; else cout << “a and b are different\n”; return 0; } We’ve used multiple values with a single cin. Press Enter following each value you type in; the three values will be assigned to a, b, and c. What happens if you enter 2, then 3, and then 3? Variable a is 2, and b is 3. They’re different, so the first test expression is false, and you would expect the else to be invoked, printing a and b are different. But in fact nothing is printed. Why not? Because the else is matched with the wrong if. The indentation would lead you to believe that the else is matched with the first if,but in fact it goes with the second if. Here’s the rule: An else is matched with the last if that doesn’t have its own else. Here’s a corrected version: if(a==b) if(b==c) cout << “a, b, and c are the same\n”; else cout << “b and c are different\n”; We changed the indentation and also the phrase printed by the else body. Now if you enter 2, 3, 3, nothing will be printed. But entering 2, 2, 3 will cause the output b and c are different If you really want to pair an else with an earlier if, you can use braces around the inner if: if(a==b) { if(b==c) cout << “a, b, and c are the same”; } else cout << “a and b are different”; Here the else is paired with the first if,as the indentation indicates. The braces make the if within them invisible to the following else. Loops and Decisions 3 L OOPS AND D ECISIONS 105 04 3087 CH03 11/29/01 2:17 PM Page 105The else...if Construction The nested if...else statements in the ADIFELSE program look clumsy and can be hard—for humans—to interpret, especially if they are nested more deeply than shown. However, there’s another approach to writing the same statements. We need only reformat the program, obtain- ing the next example, ADELSEIF. // adelseif.cpp // demonstrates ELSE...IF with adventure program #include using namespace std; #include //for getche() int main() { char dir=’a’; int x=10, y=10; cout << “Type Enter to quit\n”; while( dir != ‘\r’ ) //until Enter is typed { cout << “\nYour location is “ << x << “, “ << y; cout << “\nPress direction key (n, s, e, w): “; dir = getche(); //get character if( dir==’n’) //go north y--; else if( dir==’s’ ) //go south y++; else if( dir==’e’ ) //go east x++; else if( dir==’w’ ) //go west x--; } //end while return 0; } //end main The compiler sees this as identical to ADIFELSE,but we’ve rearranged the ifs so they directly follow the elses. The result looks almost like a new keyword: else if. The program goes down the ladder of else ifs until one of the test expressions is true. It then executes the fol- lowing statement and exits from the ladder. This format is clearer and easier to follow than the if...else approach. Chapter 3106 04 3087 CH03 11/29/01 2:17 PM Page 106The switch Statement If you have a large decision tree, and all the decisions depend on the value of the same vari- able, you will probably want to consider a switch statement instead of a ladder of if...else or else if constructions. Here’s a simple example called PLATTERS that will appeal to nostal- gia buffs: // platters.cpp // demonstrates SWITCH statement #include using namespace std; int main() { int speed; //turntable speed cout << “\nEnter 33, 45, or 78: “; cin >> speed; //user enters speed switch(speed) //selection based on speed { case 33: //user entered 33 cout << “LP album\n”; break; case 45: //user entered 45 cout << “Single selection\n”; break; case 78: //user entered 78 cout << “Obsolete format\n”; break; } return 0; } This program prints one of three possible messages, depending on whether the user inputs the number 33, 45,or 78. As old-timers may recall, long-playing records (LPs) contained many songs and turned at 33 rpm, the smaller 45’s held only a single song, and 78s were the format that preceded LPs and 45s. The keyword switch is followed by a switch variable in parentheses. switch(speed) Braces then delimit a number of case statements. Each case keyword is followed by a constant, which is not in parentheses but is followed by a colon. case 33: The data type of the case constants should match that of the switch variable. Figure 3.12 shows the syntax of the switch statement. Loops and Decisions 3 L OOPS AND D ECISIONS 107 04 3087 CH03 11/29/01 2:17 PM Page 107FIGURE 3.12 Syntax of the switch statement. Before entering the switch, the program should assign a value to the switch variable. This value will usually match a constant in one of the case statements. When this is the case (pun intended!), the statements immediately following the keyword case will be executed, until a break is reached. Here’s an example of PLATTER’s output: Enter 33, 45, or 78: 45 Single selection Chapter 3108 04 3087 CH03 11/29/01 2:18 PM Page 108The break Statement PLATTERS has a break statement at the end of each case section. The break keyword causes the entire switch statement to exit. Control goes to the first statement following the end of the switch construction, which in PLATTERS is the end of the program. Don’t forget the break; without it, control passes down (or “falls through”) to the statements for the next case, which is usually not what you want (although sometimes it’s useful). If the value of the switch variable doesn’t match any of the case constants, control passes to the end of the switch without doing anything. The operation of the switch statement is shown in Figure 3.13. The break keyword is also used to escape from loops; we’ll discuss this soon. Loops and Decisions 3 L OOPS AND D ECISIONS 109 FIGURE 3.13 Operation of the switch statement. 04 3087 CH03 11/29/01 2:18 PM Page 109switch Statement with Character Variables The PLATTERS example shows a switch statement based on a variable of type int. You can also use type char. Here’s our ADELSEIF program rewritten as ADSWITCH: // adswitch.cpp // demonstrates SWITCH with adventure program #include using namespace std; #include //for getche() int main() { char dir=’a’; int x=10, y=10; while( dir != ‘\r’ ) { cout << “\nYour location is “ << x << “, “ << y; cout << “\nEnter direction (n, s, e, w): “; dir = getche(); //get character switch(dir) //switch on it { case ‘n’: y--; break; //go north case ‘s’: y++; break; //go south case ‘e’: x++; break; //go east case ‘w’: x--; break; //go west case ‘\r’: cout << “Exiting\n”; break; //Enter key default: cout << “Try again\n”; //unknown char } //end switch } //end while return 0; } //end main A character variable dir is used as the switch variable, and character constants ‘n’, ‘s’, and so on are used as the case constants. (Note that you can use integers and characters as switch variables, as shown in the last two examples, but you can’t use floating-point numbers.) Since they are so short, the statements following each case keyword have been written on one line, which makes for a more compact listing. We’ve also added a case to print an exit mes- sage when Enter is pressed. The default Keyword In the ADSWITCH program, where you expect to see the last case at the bottom of the switch construction, you instead see the keyword default. This keyword gives the switch construc- tion a way to take an action if the value of the loop variable doesn’t match any of the case constants. Here we use it to print Try again if the user types an unknown character. No break is necessary after default, since we’re at the end of the switch anyway. Chapter 3110 04 3087 CH03 11/29/01 2:18 PM Page 110A switch statement is a common approach to analyzing input entered by the user. Each of the possible characters is represented by a case. It’s a good idea to use a default statement in all switch statements, even if you don’t think you need it. A construction such as default: cout << “Error: incorrect input to switch”; break; alerts the programmer (or the user) that something has gone wrong in the operation of the pro- gram. In the interest of brevity we don’t always include such a default statement, but you should, especially in serious programs. switch Versus if...else When do you use a series of if...else (or else if) statements, and when do you use a switch statement? In an else if construction you can use a series of expressions that involve unrelated variables and are as complex as you like. For example: if( SteamPressure*Factor > 56 ) // statements else if( VoltageIn + VoltageOut < 23000) // statements else if( day==Thursday ) // statements else // statements In a switch statement, however, all the branches are selected by the same variable; the only thing distinguishing one branch from another is the value of this variable. You can’t say case a<3: // do something break; The case constant must be an integer or character constant, like 3 or ‘a’,or an expression that evaluates to a constant, like ‘a’+32. When these conditions are met, the switch statement is very clean—easy to write and to understand. It should be used whenever possible, especially when the decision tree has more than a few possibilities. The Conditional Operator Here’s a strange sort of decision operator. It exists because of a common programming situa- tion: A variable is given one value if something is true and another value if it’s false. For example, here’s an if...else statement that gives the variable min the value of alpha or the value of beta, depending on which is smaller: Loops and Decisions 3 L OOPS AND D ECISIONS 111 04 3087 CH03 11/29/01 2:18 PM Page 111if( alpha < beta ) min = alpha; else min = beta; This sort of construction is so common that the designers of C++ (actually the designers of C, long ago) invented a compressed way to express it: the conditional operator. This operator consists of two symbols, which operate on three operands. It’s the only such operator in C++; other operators operate on one or two operands. Here’s the equivalent of the same program fragment, using a conditional operator: min = (alpha using namespace std; int main() { Loops and Decisions 3 L OOPS AND D ECISIONS 113 04 3087 CH03 11/29/01 2:18 PM Page 113for(int j=0; j<80; j++) //for every column, { //ch is ‘x’ if column is char ch = (j%8) ? ‘ ‘ : ‘x’; //multiple of 8, and cout << ch; //’ ‘ (space) otherwise } return 0; } Some of the right side of the output is lost because of the page width, but you can probably imagine it: x x x x x x x x x As j cycles through the numbers from 0 to 79, the remainder operator causes the expression (j %8)to become false—that is, 0—only when j is a multiple of 8. So the conditional expression (j%8) ? ‘ ‘ : ‘x’ has the value ‘ ‘ (the space character) when j is not a multiple of 8, and the value ‘x’ when it is. You may think this is terse, but we could have combined the two statements in the loop body into one, eliminating the ch variable: cout << ( (j%8) ? ‘ ‘ : ‘x’ ); Hotshot C++ (and C) programmers love this sort of thing—getting a lot of bang from very lit- tle code. But you don’t need to strive for concise code if you don’t want to. Sometimes it becomes so obscure it’s not worth the effort. Even using the conditional operator is optional: An if...else statement and a few extra program lines will accomplish the same thing. Logical Operators So far we’ve seen two families of operators (besides the oddball conditional operator). First are the arithmetic operators +, -, *, /, and %. Second are the relational operators <, >, <=, >=, ==, and !=. Let’s examine a third family of operators, called logical operators. These operators allow you to logically combine Boolean variables (that is, variables of type bool,with true or false val- ues). For example, today is a weekday has a Boolean value, since it’s either true or false. Another Boolean expression is Maria took the car. We can connect these expressions logically: If today is a weekday, and Maria took the car, then I’ll have to take the bus. The logical con- nection here is the word and, which provides a true or false value to the combination of the two phrases. Only if they are both true will I have to take the bus. Chapter 3114 04 3087 CH03 11/29/01 2:18 PM Page 114Logical AND Operator Let’s see how logical operators combine Boolean expressions in C++. Here’s an example, ADVENAND,that uses a logical operator to spruce up the adventure game from the ADSWITCH example. We’ll bury some treasure at coordinates (7,11) and see whether the player can find it. // advenand.cpp // demonstrates AND logical operator #include using namespace std; #include //for exit() #include //for getche() int main() { char dir=’a’; int x=10, y=10; while( dir != ‘\r’ ) { cout << “\nYour location is “ << x << “, “ << y; cout << “\nEnter direction (n, s, e, w): “; dir = getche(); //get direction switch(dir) { case ‘n’: y--; break; //update coordinates case ‘s’: y++; break; case ‘e’: x++; break; case ‘w’: x--; break; } if( x==7 && y==11 ) //if x is 7 and y is 11 { cout << “\nYou found the treasure!\n”; exit(0); //exit from program } } //end switch return 0; } //end main The key to this program is the if statement if( x==7 && y==11 ) The test expression will be true only if x is 7 and y is 11. The logical AND operator && joins the two relational expressions to achieve this result. (A relational expression is one that uses a relational operator.) Loops and Decisions 3 L OOPS AND D ECISIONS 115 04 3087 CH03 11/29/01 2:18 PM Page 115Notice that parentheses are not necessary around the relational expressions. ( (x==7) && (y==11) ) // inner parentheses not necessary This is because the relational operators have higher precedence than the logical operators. Here’s some interaction as the user arrives at these coordinates: Your location is 7, 10 Enter direction (n, s, e, w): s You found the treasure! There are three logical operators in C++: Operator Effect && Logical AND || Logical OR ! Logical NOT There is no logical XOR (exclusive OR) operator in C++. Let’s look at examples of the || and ! operators. Logical OR Operator Suppose in the adventure game you decide there will be dragons if the user goes too far east or too far west. Here’s an example, ADVENOR,that uses the logical OR operator to implement this frightening impediment to free adventuring. It’s a variation on the ADVENAND program. // advenor.cpp // demonstrates OR logical operator #include using namespace std; #include //for exit() #include //for getche() int main() { char dir=’a’; int x=10, y=10; while( dir != ‘\r’ ) //quit on Enter key { cout << “\n\nYour location is “ << x << “, “ << y; if( x<5 || x>15 ) //if x west of 5 OR east of 15 cout << “\nBeware: dragons lurk here”; Chapter 3116 04 3087 CH03 11/29/01 2:18 PM Page 116cout << “\nEnter direction (n, s, e, w): “; dir = getche(); //get direction switch(dir) { case ‘n’: y--; break; //update coordinates case ‘s’: y++; break; case ‘e’: x++; break; case ‘w’: x--; break; } //end switch } //end while return 0; } //end main() The expression x<5 || x>15 is true whenever either x is less than 5 (the player is too far west), or x is greater than 15 (the player is too far east). Again, the || operator has lower precedence than the relational opera- tors < and >,so no parentheses are needed in this expression. Logical NOT Operator The logical NOT operator ! is a unary operator—that is, it takes only one operand. (Almost all the operators we’ve seen thus far are binary operators; they take two operands. The conditional operator is the only ternary operator in C++.) The effect of the ! is that the logical value of its operand is reversed: If something is true, ! makes it false; if it is false, ! makes it true. (It would be nice if life were so easily manipulated.) For example, (x==7) is true if x is equal to 7, but !(x==7) is true if x is not equal to 7. (In this situation you could use the relational not equals operator, x != 7,to achieve the same effect.) A True/False Value for Every Integer Variable We may have given you the impression that for an expression to have a true/false value, it must involve a relational operator. But in fact, every integer expression has a true/false value, even if it is only a single variable. The expression x is true whenever x is not 0, and false when x is 0. Applying the ! operator to this situation, we can see that the !x is true whenever x is 0, since it reverses the truth value of x. Let’s put these ideas to work. Imagine in your adventure game that you want to place a mush- room on all the locations where both x and y are a multiple of 7. (As you probably know, mushrooms, when consumed by the player, confer magical powers.) The remainder when x is divided by 7, which can be calculated by x%7, is 0 only when x is a multiple of 7. So to specify the mushroom locations, we can write if( x%7==0 && y%7==0 ) cout << “There’s a mushroom here.\n”; Loops and Decisions 3 L OOPS AND D ECISIONS 117 04 3087 CH03 11/29/01 2:18 PM Page 117However, remembering that expressions are true or false even if they don’t involve relational operators, you can use the ! operator to provide a more concise format. if( !(x%7) && !(y%7) ) // if not x%7 and not y%7 This has exactly the same effect. We’ve said that the logical operators && and || have lower precedence than the relational oper- ators. Why then do we need parentheses around x%7 and y%7? Because, even though it is a log- ical operator, ! is a unary operator, which has higher precedence than relational operators. Precedence Summary Let’s summarize the precedence situation for the operators we’ve seen so far. The operators higher on the list have higher precedence than those lower down. Operators with higher prece- dence are evaluated before those with lower precedence. Operators on the same row have equal precedence. You can force an expression to be evaluated first by placing parentheses around it. You can find a more complete precedence table in Appendix B, “C++ Precedence Table and Keywords.” Operator type Operators Precedence Unary !, ++, ––, +, – Highest Arithmetic Multiplicative *, /, % Additive +, – Relational Inequality <, >, <=, >= Equality ==, != Logical And && Or || Conditional ?: Assignment =, +=, –=, *=, /=, %= Lowest We should note that if there is any possibility of confusion in a relational expression that involves multiple operators, you should use parentheses whether they are needed or not. They don’t do any harm, and they guarantee the expression does what you want, even if you’ve made a mistake with precedence. Also, they make it clear to anyone reading the listing what you intended. Other Control Statements There are several other control statements in C++. We’ve already seen one, break,used in switch statements, but it can be used other places as well. Another statement, continue,is used only in loops, and a third, goto,should be avoided. Let’s look at these statements in turn. Chapter 3118 04 3087 CH03 11/29/01 2:18 PM Page 118The break Statement The break statement causes an exit from a loop, just as it does from a switch statement. The next statement after the break is executed is the statement following the loop. Figure 3.16 shows the operation of the break statement. Loops and Decisions 3 L OOPS AND D ECISIONS 119 FIGURE 3.16 Operation of the break statement. To demonstrate break, here’s a program, SHOWPRIM,that displays the distribution of prime numbers in graphical form: // showprim.cpp // displays prime number distribution #include using namespace std; #include //for getche() int main() { const unsigned char WHITE = 219; //solid color (primes) const unsigned char GRAY = 176; //gray (non primes) unsigned char ch; //for each screen position for(int count=0; count<80*25-1; count++) { ch = WHITE; //assume it’s prime 04 3087 CH03 11/29/01 2:18 PM Page 119for(int j=2; j using namespace std; int main() { long dividend, divisor; char ch; Loops and Decisions 3 L OOPS AND D ECISIONS 121 04 3087 CH03 11/29/01 2:18 PM Page 121do { cout << “Enter dividend: “; cin >> dividend; cout << “Enter divisor: “; cin >> divisor; if( divisor == 0 ) //if attempt to { //divide by 0, cout << “Illegal divisor\n”; //display message continue; //go to top of loop } cout << “Quotient is “ << dividend / divisor; cout << “, remainder is “ << dividend % divisor; cout << “\nDo another? (y/n): “; cin >> ch; } while( ch != ‘n’ ); return 0; } Chapter 3122 FIGURE 3.18 Operation of the continue statement. If the user inputs 0 for the divisor, the program prints an error message and, using continue, returns to the top of the loop to issue the prompts again. Here’s some sample output: Enter dividend: 10 Enter divisor: 0 Illegal divisor Enter dividend: A break statement in this situation would cause an exit from the do loop and the program, an unnecessarily harsh response. 04 3087 CH03 11/29/01 2:18 PM Page 122Notice that we’ve made the format of the do loop a little more compact. The do is on the same line as the opening brace, and the while is on the same line as the closing brace. The goto Statement We’ll mention the goto statement here for the sake of completeness—not because it’s a good idea to use it. If you’ve had any exposure to structured programming principles, you know that gotos can quickly lead to “spaghetti” code that is difficult to understand and debug. There is almost never any need to use goto,as is demonstrated by its absence from the program exam- ples in this book. With that lecture out of the way, here’s the syntax. You insert a label in your code at the desired destination for the goto. The label is always terminated by a colon. The keyword goto, followed by this label name, then takes you to the label. The following code fragment demon- strates this approach. goto SystemCrash; // other statements SystemCrash: // control will begin here following goto Summary Relational operators compare two values to see whether they’re equal, whether one is larger than the other, and so on. The result is a logical or Boolean (type bool) value, which is true or false. False is indicated by 0, and true by 1 or any other non-zero number. There are three kinds of loops in C++. The for loop is most often used when you know in advance how many times you want to execute the loop. The while loop and do loops are used when the condition causing the loop to terminate arises within the loop, with the while loop not necessarily executing at all, and the do loop always executing at least once. A loop body can be a single statement or a block of multiple statements delimited by braces. A variable defined within a block is visible only within that block. There are four kinds of decision-making statements. The if statement does something if a test expression is true. The if...else statement does one thing if the test expression is true, and another thing if it isn’t. The else if construction is a way of rewriting a ladder of nested if...else statements to make it more readable. The switch statement branches to multiple sections of code, depending on the value of a single variable. The conditional operator simpli- fies returning one value if a test expression is true, and another if it’s false. The logical AND and OR operators combine two Boolean expressions to yield another one, and the logical NOT operator changes a Boolean value from true to false, or from false to true. Loops and Decisions 3 L OOPS AND D ECISIONS 123 04 3087 CH03 11/29/01 2:18 PM Page 123The break statement sends control to the end of the innermost loop or switch in which it occurs. The continue statement sends control to the top of the loop in which it occurs. The goto statement sends control to a label. Precedence specifies which kinds of operations will be carried out first. The order is unary, arithmetic, relational, logical, conditional, assignment. Questions Answers to these questions can be found in Appendix G. 1. A relational operator a. assigns one operand to another. b. yields a Boolean result. c. compares two operands. d. logically combines two operands. 2. Write an expression that uses a relational operator to return true if the variable george is not equal to sally. 3. Is –1 true or false? 4. Name and describe the usual purpose of three expressions in a for statement. 5. In a for loop with a multistatement loop body, semicolons should appear following a. the for statement itself. b. the closing brace in a multistatement loop body. c. each statement within the loop body. d. the test expression. 6. True or false: The increment expression in a for loop can decrement the loop variable. 7. Write a for loop that displays the numbers from 100 to 110. 8. A block of code is delimited by ________________. 9. A variable defined within a block is visible a. from the point of definition onward in the program. b. from the point of definition onward in the function. c. from the point of definition onward in the block. d. throughout the function. 10. Write a while loop that displays the numbers from 100 to 110. 11. True or false: Relational operators have a higher precedence than arithmetic operators. Chapter 3124 04 3087 CH03 11/29/01 2:18 PM Page 12412. How many times is the loop body executed in a do loop? 13. Write a do loop that displays the numbers from 100 to 110. 14. Write an if statement that prints Yes if a variable age is greater than 21. 15. The library function exit() causes an exit from a. the loop in which it occurs. b. the block in which it occurs. c. the function in which it occurs. d. the program in which it occurs. 16. Write an if...else statement that displays Yes if a variable age is greater than 21, and displays No otherwise. 17. The getche() library function a. returns a character when any key is pressed. b. returns a character when Enter is pressed. c. displays a character on the screen when any key is pressed. d. does not display a character on the screen. 18. What is the character obtained from cin when the user presses the Enter key? 19. An else always matches the _________ if, unless the if is _________. 20. The else...if construction is obtained from a nested if...else by ________________. 21. Write a switch statement that prints Yes if a variable ch is ‘y’,prints No if ch is ‘n’, and prints Unknown response otherwise. 22. Write a statement that uses a conditional operator to set ticket to 1 if speed is greater than 55, and to 0 otherwise. 23. The && and || operators a. compare two numeric values. b. combine two numeric values. c. compare two Boolean values. d. combine two Boolean values. 24. Write an expression involving a logical operator that is true if limit is 55 and speed is greater than 55. 25. Arrange in order of precedence (highest first) the following kinds of operators: logical, unary, arithmetic, assignment, relational, conditional. Loops and Decisions 3 L OOPS AND D ECISIONS 125 04 3087 CH03 11/29/01 2:18 PM Page 12526. The break statement causes an exit a. only from the innermost loop. b. only from the innermost switch. c. from all loops and switches. d. from the innermost loop or switch. 27. Executing the continue operator from within a loop causes control to go to ________. 28. The goto statement causes control to go to a. an operator. b. a label. c. a variable. d. a function. Exercises Answers to the starred exercises can be found in Appendix G. *1. Assume that you want to generate a table of multiples of any given number. Write a pro- gram that allows the user to enter the number and then generates the table, formatting it into 10 columns and 20 lines. Interaction with the program should look like this (only the first three lines are shown): Enter a number: 7 7 14 21 28 35 42 49 56 63 70 77 84 91 98 105 112 119 126 133 140 147 154 161 168 175 182 189 196 203 210 *2. Write a temperature-conversion program that gives the user the option of converting Fahrenheit to Celsius or Celsius to Fahrenheit. Then carry out the conversion. Use floating-point numbers. Interaction with the program might look like this: Type 1 to convert Fahrenheit to Celsius, 2 to convert Celsius to Fahrenheit: 1 Enter temperature in Fahrenheit: 70 In Celsius that’s 21.111111 *3. Operators such as >>, which read input from the keyboard, must be able to convert a series of digits into a number. Write a program that does the same thing. It should allow the user to type up to six digits, and then display the resulting number as a type long integer. The digits should be read individually, as characters, using getche(). Constructing the number involves multiplying the existing value by 10 and then adding the new digit. (Hint: Subtract 48 or ‘0’ to go from ASCII to a numerical digit.) Chapter 3126 04 3087 CH03 11/29/01 2:18 PM Page 126Here’s some sample interaction: Enter a number: 123456 Number is: 123456 *4. Create the equivalent of a four-function calculator. The program should ask the user to enter a number, an operator, and another number. (Use floating point.) It should then carry out the specified arithmetical operation: adding, subtracting, multiplying, or divid- ing the two numbers. Use a switch statement to select the operation. Finally, display the result. When it finishes the calculation, the program should ask whether the user wants to do another calculation. The response can be ‘y’ or ‘n’. Some sample interaction with the program might look like this: Enter first number, operator, second number: 10 / 3 Answer = 3.333333 Do another (y/n)? y Enter first number, operator, second number: 12 + 100 Answer = 112 Do another (y/n)? n 5. Use for loops to construct a program that displays a pyramid of Xs on the screen. The pyramid should look like this X XXX XXXXX XXXXXXX XXXXXXXXX except that it should be 20 lines high, instead of the 5 lines shown here. One way to do this is to nest two inner loops, one to print spaces and one to print Xs, inside an outer loop that steps down the screen from line to line. 6. Modify the FACTOR program in this chapter so that it repeatedly asks for a number and calculates its factorial, until the user enters 0, at which point it terminates. You can enclose the relevant statements in FACTOR in a while loop or a do loop to achieve this effect. 7. Write a program that calculates how much money you’ll end up with if you invest an amount of money at a fixed interest rate, compounded yearly. Have the user furnish the initial amount, the number of years, and the yearly interest rate in percent. Some interac- tion with the program might look like this: Enter initial amount: 3000 Enter number of years: 10 Enter interest rate (percent per year): 5.5 At the end of 10 years, you will have 5124.43 dollars. Loops and Decisions 3 L OOPS AND D ECISIONS 127 04 3087 CH03 11/29/01 2:18 PM Page 127At the end of the first year you have 3000 + (3000 * 0.055), which is 3165. At the end of the second year you have 3165 + (3165 * 0.055), which is 3339.08. Do this as many times as there are years. A for loop makes the calculation easy. 8. Write a program that repeatedly asks the user to enter two money amounts expressed in old-style British currency: pounds, shillings, and pence. (See Exercises 10 and 12 in Chapter 2, “C++ Programming Basics.”) The program should then add the two amounts and display the answer, again in pounds, shillings, and pence. Use a do loop that asks the user whether the program should be terminated. Typical interaction might be Enter first amount: £5.10.6 Enter second amount: £3.2.6 Total is £8.13.0 Do you wish to continue (y/n)? To add the two amounts, you’ll need to carry 1 shilling when the pence value is greater than 11, and carry 1 pound when there are more than 19 shillings. 9. Suppose you give a dinner party for six guests, but your table seats only four. In how many ways can four of the six guests arrange themselves at the table? Any of the six guests can sit in the first chair. Any of the remaining five can sit in the second chair. Any of the remaining four can sit in the third chair, and any of the remaining three can sit in the fourth chair. (The last two will have to stand.) So the number of possible arrange- ments of six guests in four chairs is 6*5*4*3, which is 360. Write a program that calcu- lates the number of possible arrangements for any number of guests and any number of chairs. (Assume there will never be fewer guests than chairs.) Don’t let this get too com- plicated. A simple for loop should do it. 10. Write another version of the program from Exercise 7 so that, instead of finding the final amount of your investment, you tell the program the final amount and it figures out how many years it will take, at a fixed rate of interest compounded yearly, to reach this amount. What sort of loop is appropriate for this problem? (Don’t worry about fractional years; use an integer value for the year.) 11. Create a three-function calculator for old-style English currency, where money amounts are specified in pounds, shillings, and pence. (See Exercises 10 and 12 in Chapter 2.) The calculator should allow the user to add or subtract two money amounts, or to multi- ply a money amount by a floating-point number. (It doesn’t make sense to multiply two money amounts; there is no such thing as square money. We’ll ignore division. Use the general style of the ordinary four-function calculator in Exercise 4 in this chapter.) Chapter 3128 04 3087 CH03 11/29/01 2:18 PM Page 12812. Create a four-function calculator for fractions. (See Exercise 9 in Chapter 2, and Exercise 4 in this chapter.) Here are the formulas for the four arithmetic operations applied to fractions: Addition: a/b + c/d = (a*d + b*c) / (b*d) Subtraction: a/b - c/d = (a*d - b*c) / (b*d) Multiplication: a/b * c/d = (a*c) / (b*d) Division: a/b / c/d = (a*d) / (b*c) The user should type the first fraction, an operator, and a second fraction. The program should then display the result and ask whether the user wants to continue. Loops and Decisions 3 L OOPS AND D ECISIONS 129 04 3087 CH03 11/29/01 2:18 PM Page 12904 3087 CH03 11/29/01 2:18 PM Page 130CHAPTER 4 Structures IN THIS CHAPTER •Structures 132 • Enumerations 148 05 3087 CH04 11/29/01 2:23 PM Page 131Chapter 4132 We’ve seen variables of simple data types, such as float, char, and int. Variables of such types represent one item of information: a height, an amount, a count, and so on. But just as groceries are organized into bags, employees into departments, and words into sentences, it’s often convenient to organize simple variables into more complex entities. The C++ construc- tion called the structure is one way to do this. The first part of this chapter is devoted to structures. In the second part we’ll look at a related topic: enumerations. Structures A structure is a collection of simple variables. The variables in a structure can be of different types: Some can be int,some can be float, and so on. (This is unlike the array, which we’ll meet later, in which all the variables must be the same type.) The data items in a structure are called the members of the structure. In books on C programming, structures are often considered an advanced feature and are intro- duced toward the end of the book. However, for C++ programmers, structures are one of the two important building blocks in the understanding of objects and classes. In fact, the syntax of a structure is almost identical to that of a class. A structure (as typically used) is a collection of data, while a class is a collection of both data and functions. So by learning about structures we’ll be paving the way for an understanding of classes and objects. Structures in C++ (and C) serve a similar purpose to records in some other languages such as Pascal. A Simple Structure Let’s start off with a structure that contains three variables: two integers and a floating-point number. This structure represents an item in a widget company’s parts inventory. The structure is a kind of blueprint specifying what information is necessary for a single part. The company makes several kinds of widgets, so the widget model number is the first member of the struc- ture. The number of the part itself is the next member, and the final member is the part’s cost. (Those of you who consider part numbers unexciting need to open your eyes to the romance of commerce.) The program PARTS defines the structure part, defines a structure variable of that type called part1,assigns values to its members, and then displays these values. // parts.cpp // uses parts inventory to demonstrate structures #include using namespace std; 05 3087 CH04 11/29/01 2:23 PM Page 132//////////////////////////////////////////////////////////////// struct part //declare a structure { int modelnumber; //ID number of widget int partnumber; //ID number of widget part float cost; //cost of part }; //////////////////////////////////////////////////////////////// int main() { part part1; //define a structure variable part1.modelnumber = 6244; //give values to structure members part1.partnumber = 373; part1.cost = 217.55F; //display structure members cout << “Model “ << part1.modelnumber; cout << “, part “ << part1.partnumber; cout << “, costs $” << part1.cost << endl; return 0; } The program’s output looks like this: Model 6244, part 373, costs $217.55 The PARTS program has three main aspects: defining the structure, defining a structure variable, and accessing the members of the structure. Let’s look at each of these. Defining the Structure The structure definition tells how the structure is organized: It specifies what members the structure will have. Here it is: struct part { int modelnumber; int partnumber; float cost; }; Syntax of the Structure Definition The keyword struct introduces the structure definition. Next comes the structure name or tag, which is part. The declarations of the structure members—modelnumber, partnumber, and cost—are enclosed in braces. A semicolon follows the closing brace, terminating the entire Structures 4 S TRUCTURES 133 05 3087 CH04 11/29/01 2:23 PM Page 133structure. Note that this use of the semicolon for structures is unlike the usage for a block of code. As we’ve seen, blocks of code, which are used in loops, decisions, and functions, are also delimited by braces. However, they don’t use a semicolon following the final brace. Figure 4.1 shows the syntax of the structure declaration. Chapter 4134 Keyword “struct” Structure name or “tag” Structure members Semicolon terminates definition Braces delimit structure members struct part { int modelnumber; int partnumber; float cost; } ; FIGURE 4.1 Syntax of the structure definition. Use of the Structure Definition The structure definitiondefinition serves only as a blueprint for the creation of variables of type part. It does not itself create any structure variables; that is, it does not set aside any space in memory or even name any variables. This is unlike the definition of a simple variable, which does set aside memory. A structure definition is merely a specification for how structure vari- ables will look when they are defined. This is shown in Figure 4.2. It’s not accidental that this description sounds like the distinction we noted between classes and objects in Chapter 1, “The Big Picture.” As we’ll see, an object has the same relationship to its class that a variable of a structure type has to the structure definition. Defining a Structure Variable The first statement in main() part part1; defines a variable, called part1,of type structure part. This definition reserves space in memory for part1. How much space? Enough to hold all the members of part1—namely modelnumber, partnumber, and cost. In this case there will be 4 bytes for each of the two ints (assuming a 32-bit system), and 4 bytes for the float. Figure 4.3 shows how part1 looks in memory. (The figure shows 2-byte integers.) 05 3087 CH04 11/29/01 2:23 PM Page 134FIGURE 4.2 Structures and structure variables. In some ways we can think of the part structure as the specification for a new data type. This will become more clear as we go along, but notice that the format for defining a structure vari- able is the same as that for defining a basic built-in data type such as int: part part1; int var1; This similarity is not accidental. One of the aims of C++ is to make the syntax and the opera- tion of user-defined data types as similar as possible to that of built-in data types. (In C you need to include the keyword struct in structure definitions, as in struct part part1;. In C++ the keyword is not necessary.) Structures 4 S TRUCTURES 135 Structure definition for Foo Variables of type Foo Specifications for structure Foo Foo 1 Foo 2 Foo 3 05 3087 CH04 11/29/01 2:23 PM Page 135FIGURE 4.3 Structure members in memory. Accessing Structure Members Once a structure variable has been defined, its members can be accessed using something called the dot operator. Here’s how the first member is given a value: part1.modelnumber = 6244; The structure member is written in three parts: the name of the structure variable (part1); the dot operator, which consists of a period (.); and the member name (modelnumber). This means “the modelnumber member of part1.” The real name of the dot operator is member access operator,but of course no one wants to use such a lengthy term. Remember that the first component of an expression involving the dot operator is the name of the specific structure variable (part1 in this case), not the name of the structure definition (part). The variable name must be used to distinguish one variable from another, such as part1, part2, and so on, as shown in Figure 4.4. Chapter 4136 05 3087 CH04 11/29/01 2:23 PM Page 136FIGURE 4.4 The dot operator. Structure members are treated just like other variables. In the statement part1.modelnumber = 6244;,the member is given the value 6244 using a normal assignment operator. The program also shows members used in cout statements such as cout << “\nModel “ << part1.modelnumber; These statements output the values of the structure members. Other Structure Features Structures are surprisingly versatile. Let’s look at some additional features of structure syntax and usage. Structures 4 S TRUCTURES 137 05 3087 CH04 11/29/01 2:23 PM Page 137Initializing Structure Members The next example shows how structure members can be initialized when the structure variable is defined. It also demonstrates that you can have more than one variable of a given structure type (we hope you suspected this all along). Here’s the listing for PARTINIT: // partinit.cpp // shows initialization of structure variables #include using namespace std; //////////////////////////////////////////////////////////////// struct part //specify a structure { int modelnumber; //ID number of widget int partnumber; //ID number of widget part float cost; //cost of part }; //////////////////////////////////////////////////////////////// int main() { //initialize variable part part1 = { 6244, 373, 217.55F }; part part2; //define variable //display first variable cout << “Model “ << part1.modelnumber; cout << “, part “ << part1.partnumber; cout << “, costs $” << part1.cost << endl; part2 = part1; //assign first variable to second //display second variable cout << “Model “ << part2.modelnumber; cout << “, part “ << part2.partnumber; cout << “, costs $” << part2.cost << endl; return 0; } This program defines two variables of type part: part1 and part2. It initializes part1,prints out the values of its members, assigns part1 to part2, and prints out its members. Here’s the output: Model 6244, part 373, costs $217.55 Model 6244, part 373, costs $217.55 Not surprisingly, the same output is repeated since one variable is made equal to the other. The part1 structure variable’s members are initialized when the variable is defined: part part1 = { 6244, 373, 217.55 }; Chapter 4138 05 3087 CH04 11/29/01 2:23 PM Page 138The values to be assigned to the structure members are surrounded by braces and separated by commas. The first value in the list is assigned to the first member, the second to the second member, and so on. Structure Variables in Assignment Statements As can be seen in PARTINIT, one structure variable can be assigned to another: part2 = part1; The value of each member of part1 is assigned to the corresponding member of part2. Since a large structure can have dozens of members, such an assignment statement can require the computer to do a considerable amount of work. Note that one structure variable can be assigned to another only when they are of the same structure type. If you try to assign a variable of one structure type to a variable of another type, the compiler will complain. A Measurement Example Let’s see how a structure can be used to group a different kind of information. If you’ve ever looked at an architectural drawing, you know that (at least in the United States) distances are measured in feet and inches. (As you probably know, there are 12 inches in a foot.) The length of a living room, for example, might be given as 15’–8”, meaning 15 feet plus 8 inches. The hyphen isn’t a negative sign; it merely separates the feet from the inches. This is part of the English system of measurement. (We’ll make no judgment here on the merits of English versus metric.) Figure 4.5 shows typical length measurements in the English system. Suppose you want to create a drawing or architectural program that uses the English system. It will be convenient to store distances as two numbers, representing feet and inches. The next example, ENGLSTRC,gives an idea of how this could be done using a structure. This program will show how two measurements of type Distance can be added together. // englstrc.cpp // demonstrates structures using English measurements #include using namespace std; //////////////////////////////////////////////////////////////// struct Distance //English distance { int feet; float inches; }; //////////////////////////////////////////////////////////////// Structures 4 S TRUCTURES 139 05 3087 CH04 11/29/01 2:23 PM Page 139int main() { Distance d1, d3; //define two lengths Distance d2 = { 11, 6.25 }; //define & initialize one length //get length d1 from user cout << “\nEnter feet: “; cin >> d1.feet; cout << “Enter inches: “; cin >> d1.inches; //add lengths d1 and d2 to get d3 d3.inches = d1.inches + d2.inches; //add the inches d3.feet = 0; //(for possible carry) if(d3.inches >= 12.0) //if total exceeds 12.0, { //then decrease inches by 12.0 d3.inches -= 12.0; //and d3.feet++; //increase feet by 1 } d3.feet += d1.feet + d2.feet; //add the feet //display all lengths cout << d1.feet << “\’-” << d1.inches << “\” + “; cout << d2.feet << “\’-” << d2.inches << “\” = “; cout << d3.feet << “\’-” << d3.inches << “\”\n”; return 0; } Chapter 4140 FIGURE 4.5 Measurements in the English system. 05 3087 CH04 11/29/01 2:23 PM Page 140Here the structure Distance has two members: feet and inches. The inches variable may have a fractional part, so we’ll use type float for it. Feet are always integers, so we’ll use type int for them. We define two such distances, d1 and d3,without initializing them, while we initialize another, d2,to 11'–6.25''. The program asks the user to enter a distance in feet and inches, and assigns this distance to d1. (The inches value should be smaller than 12.0.) It then adds the distance d1 to d2, obtaining the total distance d3. Finally the program displays the two initial distances and the newly calculated total distance. Here’s some output: Enter feet: 10 Enter inches: 6.75 10’-6.75” + 11’-6.25” = 22’-1” Notice that we can’t add the two distances with a program statement like d3 = d1 + d2; // can’t do this in ENGLSTRC Why not? Because there is no routine built into C++ that knows how to add variables of type Distance. The + operator works with built-in types like float,but not with types we define ourselves, like Distance. (However, one of the benefits of using classes, as we’ll see in Chapter 8, “Operator Overloading,” is the ability to add and perform other operations on user- defined data types.) Structures Within Structures You can nest structures within other structures. Here’s a variation on the ENGLSTRC program that shows how this looks. In this program we want to create a data structure that stores the dimensions of a typical room: its length and width. Since we’re working with English dis- tances, we’ll use two variables of type Distance as the length and width variables. struct Room { Distance length; Distance width; } Here’s a program, ENGLAREA,that uses the Room structure to represent a room. // englarea.cpp // demonstrates nested structures #include using namespace std; //////////////////////////////////////////////////////////////// struct Distance //English distance { int feet; Structures 4 S TRUCTURES 141 05 3087 CH04 11/29/01 2:23 PM Page 141float inches; }; //////////////////////////////////////////////////////////////// struct Room //rectangular area { Distance length; //length of rectangle Distance width; //width of rectangle }; //////////////////////////////////////////////////////////////// int main() { Room dining; //define a room dining.length.feet = 13; //assign values to room dining.length.inches = 6.5; dining.width.feet = 10; dining.width.inches = 0.0; //convert length & width float l = dining.length.feet + dining.length.inches/12; float w = dining.width.feet + dining.width.inches/12; //find area and display it cout << “Dining room area is “ << l * w << “ square feet\n” ; return 0; } This program defines a single variable—dining—of type Room,in the line Room dining; // variable dining of type Room It then assigns values to the various members of this structure. Accessing Nested Structure Members Because one structure is nested inside another, we must apply the dot operator twice to access the structure members. dining.length.feet = 13; In this statement, dining is the name of the structure variable, as before; length is the name of a member in the outer structure (Room); and feet is the name of a member of the inner struc- ture (Distance). The statement means “take the feet member of the length member of the variable dining and assign it the value 13.” Figure 4.6 shows how this works. Chapter 4142 05 3087 CH04 11/29/01 2:23 PM Page 1424 S TRUCTURES Structures 143 FIGURE 4.6 Dot operator and nested structures. Once values have been assigned to members of dining,the program calculates the floor area of the room, as shown in Figure 4.7. To find the area, the program converts the length and width from variables of type Distance to variables of type float, l, and w,representing distances in feet. The values of l and w are found by adding the feet member of Distance to the inches member divided by 12. The feet member is converted to type float automatically before the addition is performed, and the result is type float. The l and w variables are then multiplied together to obtain the area. 05 3087 CH04 11/29/01 2:23 PM Page 143FIGURE 4.7 Area in feet and inches. User-Defined Type Conversions Note that the program converts two distances of type Distance to two distances of type float: the variables l and w. In effect it also converts the room’s area, which is stored as a structure of type Room (which is defined as two structures of type Distance), to a single floating-point number representing the area in square feet. Here’s the output: Dining room area is 135.416672 square feet Converting a value of one type to a value of another is an important aspect of programs that employ user-defined data types. Initializing Nested Structures How do you initialize a structure variable that itself contains structures? The following state- ment initializes the variable dining to the same values it is given in the ENGLAREA program: Room dining = { {13, 6.5}, {10, 0.0} }; Each structure of type Distance, which is embedded in Room, is initialized separately. Remember that this involves surrounding the values with braces and separating them with commas. The first Distance is initialized to {13, 6.5} Chapter 4144 05 3087 CH04 11/29/01 2:23 PM Page 144and the second to {10, 0.0} These two Distance values are then used to initialize the Room variable; again, they are surrounded with braces and separated by commas. Depth of Nesting In theory, structures can be nested to any depth. In a program that designs apartment buildings, you might find yourself with statements like this one: apartment1.laundry_room.washing_machine.width.feet A Card Game Example Let’s examine a different kind of example. This one uses a structure to model a playing card. The program imitates a game played by cardsharps (professional gamblers) at carnivals. The cardsharp shows you three cards, then places them face down on the table and interchanges their positions several times. If you can guess correctly where a particular card is, you win. Everything is in plain sight, yet the cardsharp switches the cards so rapidly and confusingly that the player (the mark) almost always loses track of the card and loses the game, which is, of course, played for money. Here’s the structure the program uses to represent a playing card: struct card { int number; int suit; }; This structure uses separate members to hold the number of the card and the suit. The number runs from 2 to 14, where 11, 12, 13, and 14 represent the jack, queen, king, and ace, respec- tively (this is the order used in poker). The suit runs from 0 to 3, where these four numbers represent clubs, diamonds, hearts, and spades. Here’s the listing for CARDS: // cards.cpp // demonstrates structures using playing cards #include using namespace std; const int clubs = 0; //suits const int diamonds = 1; const int hearts = 2; const int spades = 3; Structures 4 S TRUCTURES 145 05 3087 CH04 11/29/01 2:23 PM Page 145const int jack = 11; //face cards const int queen = 12; const int king = 13; const int ace = 14; //////////////////////////////////////////////////////////////// struct card { int number; //2 to 10, jack, queen, king, ace int suit; //clubs, diamonds, hearts, spades }; //////////////////////////////////////////////////////////////// int main() { card temp, chosen, prize; //define cards int position; card card1 = { 7, clubs }; //initialize card1 cout << “Card 1 is the 7 of clubs\n”; card card2 = { jack, hearts }; //initialize card2 cout << “Card 2 is the jack of hearts\n”; card card3 = { ace, spades }; //initialize card3 cout << “Card 3 is the ace of spades\n”; prize = card3; //copy this card, to remember it cout << “I’m swapping card 1 and card 3\n”; temp = card3; card3 = card1; card1 = temp; cout << “I’m swapping card 2 and card 3\n”; temp = card3; card3 = card2; card2 = temp; cout << “I’m swapping card 1 and card 2\n”; temp = card2; card2 = card1; card1 = temp; cout << “Now, where (1, 2, or 3) is the ace of spades? “; cin >> position; switch (position) { case 1: chosen = card1; break; case 2: chosen = card2; break; case 3: chosen = card3; break; } Chapter 4146 05 3087 CH04 11/29/01 2:23 PM Page 146if(chosen.number == prize.number && // compare cards chosen.suit == prize.suit) cout << “That’s right! You win!\n”; else cout << “Sorry. You lose.\n”; return 0; } Here’s some sample interaction with the program: Card 1 is the 7 of clubs Card 2 is the jack of hearts Card 3 is the ace of spades I’m swapping card 1 and card 3 I’m swapping card 2 and card 3 I’m swapping card 1 and card 2 Now, where (1, 2, or 3) is the ace of spades? 3 Sorry. You lose. In this case the hapless mark chose the wrong card (the right answer is 2). The program begins by defining a number of variables of type const int for the face card and suit values. (Not all these variables are used in the program; they’re included for complete- ness.) Next the card structure is specified. The program then defines three uninitialized vari- ables of type card: temp, chosen, and prize. It also defines three cards—card1, card2,and card3—which it initializes to three arbitrary card values. It prints out the values of these cards for the user’s information. It then sets a card variable, prize,to one of these card values as a way of remembering it. This card is the one whose location the player will be asked to guess at the end of the game. Next the program rearranges the cards. It swaps the first and third cards, the second and third cards, and the first and second cards. Each time it tells the user what it’s doing. (If you find the program too easy, you can add more such statements to further shuffle the cards. Flashing the statements on the screen for a limited time would also increase the challenge.) Finally the program asks the player what position a particular card is in. It sets a card variable, chosen,to the card in this position, and then compares chosen with the prize card. If they match, it’s a win for the player; if not, it’s a loss. Notice how easy swapping cards is. temp = card3; card3 = card1; card1 = temp; Although the cards represent structures, they can be moved around very naturally, thanks to the ability of the assignment operator (=) to work with structures. Structures 4 S TRUCTURES 147 05 3087 CH04 11/29/01 2:23 PM Page 147Unfortunately, just as structures can’t be added, they also can’t be compared. You can’t say if( chosen == prize ) //not legal yet because there’s no routine built into the == operator that knows about the card structure. But, as with addition, this problem can be solved with operator overloading, as we’ll see later. Structures and Classes We must confess to having misled you slightly on the capabilities of structures. It’s true that structures are usually used to hold data only, and classes are used to hold both data and func- tions. However, in C++, structures can in fact hold both data and functions. (In C they can hold only data.) The syntactical distinction between structures and classes in C++ is minimal, so they can in theory be used almost interchangeably. But most C++ programmers use structures as we have in this chapter, exclusively for data. Classes are usually used to hold both data and functions, as we’ll see in Chapter 6, “Objects and Classes.” Enumerations As we’ve seen, structures can be looked at as a way to provide user-defined data types. A dif- ferent approach to defining your own data type is the enumeration. This feature of C++ is less crucial than structures. You can write perfectly good object-oriented programs in C++ without knowing anything about enumerations. However, they are very much in the spirit of C++, in that, by allowing you to define your own data types, they can simplify and clarify your pro- gramming. Days of the Week Enumerated types work when you know in advance a finite (usually short) list of values that a data type can take on. Here’s an example program, DAYENUM,that uses an enumeration for the days of the week: // dayenum.cpp // demonstrates enum types #include using namespace std; //specify enum type enum days_of_week { Sun, Mon, Tue, Wed, Thu, Fri, Sat }; int main() { days_of_week day1, day2; //define variables //of type days_of_week Chapter 4148 05 3087 CH04 11/29/01 2:23 PM Page 148day1 = Mon; //give values to day2 = Thu; //variables int diff = day2 - day1; //can do integer arithmetic cout << “Days between = “ << diff << endl; if(day1 < day2) //can do comparisons cout << “day1 comes before day2\n”; return 0; } An enum declaration defines the set of all names that will be permissible values of the type. These permissible values are called enumerators. The enum type days_of_week has seven enumerators: Sun, Mon, Tue, and so on, up to Sat. Figure 4.8 shows the syntax of an enum declaration. Structures 4 S TRUCTURES 149 FIGURE 4.8 Syntax of enum specifier. An enumeration is a list of all possible values. This is unlike the specification of an int,for example, which is given in terms of a range of values. In an enum you must give a specific name to every possible value. Figure 4.9 shows the difference between an int and an enum. Once you’ve declared the enum type days_of_week as shown, you can define variables of this type. DAYENUM has two such variables, day1 and day2,defined in the statement days_of_week day1, day2; (In C you must use the keyword enum before the type name, as in enum days_of_week day1, day2; In C++ this isn’t necessary.) 05 3087 CH04 11/29/01 2:23 PM Page 149FIGURE 4.9 Usage of ints and enums. Variables of an enumerated type, like day1 and day2, can be given any of the values listed in the enum declaration. In the example we give them the values Mon and Thu. You can’t use values that weren’t listed in the declaration. Such statements as day1 = halloween; are illegal. You can use the standard arithmetic operators on enum types. In the program we subtract two values. You can also use the comparison operators, as we show. Here’s the program’s output: Days between = 3 day1 comes before day2 Chapter 4150 05 3087 CH04 11/29/01 2:23 PM Page 150The use of arithmetic and relational operators doesn’t make much sense with some enum types. For example, if you have the declaration enum pets { cat, dog, hamster, canary, ocelot }; then it may not be clear what expressions like dog + canary or (cat < hamster) mean. Enumerations are treated internally as integers. This explains why you can perform arithmetic and relational operations on them. Ordinarily the first name in the list is given the value 0, the next name is given the value 1, and so on. In the DAYENUM example, the values Sun through Sat are stored as the integer values 0–6. Arithmetic operations on enum types take place on the integer values. However, although the compiler knows that your enum variables are really integers, you must be careful of trying to take advantage of this fact. If you say day1 = 5; the compiler will issue a warning (although it will compile). It’s better to forget—whenever possible—that enums are really integers. One Thing or Another Our next example counts the words in a phrase typed in by the user. Unlike the earlier CHCOUNT example, however, it doesn’t simply count spaces to determine the number of words. Instead it counts the places where a string of nonspace characters changes to a space, as shown in Figure 4.10. Structures 4 S TRUCTURES 151 FIGURE 4.10 Operation of the WDCOUNT program. This way you don’t get a false count if you type multiple spaces between words. (It still doesn’t handle tabs and other whitespace characters.) Here’s the listing for WDCOUNT:This example shows an enumeration with only two enumerators. 05 3087 CH04 11/29/01 2:23 PM Page 151// wdcount.cpp // demonstrates enums, counts words in phrase #include using namespace std; #include //for getche() enum itsaWord { NO, YES }; //NO=0, YES=1 int main() { itsaWord isWord = NO; //YES when in a word, //NO when in whitespace char ch = ‘a’; //character read from keyboard int wordcount = 0; //number of words read cout << “Enter a phrase:\n”; do { ch = getche(); //get character if(ch==’ ‘ || ch==’\r’) //if white space, { if( isWord == YES ) //and doing a word, { //then it’s end of word wordcount++; //count the word isWord = NO; //reset flag } } //otherwise, it’s else //normal character if( isWord == NO ) //if start of word, isWord = YES; //then set flag } while( ch != ‘\r’ ); //quit on Enter key cout << “\n---Word count is “ << wordcount << “---\n”; return 0; } The program cycles in a do loop, reading characters from the keyboard. It passes over (non- space) characters until it finds a space. At this point it counts a word. Then it passes over spaces until it finds a character, and again counts characters until it finds a space. Doing this requires the program to remember whether it’s in the middle of a word, or in the middle of a string of spaces. It remembers this with the enum variable isWord. This variable is defined to be of type itsaWord. This type is specified in the statement enum itsaWord { NO, YES }; Variables of type itsaWord have only two possible values: NO and YES. Notice that the list starts with NO,so this value will be given the value 0—the value that indicates false. (We could also use a variable of type bool for this purpose.) Chapter 4152 05 3087 CH04 11/29/01 2:23 PM Page 152The isWord variable is set to NO when the program starts. When the program encounters the first nonspace character, it sets isWord to YES to indicate that it’s in the middle of a word. It keeps this value until the next space is found, at which point it’s set back to NO. Behind the scenes, NO has the value 0 and YES has the value 1, but we avoid making use of this fact. We could have used if(isWord) instead of if(isWord == YES), and if(!isWord) instead of if(isWord == NO),but this is not good style. Note also that we need an extra set of braces around the second if statement in the program, so that the else will match the first if. Another approach to a yes/no situation such as that in WDCOUNT is to use a variable of type bool. This may be a little more straightforward, depending on the situation. Organizing the Cards Here’s our final example of enum types. Remember that in the CARDS program earlier in this chapter we defined a group of constants of type const int to represent a card’s suits. const int clubs = 0; const int diamonds = 1; const int hearts = 2; const int spades = 3; This sort of list is somewhat clumsy. Let’s revise the CARDS program to use enumerations instead. Here’s the listing for CARDENUM: // cardenum.cpp // demonstrates enumerations #include using namespace std; const int jack = 11; //2 through 10 are unnamed integers const int queen = 12; const int king = 13; const int ace = 14; enum Suit { clubs, diamonds, hearts, spades }; //////////////////////////////////////////////////////////////// struct card { int number; //2 to 10, jack, queen, king, ace Suit suit; //clubs, diamonds, hearts, spades }; //////////////////////////////////////////////////////////////// int main() { Structures 4 S TRUCTURES 153 05 3087 CH04 11/29/01 2:23 PM Page 153card temp, chosen, prize; //define cards int position; card card1 = { 7, clubs }; //initialize card1 cout << “Card 1 is the seven of clubs\n”; card card2 = { jack, hearts }; //initialize card2 cout << “Card 2 is the jack of hearts\n”; card card3 = { ace, spades }; //initialize card3 cout << “Card 3 is the ace of spades\n”; prize = card3; //copy this card, to remember it cout << “I’m swapping card 1 and card 3\n”; temp = card3; card3 = card1; card1 = temp; cout << “I’m swapping card 2 and card 3\n”; temp = card3; card3 = card2; card2 = temp; cout << “I’m swapping card 1 and card 2\n”; temp = card2; card2 = card1; card1 = temp; cout << “Now, where (1, 2, or 3) is the ace of spades? “; cin >> position; switch (position) { case 1: chosen = card1; break; case 2: chosen = card2; break; case 3: chosen = card3; break; } if(chosen.number == prize.number && //compare cards chosen.suit == prize.suit) cout << “That’s right! You win!\n”; else cout << “Sorry. You lose.\n”; return 0; } Here the set of definitions for suits used in the CARDS program has been replaced by an enum declaration: enum Suit { clubs, diamonds, hearts, spades }; Chapter 4154 05 3087 CH04 11/29/01 2:23 PM Page 154This is a cleaner approach than using const variables. We know exactly what the possible val- ues of the suit are; attempts to use other values, as in card1.suit = 5; result in warnings from the compiler. Specifying Integer Values We said that in an enum declaration the first enumerator was given the integer value 0, the sec- ond the value 1, and so on. This ordering can be altered by using an equal sign to specify a starting point other than 0. For example, if you want the suits to start with 1 instead of 0, you can say enum Suit { clubs=1, diamonds, hearts, spades }; Subsequent names are given values starting at this point, so diamonds is 2, hearts is 3, and spades is 4. Actually you can use an equal sign to give a specified value to any enumerator. Not Perfect One annoying aspect of enum types is that they are not recognized by C++ input/output (I/O) statements. As an example, what do you think the following code fragment will cause to be displayed? enum direction { north, south, east, west }; direction dir1 = south; cout << dir1; Did you guess the output would be south? That would be nice, but C++ I/O treats variables of enum types as integers, so the output would be 1. Other Examples Here are some other examples of enumerated data declarations, to give you a feeling for possi- ble uses of this feature: enum months { Jan, Feb, Mar, Apr, May, Jun, Jul, Aug, Sep, Oct, Nov, Dec }; enum switch { off, on }; enum meridian { am, pm }; enum chess { pawn, knight, bishop, rook, queen, king }; enum coins { penny, nickel, dime, quarter, half-dollar, dollar }; We’ll see other examples in future programs. Structures 4 S TRUCTURES 155 05 3087 CH04 11/29/01 2:23 PM Page 155Summary We’ve covered two topics in this chapter: structures and enumerations. Structures are an impor- tant component of C++, since their syntax is the same as that of classes. In fact, classes are (syntactically, at least) nothing more than structures that include functions. Structures are typi- cally used to group several data items together to form a single entity. A structure definition lists the variables that make up the structure. Other definitions then set aside memory for struc- ture variables. Structure variables are treated as indivisible units in some situations (such as setting one structure variable equal to another), but in other situations their members are accessed individually (often using the dot operator). An enumeration is a programmer-defined type that is limited to a fixed list of values. A decla- ration gives the type a name and specifies the permissible values, which are called enumerators. Definitions can then create variables of this type. Internally the compiler treats enumeration variables as integers. Structures should not be confused with enumerations. Structures are a powerful and flexible way of grouping a diverse collection of data into a single entity. An enumeration allows the definition of variables that can take on a fixed set of values that are listed (enumerated) in the type’s declaration. Questions Answers to these questions can be found in Appendix G. 1. A structure brings together a group of a. items of the same data type. b. related data items. c. integers with user-defined names. d. variables. 2. True or false: A structure and a class use similar syntax. 3. The closing brace of a structure is followed by a __________. 4. Write a structure specification that includes three variables—all of type int—called hrs, mins, and secs. Call this structure time. 5. True or false: A structure definition creates space in memory for a variable. Chapter 4156 05 3087 CH04 11/29/01 2:23 PM Page 1566. When accessing a structure member, the identifier to the left of the dot operator is the name of a. a structure member. b. a structure tag. c. a structure variable. d. the keyword struct. 7. Write a statement that sets the hrs member of the time2 structure variable equal to 11. 8. If you have three variables defined to be of type struct time, and this structure contains three int members, how many bytes of memory do the variables use together? 9. Write a definition that initializes the members of time1—which is a variable of type struct time,as defined in Question 4—to hrs = 11, mins = 10, secs = 59. 10. True or false: You can assign one structure variable to another, provided they are of the same type. 11. Write a statement that sets the variable temp equal to the paw member of the dogs mem- ber of the fido variable. 12. An enumeration brings together a group of a. items of different data types. b. related data variables. c. integers with user-defined names. d. constant values. 13. Write a statement that declares an enumeration called players with the values B1, B2, SS, B3, RF, CF, LF, P, and C. 14. Assuming the enum type players as declared in Question 13, define two variables joe and tom, and assign them the values LF and P, respectively. 15. Assuming the statements of Questions 13 and 14, state whether each of the following statements is legal. a. joe = QB; b. tom = SS; c. LF = tom; d. difference = joe - tom; 16. The first three enumerators of an enum type are normally represented by the values _________, _________, and _________. 17. Write a statement that declares an enumeration called speeds with the enumerators obsolete, single, and album. Give these three names the integer values 78, 45, and 33. Structures 4 S TRUCTURES 157 05 3087 CH04 11/29/01 2:23 PM Page 15718. State the reason that enum isWord{ NO, YES }; is better than enum isWord{ YES, NO }; Exercises Answers to the starred exercises can be found in Appendix G. *1. A phone number, such as (212) 767-8900, can be thought of as having three parts: the area code (212), the exchange (767), and the number (8900). Write a program that uses a structure to store these three parts of a phone number separately. Call the structure phone. Create two structure variables of type phone. Initialize one, and have the user input a number for the other one. Then display both numbers. The interchange might look like this: Enter your area code, exchange, and number: 415 555 1212 My number is (212) 767-8900 Your number is (415) 555-1212 *2. A point on the two-dimensional plane can be represented by two numbers: an x coordi- nate and a y coordinate. For example, (4,5) represents a point 4 units to the right of the vertical axis, and 5 units up from the horizontal axis. The sum of two points can be defined as a new point whose x coordinate is the sum of the x coordinates of the two points, and whose y coordinate is the sum of the y coordinates. Write a program that uses a structure called point to model a point. Define three points, and have the user input values to two of them. Then set the third point equal to the sum of the other two, and display the value of the new point. Interaction with the program might look like this: Enter coordinates for p1: 3 4 Enter coordinates for p2: 5 7 Coordinates of p1+p2 are: 8, 11 *3. Create a structure called Volume that uses three variables of type Distance (from the ENGLSTRC example) to model the volume of a room. Initialize a variable of type Volume to specific dimensions, then calculate the volume it represents, and print out the result. To calculate the volume, convert each dimension from a Distance variable to a variable of type float representing feet and fractions of a foot, and then multiply the resulting three numbers. 4. Create a structure called employee that contains two members: an employee number (type int) and the employee’s compensation (in dollars; type float). Ask the user to fill in this data for three employees, store it in three variables of type struct employee, and then display the information for each employee. Chapter 4158 05 3087 CH04 11/29/01 2:23 PM Page 1585. Create a structure of type date that contains three members: the month, the day of the month, and the year, all of type int. (Or use day-month-year order if you prefer.) Have the user enter a date in the format 12/31/2001, store it in a variable of type struct date, then retrieve the values from the variable and print them out in the same format. 6. We said earlier that C++ I/O statements don’t automatically understand the data types of enumerations. Instead, the (>>) and (<<) operators think of such variables simply as inte- gers. You can overcome this limitation by using switch statements to translate between the user’s way of expressing an enumerated variable and the actual values of the enumer- ated variable. For example, imagine an enumerated type with values that indicate an employee type within an organization: enum etype { laborer, secretary, manager, accountant, executive, researcher }; Write a program that first allows the user to specify a type by entering its first letter (‘l’, ‘s’, ‘m’, and so on), then stores the type chosen as a value of a variable of type enum etype, and finally displays the complete word for this type. Enter employee type (first letter only) laborer, secretary, manager, accountant, executive, researcher): a Employee type is accountant. You’ll probably need two switch statements: one for input and one for output. 7. Add a variable of type enum etype (see Exercise 6), and another variable of type struct date (see Exercise 5) to the employee class of Exercise 4. Organize the resulting pro- gram so that the user enters four items of information for each of three employees: an employee number, the employee’s compensation, the employee type, and the date of first employment. The program should store this information in three variables of type employee, and then display their contents. 8. Start with the fraction-adding program of Exercise 9 in Chapter 2, “C++ Programming Basics.” This program stores the numerator and denominator of two fractions before adding them, and may also store the answer, which is also a fraction. Modify the pro- gram so that all fractions are stored in variables of type struct fraction, whose two members are the fraction’s numerator and denominator (both type int). All fraction- related data should be stored in structures of this type. 9. Create a structure called time. Its three members, all type int,should be called hours, minutes, and seconds. Write a program that prompts the user to enter a time value in hours, minutes, and seconds. This can be in 12:59:59 format, or each number can be entered at a separate prompt (“Enter hours:”, and so forth). The program should then store the time in a variable of type struct time, and finally print out the total number of seconds represented by this time value: long totalsecs = t1.hours*3600 + t1.minutes*60 + t1.seconds Structures 4 S TRUCTURES 159 05 3087 CH04 11/29/01 2:23 PM Page 15910. Create a structure called sterling that stores money amounts in the old-style British system discussed in Exercises 8 and 11 in Chapter 3, “Loops and Decisions.” The mem- bers could be called pounds, shillings, and pence,all of type int. The program should ask the user to enter a money amount in new-style decimal pounds (type double), con- vert it to the old-style system, store it in a variable of type struct sterling, and then display this amount in pounds-shillings-pence format. 11. Use the time structure from Exercise 9, and write a program that obtains two time val- ues from the user in 12:59:59 format, stores them in struct time variables, converts each one to seconds (type int), adds these quantities, converts the result back to hours- minutes-seconds, stores the result in a time structure, and finally displays the result in 12:59:59 format. 12. Revise the four-function fraction calculator program of Exercise 12 in Chapter 3 so that each fraction is stored internally as a variable of type struct fraction,as discussed in Exercise 8 in this chapter. Chapter 4160 05 3087 CH04 11/29/01 2:23 PM Page 160CHAPTER 5 Functions IN THIS CHAPTER •Simple Functions 162 • Passing Arguments to Functions 167 • Returning Values from Functions 176 • Reference Arguments 182 • Overloaded Functions 188 • Recursion 193 •Inline Functions 195 • Default Arguments 197 • Scope and Storage Class 199 • Returning by Reference 206 • const Function Arguments 208 06 3087 CH05 11/29/01 2:22 PM Page 161Chapter 5162 A function groups a number of program statements into a unit and gives it a name. This unit can then be invoked from other parts of the program. The most important reason to use functions is to aid in the conceptual organization of a pro- gram. Dividing a program into functions is, as we discussed in Chapter 1, “The Big Picture,” one of the major principles of structured programming. (However, object-oriented program- ming provides additional, more powerful ways to organize programs.) Another reason to use functions (and the reason they were invented, long ago) is to reduce pro- gram size. Any sequence of instructions that appears in a program more than once is a candi- date for being made into a function. The function’s code is stored in only one place in memory, even though the function is executed many times in the course of the program. Figure 5.1 shows how a function is invoked from different sections of a program. FIGURE 5.1 Flow of control to a function. Functions in C++ (and C) are similar to subroutines and procedures in various other languages. Simple Functions Our first example demonstrates a simple function whose purpose is to print a line of 45 aster- isks. The example program generates a table, and lines of asterisks are used to make the table more readable. Here’s the listing for TABLE: 06 3087 CH05 11/29/01 2:22 PM Page 162// table.cpp // demonstrates simple function #include using namespace std; void starline(); //function declaration // (prototype) int main() { starline(); //call to function cout << “Data type Range” << endl; starline(); //call to function cout << “char -128 to 127” << endl << “short -32,768 to 32,767” << endl << “int System dependent” << endl << “long -2,147,483,648 to 2,147,483,647” << endl; starline(); //call to function return 0; } //-------------------------------------------------------------- // starline() // function definition void starline() //function declarator { for(int j=0; j<45; j++) //function body cout << ‘*’; cout << endl; } The output from the program looks like this: ********************************************* Data type Range ********************************************* char -128 to 127 short -32,768 to 32,767 int System dependent long -2,147,483,648 to 2,147,483,647 ********************************************* The program consists of two functions: main() and starline(). You’ve already seen many programs that use main() alone. What other components are necessary to add a function to the program? There are three: the function declaration,the calls to the function, and the function definition. 5 F UNCTIONS 163Functions 06 3087 CH05 11/29/01 2:22 PM Page 163The Function Declaration Just as you can’t use a variable without first telling the compiler what it is, you also can’t use a function without telling the compiler about it. There are two ways to do this. The approach we show here is to declare the function before it is called. (The other approach is to define it before it’s called; we’ll examine that next.) In the TABLE program, the function starline() is declared in the line void starline(); The declaration tells the compiler that at some later point we plan to present a function called starline. The keyword void specifies that the function has no return value, and the empty parentheses indicate that it takes no arguments. (You can also use the keyword void in paren- theses to indicate that the function takes no arguments, as is often done in C, but leaving them empty is the more common practice in C++.) We’ll have more to say about arguments and return values soon. Notice that the function declaration is terminated with a semicolon. It is a complete statement in itself. Function declarations are also called prototypes, since they provide a model or blueprint for the function. They tell the compiler, “a function that looks like this is coming up later in the pro- gram, so it’s all right if you see references to it before you see the function itself.” The infor- mation in the declaration (the return type and the number and types of any arguments) is also sometimes referred to as the function signature. Calling the Function The function is called (or invoked,or executed) three times from main(). Each of the three calls looks like this: starline(); This is all we need to call the function: the function name, followed by parentheses. The syn- tax of the call is very similar to that of the declaration, except that the return type is not used. The call is terminated by a semicolon. Executing the call statement causes the function to exe- cute; that is, control is transferred to the function, the statements in the function definition (which we’ll examine in a moment) are executed, and then control returns to the statement fol- lowing the function call. The Function Definition Finally we come to the function itself, which is referred to as the function definition. The defi- nition contains the actual code for the function. Here’s the definition for starline(): Chapter 5164 06 3087 CH05 11/29/01 2:22 PM Page 164void starline() //declarator { for(int j=0; j<45; j++) //function body cout << ‘*’; cout << endl; } The definition consists of a line called the declarator,followed by the function body. The function body is composed of the statements that make up the function, delimited by braces. The declarator must agree with the declaration: It must use the same function name, have the same argument types in the same order (if there are arguments), and have the same return type. Notice that the declarator is not terminated by a semicolon. Figure 5.2 shows the syntax of the function declaration, function call, and function definition. 5 F UNCTIONS 165Functions FIGURE 5.2 Function syntax. 06 3087 CH05 11/29/01 2:22 PM Page 165When the function is called, control is transferred to the first statement in the function body. The other statements in the function body are then executed, and when the closing brace is encountered, control returns to the calling program. Table 5.1 summarizes the different function components. TABLE 5.1 Function Components Component Purpose Example Declaration Specifies function name, argument void func(); (prototype) types, and return value. Alerts compiler (and programmer) that a function is coming up later. Call Causes the function to be executed. func(); Definition The function itself. Contains the void func() lines of code that constitute { the function. // lines of code } Declarator First line of definition. void func() Comparison with Library Functions We’ve already seen some library functions in use. We have embedded calls to library functions, such as ch = getche(); in our program code. Where are the declaration and definition for this library function? The declaration is in the header file specified at the beginning of the program (CONIO.H,for getche()). The definition (compiled into executable code) is in a library file that’s linked auto- matically to your program when you build it. When we use a library function we don’t need to write the declaration or definition. But when we write our own functions, the declaration and definition are part of our source file, as we’ve shown in the TABLE example. (Things get more complicated in multifile programs, as we’ll dis- cuss in Chapter 13, “Multifile Programs.”) Eliminating the Declaration The second approach to inserting a function into a program is to eliminate the function declara- tion and place the function definition (the function itself) in the listing before the first call to the function. For example, we could rewrite TABLE to produce TABLE2, in which the definition for starline() appears first. Chapter 5166 06 3087 CH05 11/29/01 2:22 PM Page 166// table2.cpp // demonstrates function definition preceding function calls #include using namespace std; //no function declaration //-------------------------------------------------------------- // starline() //function definition void starline() { for(int j=0; j<45; j++) cout << ‘*’; cout << endl; } //-------------------------------------------------------------- int main() //main() follows function { starline(); //call to function cout << “Data type Range” << endl; starline(); //call to function cout << “char -128 to 127” << endl << “short -32,768 to 32,767” << endl << “int System dependent” << endl << “long -2,147,483,648 to 2,147,483,647” << endl; starline(); //call to function return 0; } This approach is simpler for short programs, in that it removes the declaration, but it is less flexible. To use this technique when there are more than a few functions, the programmer must give considerable thought to arranging the functions so that each one appears before it is called by any other. Sometimes this is impossible. Also, many programmers prefer to place main() first in the listing, since it is where execution begins. In general we’ll stick with the first approach, using declarations and starting the listing with main(). Passing Arguments to Functions An argument is a piece of data (an int value, for example) passed from a program to the func- tion. Arguments allow a function to operate with different values, or even to do different things, depending on the requirements of the program calling it. Passing Constants As an example, let’s suppose we decide that the starline() function in the last example is too rigid. Instead of a function that always prints 45 asterisks, we want a function that will print any character any number of times. 5 F UNCTIONS 167Functions 06 3087 CH05 11/29/01 2:23 PM Page 167Here’s a program, TABLEARG,that incorporates just such a function. We use arguments to pass the character to be printed and the number of times to print it. // tablearg.cpp // demonstrates function arguments #include using namespace std; void repchar(char, int); //function declaration int main() { repchar(‘-’, 43); //call to function cout << “Data type Range” << endl; repchar(‘=’, 23); //call to function cout << “char -128 to 127” << endl << “short -32,768 to 32,767” << endl << “int System dependent” << endl << “double -2,147,483,648 to 2,147,483,647” << endl; repchar(‘-’, 43); //call to function return 0; } //-------------------------------------------------------------- // repchar() // function definition void repchar(char ch, int n) //function declarator { for(int j=0; j using namespace std; void repchar(char, int); //function declaration 5 F UNCTIONS 169Functions 06 3087 CH05 11/29/01 2:23 PM Page 169int main() { char chin; int nin; cout << “Enter a character: “; cin >> chin; cout << “Enter number of times to repeat it: “; cin >> nin; repchar(chin, nin); return 0; } //-------------------------------------------------------------- // repchar() // function definition void repchar(char ch, int n) //function declarator { for(int j=0; j using namespace std; //////////////////////////////////////////////////////////////// struct Distance //English distance { int feet; float inches; }; //////////////////////////////////////////////////////////////// void engldisp( Distance ); //declaration int main() { Distance d1, d2; //define two lengths //get length d1 from user cout << “Enter feet: “; cin >> d1.feet; cout << “Enter inches: “; cin >> d1.inches; //get length d2 from user cout << “\nEnter feet: “; cin >> d2.feet; cout << “Enter inches: “; cin >> d2.inches; cout << “\nd1 = “; engldisp(d1); //display length 1 cout << “\nd2 = “; engldisp(d2); //display length 2 cout << endl; return 0; } //-------------------------------------------------------------- // engldisp() // display structure of type Distance in feet and inches void engldisp( Distance dd ) //parameter dd of type Distance { cout << dd.feet << “\’-” << dd.inches << “\””; } The main() part of this program accepts two distances in feet-and-inches format from the user, and places these values in two structures, d1 and d2. It then calls a function, engldisp(),that takes a Distance structure variable as an argument. The purpose of the function is to display the distance passed to it in the standard format, such as 10'–2.25''. Here’s some sample interac- tion with the program: Enter feet: 6 Enter inches: 4 Chapter 5172 06 3087 CH05 11/29/01 2:23 PM Page 172Enter feet: 5 Enter inches: 4.25 d1 = 6’-4” d2 = 5’-4.25” The function declaration and the function calls in main(),and the declarator in the function body, treat the structure variables just as they would any other variable used as an argument; this one just happens to be type Distance,rather than a basic type like char or int. In main() there are two calls to the function engldisp(). The first passes the structure d1; the second passes d2. The function engldisp() uses a parameter that is a structure of type Distance, which it names dd. As with simple variables, this structure variable is automatically initialized to the value of the structure passed from main(). Statements in engldisp() can then access the members of dd in the usual way, with the expressions dd.feet and dd.inches. Figure 5.4 shows a structure being passed as an argument to a function. 5 F UNCTIONS 173Functions FIGURE 5.4 Structure passed as an argument. 06 3087 CH05 11/29/01 2:23 PM Page 173As with simple variables, the structure parameter dd in engldisp() is not the same as the argu- ments passed to it (d1 and d2). Thus, engldisp() could (although it doesn’t do so here) modify dd without affecting d1 and d2. That is, if engldisp() contained statements like dd.feet = 2; dd.inches = 3.25; this would have no effect on d1 or d2 in main(). Passing a circle Structure The next example of passing a structure to a function makes use of the Console Graphics Lite functions. The source and header files for these functions are shown in Appendix E, “Console Graphics Lite,” and can be downloaded from the publisher’s Web site as described in the Introduction. You’ll need to include the appropriate header file (MSOFTCON.H or BORLACON.H, depending on your compiler), and add the source file (MSOFTCON.CPP or BORLACON.CPP) to your project. The Console Graphics Lite functions are described in Appendix E, and the procedure for adding files to projects is described in Appendix C, “Microsoft Visual C++,” and Appendix D, “Borland C++Builder.” In this example a structure called circle represents a circular shape. Circles are positioned at a certain place on the console screen, and have a certain radius. They also have a color and a fill pattern. Possible values for the colors and fill patterns can be found in Appendix E. Here’s the listing for CIRCSTRC: // circstrc.cpp // circles as graphics objects #include “msoftcon.h” // for graphics functions //////////////////////////////////////////////////////////////// struct circle //graphics circle { int xCo, yCo; //coordinates of center int radius; color fillcolor; //color fstyle fillstyle; //fill pattern }; //////////////////////////////////////////////////////////////// void circ_draw(circle c) { set_color(c.fillcolor); //set color set_fill_style(c.fillstyle); //set fill pattern draw_circle(c.xCo, c.yCo, c.radius); //draw solid circle } //-------------------------------------------------------------- int main() { init_graphics(); //initialize graphics system //create circles Chapter 5174 06 3087 CH05 11/29/01 2:23 PM Page 174circle c1 = { 15, 7, 5, cBLUE, X_FILL }; circle c2 = { 41, 12, 7, cRED, O_FILL }; circle c3 = { 65, 18, 4, cGREEN, MEDIUM_FILL }; circ_draw(c1); //draw circles circ_draw(c2); circ_draw(c3); set_cursor_pos(1, 25); //cursor to lower left corner return 0; } The variables of type circle, which are c1, c2, and c3,are initialized to different sets of val- ues. Here’s how that looks for c1: circle c1 = { 15, 7, 5, cBLUE, X_FILL }; We assume that your console screen has 80 columns and 25 rows. The first value in this defini- tion, 15, is the column number (the x coordinate) and the 7 is the row number (the y coordi- nate, starting at the top of the screen) where the center of the circle will be located. The 5 is the radius of the circle, the cBLUE is its color, and the X_FILL constant means it will be filled with the letter X. The two other circles are initialized similarly. Once all the circles are created and initialized, we draw them by calling the circ_draw() func- tion three times, once for each circle. Figure 5.5 shows the output of the CIRCSTRC program. Admittedly the circles are a bit ragged; a result of the limited number of pixels in console- mode graphics. 5 F UNCTIONS 175Functions FIGURE 5.5 Output of the CIRCSTRC program. 06 3087 CH05 11/29/01 2:23 PM Page 175Notice how the structure holds the characteristics of the circles, while the circ_draw() func- tion causes them to actually do something (draw themselves). As we’ll see in Chapter 6, “Objects and Classes,” objects are formed by combining structures and functions to create enti- ties that both possess characteristics and perform actions. Names in the Declaration Here’s a way to increase the clarity of your function declarations. The idea is to insert mean- ingful names in the declaration, along with the data types. For example, suppose you were using a function that displayed a point on the screen. You could use a declaration with only data types void display_point(int, int); //declaration but a better approach is void display_point(int horiz, int vert); //declaration These two declarations mean exactly the same thing to the compiler. However, the first approach, with (int, int), doesn’t contain any hint about which argument is for the vertical coordinate and which is for the horizontal coordinate. The advantage of the second approach is clarity for the programmer: Anyone seeing this declaration is more likely to use the correct arguments when calling the function. Note that the names in the declaration have no effect on the names you use when calling the function. You are perfectly free to use any argument names you want: display_point(x, y); // function call We’ll use this name-plus-datatype approach when it seems to make the listing clearer. Returning Values from Functions When a function completes its execution, it can return a single value to the calling program. Usually this return value consists of an answer to the problem the function has solved. The next example demonstrates a function that returns a weight in kilograms after being given a weight in pounds. Here’s the listing for CONVERT: // convert.cpp // demonstrates return values, converts pounds to kg #include using namespace std; float lbstokg(float); //declaration int main() { float lbs, kgs; Chapter 5176 06 3087 CH05 11/29/01 2:23 PM Page 176cout << “\nEnter your weight in pounds: “; cin >> lbs; kgs = lbstokg(lbs); cout << “Your weight in kilograms is “ << kgs << endl; return 0; } //-------------------------------------------------------------- // lbstokg() // converts pounds to kilograms float lbstokg(float pounds) { float kilograms = 0.453592 * pounds; return kilograms; } Here’s some sample interaction with this program: Enter your weight in pounds: 182 Your weight in kilograms is 82.553741 When a function returns a value, the data type of this value must be specified. The function declaration does this by placing the data type, float in this case, before the function name in the declaration and the definition. Functions in earlier program examples returned no value, so the return type was void. In the CONVERT program, the function lbstokg() (pounds to kilo- grams, where lbs means pounds) returns type float,so the declaration is float lbstokg(float); The first float specifies the return type. The float in parentheses specifies that an argument to be passed to lbstokg() is also of type float. When a function returns a value, the call to the function lbstokg(lbs) is considered to be an expression that takes on the value returned by the function. We can treat this expression like any other variable; in this case we use it in an assignment statement: kgs = lbstokg(lbs); This causes the variable kgs to be assigned the value returned by lbstokg(). The return Statement The function lbstokg() is passed an argument representing a weight in pounds, which it stores in the parameter pounds. It calculates the corresponding weight in kilograms by multi- plying this pounds value by a constant; the result is stored in the variable kilograms. The value of this variable is then returned to the calling program using a return statement: return kilograms; 5 F UNCTIONS 177Functions 06 3087 CH05 11/29/01 2:23 PM Page 177Notice that both main() and lbstokg() have a place to store the kilogram variable: kgs in main(), and kilograms in lbstokg(). When the function returns, the value in kilograms is copied into kgs. The calling program does not access the kilograms variable in the function; only the value is returned. This process is shown in Figure 5.6. Chapter 5178 FIGURE 5.6 Returning a value. While many arguments may be sent to a function, only one argument may be returned from it. This is a limitation when you need to return more information. However, there are other approaches to returning multiple variables from functions. One is to pass arguments by refer- ence, which we’ll look at later in this chapter. Another is to return a structure with the multiple values as members, as we’ll see soon. You should always include a function’s return type in the function declaration. If the function doesn’t return anything, use the keyword void to indicate this fact. If you don’t use a return type in the declaration, the compiler will assume that the function returns an int value. For example, the declaration somefunc(); // declaration -- assumes return type is int tells the compiler that somefunc() has a return type of int. 06 3087 CH05 11/29/01 2:23 PM Page 178The reason for this is historical, based on usage in early versions of C. In practice, you shouldn’t take advantage of this default type. Always specify the return type explicitly, even if it actually is int. This keeps the listing consistent and readable. Eliminating Unnecessary Variables The CONVERT program contains several variables that are used in the interest of clarity but are not really necessary. A variation of this program, CONVERT2, shows how expressions containing functions can often be used in place of variables. // convert2.cpp // eliminates unnecessary variables #include using namespace std; float lbstokg(float); //declaration int main() { float lbs; cout << “\nEnter your weight in pounds: “; cin >> lbs; cout << “Your weight in kilograms is “ << lbstokg(lbs) << endl; return 0; } //-------------------------------------------------------------- // lbstokg() // converts pounds to kilograms float lbstokg(float pounds) { return 0.453592 * pounds; } In main() the variable kgs from the CONVERT program has been eliminated. Instead the func- tion lbstokg(lbs) is inserted directly into the cout statement: cout << “Your weight in kilograms is “ << lbstokg(lbs) << endl; Also in the lbstokg() function, the variable kilograms is no longer used. The expression 0.453592*pounds is inserted directly into the return statement: return 0.453592 * pounds; The calculation is carried out and the resulting value is returned to the calling program, just as the value of a variable would be. 5 F UNCTIONS 179Functions 06 3087 CH05 11/29/01 2:23 PM Page 179For clarity, programmers often put parentheses around the expression used in a return state- ment: return (0.453592 * pounds); Even when not required by the compiler, extra parentheses in an expression don’t do any harm, and they may help make the listing easier for us poor humans to read. Experienced C++ (and C) programmers will probably prefer the concise form of CONVERT2to the more verbose CONVERT. However, CONVERT2 is not so easy to understand, especially for the non-expert. The brevity-versus-clarity issue is a question of style, depending on your personal preference and on the expectations of those who will be reading your code. Returning Structure Variables We’ve seen that structures can be used as arguments to functions. You can also use them as return values. Here’s a program, RETSTRC,that incorporates a function that adds variables of type Distance and returns a value of this same type: // retstrc.cpp // demonstrates returning a structure #include using namespace std; //////////////////////////////////////////////////////////////// struct Distance //English distance { int feet; float inches; }; //////////////////////////////////////////////////////////////// Distance addengl(Distance, Distance); //declarations void engldisp(Distance); int main() { Distance d1, d2, d3; //define three lengths //get length d1 from user cout << “\nEnter feet: “; cin >> d1.feet; cout << “Enter inches: “; cin >> d1.inches; //get length d2 from user cout << “\nEnter feet: “; cin >> d2.feet; cout << “Enter inches: “; cin >> d2.inches; d3 = addengl(d1, d2); //d3 is sum of d1 and d2 cout << endl; engldisp(d1); cout << “ + “; //display all lengths Chapter 5180 06 3087 CH05 11/29/01 2:23 PM Page 180engldisp(d2); cout << “ = “; engldisp(d3); cout << endl; return 0; } //-------------------------------------------------------------- // addengl() // adds two structures of type Distance, returns sum Distance addengl( Distance dd1, Distance dd2 ) { Distance dd3; //define a new structure for sum dd3.inches = dd1.inches + dd2.inches; //add the inches dd3.feet = 0; //(for possible carry) if(dd3.inches >= 12.0) //if inches >= 12.0, { //then decrease inches dd3.inches -= 12.0; //by 12.0 and dd3.feet++; //increase feet } //by 1 dd3.feet += dd1.feet + dd2.feet; //add the feet return dd3; //return structure } //-------------------------------------------------------------- // engldisp() // display structure of type Distance in feet and inches void engldisp( Distance dd ) { cout << dd.feet << “\’-” << dd.inches << “\””; } The program asks the user for two lengths, in feet-and-inches format, adds them together by calling the function addengl(), and displays the results using the engldisp() function intro- duced in the ENGLDISP program. Here’s some output from the program: Enter feet: 4 Enter inches: 5.5 Enter feet: 5 Enter inches: 6.5 4’-5.5” + 5’-6.5” = 10’-0” The main() part of the program adds the two lengths, each represented by a structure of type Distance, by calling the function addengl(): d3 = addengl(d1, d2); This function returns the sum of d1 and d2,in the form of a structure of type Distance. In main() the result is assigned to the structure d3. 5 F UNCTIONS 181Functions 06 3087 CH05 11/29/01 2:23 PM Page 181Besides showing how structures are used as return values, this program also shows two func- tions (three if you count main()) used in the same program. You can arrange the functions in any order. The only rule is that the function declarations must appear in the listing before any calls are made to the functions. Reference Arguments A reference provides an alias—a different name—for a variable. One of the most important uses for references is in passing arguments to functions. We’ve seen examples of function arguments passed by value. When arguments are passed by value, the called function creates a new variable of the same type as the argument and copies the argument’s value into it. As we noted, the function cannot access the original variable in the calling program, only the copy it created. Passing arguments by value is useful when the function does not need to modify the original variable in the calling program. In fact, it offers insurance that the function cannot harm the original variable. Passing arguments by reference uses a different mechanism. Instead of a value being passed to the function, a reference to the original variable, in the calling program, is passed. (It’s actually the memory address of the variable that is passed, although you don’t need to know this.) An important advantage of passing by reference is that the function can access the actual vari- ables in the calling program. Among other benefits, this provides a mechanism for passing more than one value from the function back to the calling program. Passing Simple Data Types by Reference The next example, REF,shows a simple variable passed by reference. // ref.cpp // demonstrates passing by reference #include using namespace std; int main() { void intfrac(float, float&, float&); //declaration float number, intpart, fracpart; //float variables do { cout << “\nEnter a real number: “; //number from user cin >> number; intfrac(number, intpart, fracpart); //find int and frac cout << “Integer part is “ << intpart //print them << “, fraction part is “ << fracpart << endl; Chapter 5182 06 3087 CH05 11/29/01 2:23 PM Page 182} while( number != 0.0 ); //exit loop on 0.0 return 0; } //-------------------------------------------------------------- // intfrac() // finds integer and fractional parts of real number void intfrac(float n, float& intp, float& fracp) { long temp = static_cast(n); //convert to long, intp = static_cast(temp); //back to float fracp = n - intp; //subtract integer part } The main() part of this program asks the user to enter a number of type float. The program will separate this number into an integer and a fractional part. That is, if the user’s number is 12.456, the program should report that the integer part is 12.0 and the fractional part is 0.456. To find these two values, main() calls the function intfrac(). Here’s some sample interac- tion: Enter a real number: 99.44 Integer part is 99, fractional part is 0.44 Some compilers may generate spurious digits in the fractional part, such as 0.440002. This is an error in the compiler’s conversion routine and can be ignored. Refer to Figure 5.7 in the fol- lowing discussion. The intfrac() function finds the integer part by converting the number (which was passed to the parameter n) into a variable of type long with a cast, using the expression long temp = static_cast(n); This effectively chops off the fractional part of the number, since integer types (of course) store only the integer part. The result is then converted back to type float with another cast: intp = static_cast(temp); The fractional part is simply the original number less the integer part. (We should note that a library function, fmod(), performs a similar task for type double.) The intfrac() function can find the integer and fractional parts, but how does it pass them back to main()? It could use a return statement to return one value, but not both. The problem is solved using reference arguments. Here’s the declarator for the function: void intfrac(float n, float& intp, float& fracp) 5 F UNCTIONS 183Functions 06 3087 CH05 11/29/01 2:23 PM Page 183FIGURE 5.7 Passing by reference in the REF program. Reference arguments are indicated by the ampersand (&) following the data type: float& intp The & indicates that intp is an alias—another name—for whatever variable is passed as an argument. In other words, when you use the name intp in the intfrac() function, you are really referring to intpart in main(). The & can be taken to mean reference to,so float& intp means intp is a reference to the float variable passed to it. Similarly, fracp is an alias for— or a reference to—fracpart. The function declaration echoes the usage of the ampersand in the definition: void intfrac(float, float&, float&); // ampersands As in the definition, the ampersand follows those arguments that are passed by reference. Chapter 5184 intfrac(number,intpart,fracpart); main() number intpart fracpart intfrac() n intp fracp fracp = n-intp; long temp = static_cast(n); This statement in main() causes this variable to be copied into this parameter. It also sets up aliases for these variables with these names. These statements in intfrac() operate on these variables as if they were in intfrac(). 06 3087 CH05 11/29/01 2:23 PM Page 184The ampersand is not used in the function call: intfrac(number, intpart, fracpart); // no ampersands From the function call alone, there’s no way to tell whether an argument will be passed by ref- erence or by value. While intpart and fracpart are passed by reference, the variable number is passed by value. intp and intpart are different names for the same place in memory, as are fracp and fracpart. On the other hand, since it is passed by value, the parameter n in intfrac() is a separate variable into which the value of number is copied. It can be passed by value because the intfrac() function doesn’t need to modify number. (C programmers should not confuse the ampersand that is used to mean reference to with the same symbol used to mean address of. These are different usages. We’ll discuss the address of meaning of & in Chapter 10, “Pointers.”) A More Complex Pass by Reference Here’s a somewhat more complex example of passing simple arguments by reference. Suppose you have pairs of numbers in your program and you want to be sure that the smaller one always precedes the larger one. To do this you call a function, order(), which checks two numbers passed to it by reference and swaps the originals if the first is larger than the second. Here’s the listing for REFORDER: // reforder.cpp // orders two arguments passed by reference #include using namespace std; int main() { void order(int&, int&); //prototype int n1=99, n2=11; //this pair not ordered int n3=22, n4=88; //this pair ordered order(n1, n2); //order each pair of numbers order(n3, n4); cout << “n1=” << n1 << endl; //print out all numbers cout << “n2=” << n2 << endl; cout << “n3=” << n3 << endl; cout << “n4=” << n4 << endl; return 0; } 5 F UNCTIONS 185Functions 06 3087 CH05 11/29/01 2:23 PM Page 185//-------------------------------------------------------------- void order(int& numb1, int& numb2) //orders two numbers { if(numb1 > numb2) //if 1st larger than 2nd, { int temp = numb1; //swap them numb1 = numb2; numb2 = temp; } } In main() there are two pairs of numbers—the first pair is not ordered and the second pair is ordered. The order() function is called once for each pair, and then all the numbers are printed out. The output reveals that the first pair has been swapped while the second pair hasn’t. Here it is: n1=11 n2=99 n3=22 n4=88 In the order() function the first variable is called numb1 and the second is numb2. If numb1 is greater than numb2 the function stores numb1 in temp, puts numb2 in numb1, and finally puts temp back in numb2. Remember that numb1 and numb2 are simply different names for whatever arguments were passed; in this case, n1 and n2 on the first call to the function, and n2 and n3 on the second call. The effect is to check the ordering of the original arguments in the calling program and swap them if necessary. Using reference arguments in this way is a sort of remote-control operation. The calling pro- gram tells the function what variables in the calling program to operate on, and the function modifies these variables without ever knowing their real names. It’s as if you called the house painters and, although they never left their office, you sat back and watched as your dining room walls mysteriously changed color. Passing Structures by Reference You can pass structures by reference just as you can simple data types. Here’s a program, REFERST,that performs scale conversions on values of type Distance. A scale conversion involves multiplying a group of distances by a factor. If a distance is 6'–8'', and a scale factor is 0.5, the new distance is 3'–4''. Such a conversion might be applied to all the dimensions of a building to make the building shrink but remain in proportion. // referst.cpp // demonstrates passing structure by reference #include Chapter 5186 06 3087 CH05 11/29/01 2:23 PM Page 186using namespace std; //////////////////////////////////////////////////////////////// struct Distance //English distance { int feet; float inches; }; //////////////////////////////////////////////////////////////// void scale( Distance&, float ); //function void engldisp( Distance ); //declarations int main() { Distance d1 = { 12, 6.5 }; //initialize d1 and d2 Distance d2 = { 10, 5.5 }; cout << “d1 = “; engldisp(d1); //display old d1 and d2 cout << “\nd2 = “; engldisp(d2); scale(d1, 0.5); //scale d1 and d2 scale(d2, 0.25); cout << “\nd1 = “; engldisp(d1); //display new d1 and d2 cout << “\nd2 = “; engldisp(d2); cout << endl; return 0; } //-------------------------------------------------------------- // scale() // scales value of type Distance by factor void scale( Distance& dd, float factor) { float inches = (dd.feet*12 + dd.inches) * factor; dd.feet = static_cast(inches / 12); dd.inches = inches - dd.feet * 12; } //-------------------------------------------------------------- // engldisp() // display structure of type Distance in feet and inches void engldisp( Distance dd ) //parameter dd of type Distance { cout << dd.feet << “\’-” << dd.inches << “\””; } 5 F UNCTIONS 187Functions 06 3087 CH05 11/29/01 2:23 PM Page 187REFERST initializes two Distance variables—d1 and d2—to specific values, and displays them. Then it calls the scale() function to multiply d1 by 0.5 and d2 by 0.25. Finally, it displays the resulting values of the distances. Here’s the program’s output: d1 = 12’-6.5” d2 = 10’-5.5” d1 = 6’-3.25” d2 = 2’-7.375” Here are the two calls to the function scale(): scale(d1, 0.5); scale(d2, 0.25); The first call causes d1 to be multiplied by 0.5 and the second causes d2 to be multiplied by 0.25. Notice that these changes take place directly to d1 and d2. The function doesn’t return anything; the operation is performed directly on the Distance argument, which is passed by reference to scale(). (Since only one value is changed in the calling program, you could rewrite the function to pass the argument by value and return the scaled value. Calling such a function would look like this: d1 = scale(d1, 0.5); However, this is unnecessarily verbose.) Notes on Passing by Reference References don’t exist in C, where pointers serve a somewhat similar purpose, although often less conveniently. Reference arguments were introduced into C++ to provide flexibility in a variety of situations involving objects as well as simple variables. The third way to pass arguments to functions, besides by value and by reference, is to use pointers. We’ll explore this in Chapter 10. Overloaded Functions An overloaded function appears to perform different activities depending on the kind of data sent to it. Overloading is like the joke about the famous scientist who insisted that the thermos bottle was the greatest invention of all time. Why? “It’s a miracle device,” he said. “It keeps hot things hot, but cold things it keeps cold. How does it know?” It may seem equally mysterious how an overloaded function knows what to do. It performs one operation on one kind of data but another operation on a different kind. Let’s clarify matters with some examples. Chapter 5188 06 3087 CH05 11/29/01 2:23 PM Page 188Different Numbers of Arguments Recall the starline() function in the TABLE example and the repchar() function from the TABLEARG example, both shown earlier in this chapter. The starline() function printed a line using 45 asterisks, while repchar() used a character and a line length that were both specified when the function was called. We might imagine a third function, charline(),that always prints 45 characters but that allows the calling program to specify the character to be printed. These three functions—starline(), repchar(), and charline()—perform similar activities but have different names. For programmers using these functions, that means three names to remember and three places to look them up if they are listed alphabetically in an application’s Function Reference documentation. It would be far more convenient to use the same name for all three functions, even though they each have different arguments. Here’s a program, OVERLOAD,that makes this possible: // overload.cpp // demonstrates function overloading #include using namespace std; void repchar(); //declarations void repchar(char); void repchar(char, int); int main() { repchar(); repchar(‘=’); repchar(‘+’, 30); return 0; } //-------------------------------------------------------------- // repchar() // displays 45 asterisks void repchar() { for(int j=0; j<45; j++) // always loops 45 times cout << ‘*’; // always prints asterisk cout << endl; } //-------------------------------------------------------------- // repchar() // displays 45 copies of specified character void repchar(char ch) { for(int j=0; j<45; j++) // always loops 45 times cout << ch; // prints specified character 5 F UNCTIONS 189Functions 06 3087 CH05 11/29/01 2:23 PM Page 189cout << endl; } //-------------------------------------------------------------- // repchar() // displays specified number of copies of specified character void repchar(char ch, int n) { for(int j=0; j using namespace std; //////////////////////////////////////////////////////////////// struct Distance //English distance { int feet; float inches; }; //////////////////////////////////////////////////////////////// 5 F UNCTIONS 191Functions 06 3087 CH05 11/29/01 2:23 PM Page 191void engldisp( Distance ); //declarations void engldisp( float ); int main() { Distance d1; //distance of type Distance float d2; //distance of type float //get length d1 from user cout << “\nEnter feet: “; cin >> d1.feet; cout << “Enter inches: “; cin >> d1.inches; //get length d2 from user cout << “Enter entire distance in inches: “; cin >> d2; cout << “\nd1 = “; engldisp(d1); //display length 1 cout << “\nd2 = “; engldisp(d2); //display length 2 cout << endl; return 0; } //-------------------------------------------------------------- // engldisp() // display structure of type Distance in feet and inches void engldisp( Distance dd ) //parameter dd of type Distance { cout << dd.feet << “\’-” << dd.inches << “\””; } //-------------------------------------------------------------- // engldisp() // display variable of type float in feet and inches void engldisp( float dd ) //parameter dd of type float { int feet = static_cast(dd / 12); float inches = dd - feet*12; cout << feet << “\’-” << inches << “\””; } The user is invited to enter two distances, the first with separate feet and inches inputs, the sec- ond with a single large number for inches (109.5 inches, for example, instead of 9'–1.5''). The program calls the overloaded function engldisp() to display a value of type Distance for the first distance and of type float for the second. Here’s some sample interaction with the pro- gram: Enter feet: 5 Enter inches: 10.5 Enter entire distance in inches: 76.5 d1 = 5’-10.5” d2 = 6’-4.5” Chapter 5192 06 3087 CH05 11/29/01 2:23 PM Page 192Notice that, while the different versions of engldisp() do similar things, the code is quite dif- ferent. The version that accepts the all-inches input has to convert to feet and inches before dis- playing the result. Overloaded functions can simplify the programmer’s life by reducing the number of function names to be remembered. As an example of the complexity that arises when overloading is not used, consider the C++ library routines for finding the absolute value of a number. Because these routines must work with C (which does not allow overloading) as well as with C++, there must be separate versions of the absolute value routine for each data type. There are four of them: abs() for type int, cabs() for complex numbers, fabs() for type double, and labs() for type long. In C++, a single name, abs(),would suffice for all these data types. As we’ll see later, overloaded functions are also useful for handling different types of objects. Recursion The existence of functions makes possible a programming technique called recursion. Recursion involves a function calling itself. This sounds rather improbable, and indeed a func- tion calling itself is often a bug. However, when used correctly this technique can be surpris- ingly powerful. Recursion is much easier to understand with an example than with lengthy explanations, so let’s apply it to a program we’ve seen before: the FACTOR program of Chapter 3, “Loops and Decisions.” That program used a for loop to calculate the factorial of a number. (See that example for an explanation of factorials.) Our new program, FACTOR2, uses recursion instead of a loop. //factor2.cpp //calculates factorials using recursion #include using namespace std; unsigned long factfunc(unsigned long); //declaration int main() { int n; //number entered by user unsigned long fact; //factorial cout << “Enter an integer: “; cin >> n; fact = factfunc(n); cout << “Factorial of “ << n << “ is “ << fact << endl; return 0; } 5 F UNCTIONS 193Functions 06 3087 CH05 11/29/01 2:23 PM Page 193//------------------------------------------------------------- // factfunc() // calls itself to calculate factorials unsigned long factfunc(unsigned long n) { if(n > 1) return n * factfunc(n-1); //self call else return 1; } The output of this program is the same as the FACTOR program in Chapter 3. The main() part of FACTOR2looks reasonable: it calls a function, factfunc(),with an argu- ment that is a number entered by the user. This function then returns the factorial of that num- ber to main(). The function factfunc() is another story. What’s it doing? If n is greater than 1, the function calls itself. Notice that when it does this it uses an argument one less than the argument it was called with. Suppose it was called from main() with an argument of 5. It will call a second version of itself with an argument of 4. Then this function will call a third version with an argument of 3, and so on. Notice that each version of the function stores its own value of n while it’s busy calling another version of itself. After factfunc() calls itself four times, the fifth version of the function is called with an argu- ment of 1. It discovers this with the if statement, and instead of calling itself, as previous ver- sions have, it returns 1 to the fourth version. The fourth version has stored a value of 2, so it multiplies the stored 2 by the returned 1, and returns 2 to the third version. The third version has stored 3, so it multiplies 3 by the returned 2, and returns 6 to the second version. The second version has stored 4, so it multiplies this by the returned 6 and returns 24 to the first version. The first version has stored 5, so it multiplies this by the returned 24 and returns 120 to main(). Thus in this example we have five function calls followed by five function returns. Here’s a summary of this process: Argument or Version Action Return Value 1 Call 5 2 Call 4 3 Call 3 4 Call 2 Chapter 5194 06 3087 CH05 11/29/01 2:23 PM Page 1945 Call 1 5 Return 1 4 Return 2 3 Return 6 2 Return 24 1 Return 120 Every recursive function must be provided with a way to end the recursion. Otherwise it will call itself forever and crash the program. The if statement in factfunc() plays this role, ter- minating the recursion when n is 1. Is it true that many versions of a recursive function are stored in memory while it’s calling itself? Not really. Each version’s variables are stored, but there’s only one copy of the func- tion’s code. Even so, a deeply-nested recursion can create a great many stored variables, which can pose a problem to the system if it doesn’t have enough space for them. Inline Functions We mentioned that functions save memory space because all the calls to the function cause the same code to be executed; the function body need not be duplicated in memory. When the compiler sees a function call, it normally generates a jump to the function. At the end of the function it jumps back to the instruction following the call, as shown in Figure 5.1 earlier in this chapter. While this sequence of events may save memory space, it takes some extra time. There must be an instruction for the jump to the function (actually the assembly-language instruction CALL or something similar), instructions for saving registers, instructions for pushing argu- ments onto the stack in the calling program and removing them from the stack in the function (if there are arguments), instructions for restoring registers, and an instruction to return to the calling program. The return value (if any) must also be dealt with. All these instructions slow down the program. To save execution time in short functions, you may elect to put the code in the function body directly inline with the code in the calling program. That is, each time there’s a function call in the source file, the actual code from the function is inserted, instead of a jump to the function. The difference between a function and inline code is shown in Figure 5.9. 5 F UNCTIONS 195Functions Argument or Version Action Return Value 06 3087 CH05 11/29/01 2:23 PM Page 195FIGURE 5.9 Functions versus inline code. Long sections of repeated code are generally better off as normal functions: The savings in memory space is worth the comparatively small sacrifice in execution speed. But making a short section of code into an ordinary function may result in little savings in memory space, while imposing just as much time penalty as a larger function. In fact, if a function is very short, the instructions necessary to call it may take up as much space as the instructions within the function body, so that there is not only a time penalty but a space penalty as well. In such cases you could simply repeat the necessary code in your program, inserting the same group of statements wherever it was needed. The trouble with repeatedly inserting the same code is that you lose the benefits of program organization and clarity that come with using functions. The program may run faster and take less space, but the listing is longer and more complex. The solution to this quandary is the inline function. This kind of function is written like a nor- mal function in the source file but compiles into inline code instead of into a function. The source file remains well organized and easy to read, since the function is shown as a separate entity. However, when the program is compiled, the function body is actually inserted into the program wherever a function call occurs. Chapter 5196 06 3087 CH05 11/29/01 2:23 PM Page 196Functions that are very short, say one or two statements, are candidates to be inlined. Here’s INLINE,a variation on the CONVERT2program. It inlines the lbstokg() function. // inliner.cpp // demonstrates inline functions #include using namespace std; // lbstokg() // converts pounds to kilograms inline float lbstokg(float pounds) { return 0.453592 * pounds; } //-------------------------------------------------------------- int main() { float lbs; cout << “\nEnter your weight in pounds: “; cin >> lbs; cout << “Your weight in kilograms is “ << lbstokg(lbs) << endl; return 0; } It’s easy to make a function inline: All you need is the keyword inline in the function defini- tion: inline float lbstokg(float pounds) You should be aware that the inline keyword is actually just a request to the compiler. Sometimes the compiler will ignore the request and compile the function as a normal function. It might decide the function is too long to be inline, for instance. (C programmers should note that inline functions largely take the place of #define macros in C. They serve the same purpose but provide better type checking and do not need special care with parentheses, as macros do.) Default Arguments Surprisingly, a function can be called without specifying all its arguments. This won’t work on just any function: The function declaration must provide default values for those arguments that are not specified. 5 F UNCTIONS 197Functions 06 3087 CH05 11/29/01 2:23 PM Page 197Here’s an example, a variation on the OVERLOAD program that demonstrates this effect. In OVERLOAD we used three different functions with the same name to handle different numbers of arguments. The present example, MISSARG,achieves the same effect in a different way. // missarg.cpp // demonstrates missing and default arguments #include using namespace std; void repchar(char=’*’, int=45); //declaration with //default arguments int main() { repchar(); //prints 45 asterisks repchar(‘=’); //prints 45 equal signs repchar(‘+’, 30); //prints 30 plus signs return 0; } //-------------------------------------------------------------- // repchar() // displays line of characters void repchar(char ch, int n) //defaults supplied { // if necessary for(int j=0; j using namespace std; #include //for getch() char ch = ‘a’; //global variable ch void getachar(); //function declarations void putachar(); int main() { while( ch != ‘\r’ ) //main() accesses ch { getachar(); putachar(); } cout << endl; return 0; } //-------------------------------------------------------------- void getachar() //getachar() accesses ch { ch = getch(); } //-------------------------------------------------------------- void putachar() //putachar() accesses ch { cout << ch; } Chapter 5202 06 3087 CH05 11/29/01 2:23 PM Page 202One function in EXTERN, getachar(),reads characters from the keyboard. It uses the library function getch(), which is like getche() except that it doesn’t echo the character typed to the screen (hence the absence of the final e in the name). A second EXTERN function, putachar(), displays each character on the screen. The effect is that what you type is displayed in the nor- mal way: I’m typing in this line of text The significant thing about this program is that the variable ch is not defined in any of the functions. Instead it is defined at the beginning of the file, before the first function. It is a global (external) variable. Any function that follows the definition of ch in the listing can access it—in this case all the functions in EXTERN: main(), getachar(),and putachar(). Thus the visibility of ch is the entire source file. Role of Global Variables A global variable is used when it must be accessible to more than one function in a program. Global variables are often the most important variables in procedural programs. However, as we noted in Chapter 1, global variables create organizational problems because they can be accessed by any function. The wrong functions may access them, or functions may access them incorrectly. In an object-oriented program, there is much less necessity for global variables. Initialization If a global variable is initialized, as in int exvar = 199; this initialization takes place when the program is first loaded. If a global variable is not initial- ized explicitly by the program—for example, if it is defined as int exvar; then it is initialized automatically to 0 when it is created. (This is unlike local variables, which are not initialized and probably contain random or garbage values when they are created.) Lifetime and Visibility Global variables have storage class static, which means they exist for the life of the program. Memory space is set aside for them when the program begins, and continues to exist until the program ends. You don’t need to use the keyword static when declaring global variables; they are given this storage class automatically. As we noted, global variables are visible in the file in which they are defined, starting at the point where they are defined. If ch were defined following main() but before getachar(),it would be visible in getachar() and putachar(),but not in main(). 5 F UNCTIONS 203Functions 06 3087 CH05 11/29/01 2:23 PM Page 203Static Local Variables Let’s look at another kind of variable: the static local variable. There are static global vari- ables, but they are meaningful only in multifile programs, which we don’t examine until Chapter 13. A static local variable has the visibility of an automatic local variable (that is, inside the func- tion containing it). However, its lifetime is the same as that of a global variable, except that it doesn’t come into existence until the first call to the function containing it. Thereafter it remains in existence for the life of the program. Static local variables are used when it’s necessary for a function to remember a value when it is not being executed; that is, between calls to the function. In the next example, a function, getavg(), calculates a running average. It remembers the total of the numbers it has averaged before, and how many there were. Each time it receives a new number, sent as an argument from the calling program, it adds this number to the total, adds 1 to a count, and returns the new average by dividing the total by the count. Here’s the listing for STATIC: // static.cpp // demonstrates static variables #include using namespace std; float getavg(float); //declaration int main() { float data=1, avg; while( data != 0 ) { cout << “Enter a number: “; cin >> data; avg = getavg(data); cout << “New average is “ << avg << endl; } return 0; } //-------------------------------------------------------------- // getavg() // finds average of old plus new data float getavg(float newdata) { static float total = 0; //static variables are initialized static int count = 0; // only once per program Chapter 5204 06 3087 CH05 11/29/01 2:23 PM Page 204count++; //increment count total += newdata; //add new data to total return total / count; //return the new average } Here’s some sample interaction: Enter a number: 10 New average is 10 ← total is 10, count is 1 Enter a number: 20 New average is 15 ← total is 30, count is 2 Enter a number: 30 New average is 20 ← total is 60, count is 3 The static variables total and count in getavg() retain their values after getavg() returns, so they’re available the next time it’s called. Initialization When static variables are initialized, as total and count are in getavg(),the initialization takes place only once—the first time their function is called. They are not reinitialized on sub- sequent calls to the function, as ordinary local variables are. Storage If you’re familiar with operating system architecture, you might be interested to know that local variables and function arguments are stored on the stack, while global and static variables are stored on the heap. Table 5.2 summarizes the lifetime, visibility, and some other aspects of local, static local, and global variables. TABLE 5.2 Storage Types Local Static Local Global Visibility function function file Lifetime function program program Initialized value not initialized 0 0 Storage stack heap heap Purpose Variables used by Same as local, but Variables a single function retains value used by when function several terminates functions 5 F UNCTIONS 205Functions 06 3087 CH05 11/29/01 2:23 PM Page 205Returning by Reference Now that we know about global variables, we can examine a rather odd-looking C++ feature. Besides passing values by reference, you can also return a value by reference. Why you would want to do this may seem obscure. One reason is to avoid copying a large object, as we’ll see in Chapter 11, “Virtual Functions.” Another reason is to allow you to use a function call on the left side of the equal sign. This is a somewhat bizarre concept, so let’s look at an example. The RETREF program shows the mechanism. // retref.cpp // returning reference values #include using namespace std; int x; // global variable int& setx(); // function declaration int main() { // set x to a value, using setx() = 92; // function call on left side cout << “x=” << x << endl; // display new value in x return 0; } //-------------------------------------------------------------- int& setx() { return x; // returns the value to be modified } In this program the function setx() is declared with a reference type, int&,as the return type: int& setx(); This function contains the statement return x; where x has been defined as a global variable. Now—and this is what looks so strange—you can put a call to this function on the left side of the equal sign: setx() = 92; The result is that the variable returned by the function is assigned the value on the right side of the equal sign. That is, x is given the value 92. The output from the program x=92 verifies that this assignment has taken place. Chapter 5206 06 3087 CH05 11/29/01 2:23 PM Page 206Function Calls on the Left of the Equal Sign Does this still sound obscure? Remember that an ordinary function—one that returns a value— can be used as if it were a value: y=squareroot(x); Here, whatever value squareroot(x) has (for instance, 27.2) is assigned to y. The function is treated as if it were a value. A function that returns a reference, on the other hand, is treated as if it were a variable. It returns an alias to a variable, namely the variable in the function’s return statement. In RETREF.C the function setx() returns a reference to the variable x. When this function is called, it’s treated as if it were the variable x. Thus it can be used on the left side of an equal sign. There are two corollaries to this. One is that you can’t return a constant from a function that returns by reference. In setx(), you can’t say int& setx() { return 3; } If you try this the compiler will complain that you need an lvalue,that is, something that can go on the left side of the equal sign: a variable and not a constant. More subtly, you can’t return a reference to a local variable: int& setx() { int x = 3; return x; // error } What’s wrong with this? The problem is that a function’s local variables are probably destroyed when the function returns, and it doesn’t make sense to return a reference to some- thing that no longer exists. Don’t Worry Yet Of course, the question remains why one would ever want to use a function call on the left of an equal sign. In procedural programming there probably isn’t too much use for this technique. As in the above example, there are easier ways to achieve the same result. However, in Chapter 8, “Operator Overloading,” we’ll find that returning by reference is an indispensable technique. Until then, keep it in the back of your mind. 5 F UNCTIONS 207Functions 06 3087 CH05 11/29/01 2:23 PM Page 207const Function Arguments We’ve seen that passing an argument by reference can be used to allow a function to modify a variable in the calling program. However, there are other reasons to pass by reference. One is efficiency. Some variables used for function arguments can be very large; a large structure would be an example. If an argument is large, passing by reference is more efficient because, behind the scenes, only an address is really passed, not the entire variable. Suppose you want to pass an argument by reference for efficiency, but not only do you want the function not to modify it, you want a guarantee that the function cannot modify it. To obtain such a guarantee, you can apply the const modifier to the variable in the function declaration. The CONSTARG program shows how this looks. //constarg.cpp //demonstrates constant function arguments void aFunc(int& a, const int& b); //declaration int main() { int alpha = 7; int beta = 11; aFunc(alpha, beta); return 0; } //-------------------------------------------------------------- void aFunc(int& a, const int& b) //definition { a = 107; //OK b = 111; //error: can’t modify constant argument } Here we want to be sure that aFunc() can’t modify the variable beta. (We don’t care if it mod- ifies alpha.) So we use the const modifier with beta in the function declaration (and defini- tion): void aFunc(int& alpha, const int& beta); Now the attempt to modify the beta in aFunc() is flagged as an error by the compiler. One of the design philosophies in C++ is that it’s better for the compiler to find errors than to wait for them to surface at runtime. The use of const function arguments is an example of this approach in action. If you want to pass a const variable to a function as a reference argument, you don’t have a choice: It must be declared const in the function declaration. (There’s no problem passing a const argument by value, because the function can’t modify the original variable anyway.) Chapter 5208 06 3087 CH05 11/29/01 2:23 PM Page 208Many library functions use constant arguments in a similar way. We’ll see examples as we go along. Summary Functions provide a way to help organize programs, and to reduce program size, by giving a block of code a name and allowing it to be executed from other parts of the program. Function declarations (prototypes) specify what the function looks like, function calls transfer control to the function, and function definitions contain the statements that make up the function. The function declarator is the first line of the definition. Arguments can be sent to functions either by value, where the function works with a copy of the argument, or by reference, where the function works with the original argument in the call- ing program. Functions can return only one value. Functions ordinarily return by value, but they can also return by reference, which allows the function call to be used on the left side of an assignment statement. Arguments and return values can be either simple data types or structures. An overloaded function is actually a group of functions with the same name. Which of them is executed when the function is called depends on the type and number of arguments supplied in the call. An inline function looks like a normal function in the source file but inserts the function’s code directly into the calling program. Inline functions execute faster but may require more memory than normal functions unless they are very small. If a function uses default arguments, calls to it need not include all the arguments shown in the declaration. Default values supplied by the function are used for the missing arguments. Variables possess a characteristic called the storage class. The most common storage class is automatic. Local variables have the automatic storage class: they exist only while the function in which they are defined is executing. They are also visible only within that function. Global variables have static storage class: they exist for the life of a program. They are also visible throughout an entire file. Static local variables exist for the life of a program but are visible only in their own function. A function cannot modify any of its arguments that are given the const modifier. A variable already defined as const in the calling program must be passed as a const argument. In Chapter 4 we examined one of the two major parts of objects: structures, which are collec- tions of data. In this chapter we explored the second part: functions. Now we’re ready to put these two components together to create objects, the subject of Chapter 6. 5 F UNCTIONS 209Functions 06 3087 CH05 11/29/01 2:23 PM Page 209Questions Answers to these questions can be found in Appendix G. 1. A function’s single most important role is to a. give a name to a block of code. b. reduce program size. c. accept arguments and provide a return value. d. help organize a program into conceptual units. 2. A function itself is called the function d_________. 3. Write a function called foo() that displays the word foo. 4. A one-statement description of a function is referred to as a function d_________ or a p_________. 5. The statements that carry out the work of the function constitute the function _________. 6. A program statement that invokes a function is a function _________. 7. The first line of a function definition is referred to as the _________. 8. A function argument is a. a variable in the function that receives a value from the calling program. b. a way that functions resist accepting the calling program’s values. c. a value sent to the function by the calling program. d. a value returned by the function to the calling program. 9. True or false: When arguments are passed by value, the function works with the original arguments in the calling program. 10. What is the purpose of using argument names in a function declaration? 11. Which of the following can legitimately be passed to a function? a. A constant b. A variable c. A structure d. A header file 12. What is the significance of empty parentheses in a function declaration? 13. How many values can be returned from a function? 14. True or false: When a function returns a value, the entire function call can appear on the right side of the equal sign and be assigned to another variable. 15. Where is a function’s return type specified? Chapter 5210 06 3087 CH05 11/29/01 2:23 PM Page 21016. A function that doesn’t return anything has return type _________. 17. Here’s a function: int times2(int a) { return (a*2); } Write a main() program that includes everything necessary to call this function. 18. When an argument is passed by reference a. a variable is created in the function to hold the argument’s value. b. the function cannot access the argument’s value. c. a temporary variable is created in the calling program to hold the argument’s value. d. the function accesses the argument’s original value in the calling program. 19. What is a principal reason for passing arguments by reference? 20. Overloaded functions a. are a group of functions with the same name. b. all have the same number and types of arguments. c. make life simpler for programmers. d. may fail unexpectedly due to stress. 21. Write declarations for two overloaded functions named bar(). They both return type int. The first takes one argument of type char, and the second takes two arguments of type char. If this is impossible, say why. 22. In general, an inline function executes _________ than a normal function, but requires _________ memory. 23. Write the declarator for an inline function named foobar() that takes one argument of type float and returns type float. 24. A default argument has a value that a. may be supplied by the calling program. b. may be supplied by the function. c. must have a constant value. d. must have a variable value. 25. Write a declaration for a function called blyth() that takes two arguments and returns type char. The first argument is type int, and the second is type float with a default value of 3.14159. 26. Scope and storage class are concerned with the _________ and _________ of a variable. 5 F UNCTIONS 211Functions 06 3087 CH05 11/29/01 2:23 PM Page 21127. What functions can access a global variable that appears in the same file with them? 28. What functions can access a local variable? 29. A static local variable is used to a. make a variable visible to several functions. b. make a variable visible to only one function. c. conserve memory when a function is not executing. d. retain a value when a function is not executing. 30. In what unusual place can you use a function call when a function returns a value by ref- erence? Exercises Answers to the starred exercises can be found in Appendix G. *1. Refer to the CIRCAREA program in Chapter 2, “C++ Programming Basics.” Write a func- tion called circarea() that finds the area of a circle in a similar way. It should take an argument of type float and return an argument of the same type. Write a main() func- tion that gets a radius value from the user, calls circarea(), and displays the result. *2. Raising a number n to a power p is the same as multiplying n by itself p times. Write a function called power() that takes a double value for n and an int value for p, and returns the result as a double value. Use a default argument of 2 for p,so that if this argument is omitted, the number n will be squared. Write a main() function that gets val- ues from the user to test this function. *3. Write a function called zeroSmaller() that is passed two int arguments by reference and then sets the smaller of the two numbers to 0. Write a main() program to exercise this function. *4. Write a function that takes two Distance values as arguments and returns the larger one. Include a main() program that accepts two Distance values from the user, compares them, and displays the larger. (See the RETSTRC program for hints.) 5. Write a function called hms_to_secs() that takes three int values—for hours, minutes, and seconds—as arguments, and returns the equivalent time in seconds (type long). Create a program that exercises this function by repeatedly obtaining a time value in hours, minutes, and seconds from the user (format 12:59:59), calling the function, and displaying the value of seconds it returns. 6. Start with the program from Exercise 11 in Chapter 4, “Structures,” which adds two struct time values. Keep the same functionality, but modify the program so that it uses two functions. The first, time_to_secs(),takes as its only argument a structure of type Chapter 5212 06 3087 CH05 11/29/01 2:23 PM Page 212time, and returns the equivalent in seconds (type long). The second function, secs_to_time(),takes as its only argument a time in seconds (type long), and returns a structure of type time. 7. Start with the power() function of Exercise 2, which works only with type double. Create a series of overloaded functions with the same name that, in addition to double, also work with types char, int, long, and float. Write a main() program that exercises these overloaded functions with all argument types. 8. Write a function called swap() that interchanges two int values passed to it by the call- ing program. (Note that this function swaps the values of the variables in the calling pro- gram, not those in the function.) You’ll need to decide how to pass the arguments. Create a main() program to exercise the function. 9. Repeat Exercise 8, but instead of two int variables, have the swap() function inter- change two struct time values (see Exercise 6). 10. Write a function that, when you call it, displays a message telling how many times it has been called: “I have been called 3 times”, for instance. Write a main() program that calls this function at least 10 times. Try implementing this function in two different ways. First, use a global variable to store the count. Second, use a local static variable. Which is more appropriate? Why can’t you use a local variable? 11. Write a program, based on the sterling structure of Exercise 10 in Chapter 4, that obtains from the user two money amounts in old-style British format (£9:19:11), adds them, and displays the result, again in old-style format. Use three functions. The first should obtain a pounds-shillings-pence value from the user and return the value as a structure of type sterling. The second should take two arguments of type sterling and return a value of the same type, which is the sum of the arguments. The third should take a sterling structure as its argument and display its value. 12. Revise the four-function fraction calculator from Exercise 12, Chapter 4, so that it uses functions for each of the four arithmetic operations. They can be called fadd(), fsub(), fmul(), and fdiv(). Each of these functions should take two arguments of type struct fraction, and return an argument of the same type. 5 F UNCTIONS 213Functions 06 3087 CH05 11/29/01 2:23 PM Page 21306 3087 CH05 11/29/01 2:23 PM Page 214CHAPTER 6 Objects and Classes IN THIS CHAPTER •A Simple Class 216 • C++ Objects as Physical Objects 223 • C++ Objects as Data Types 226 • Constructors 227 • Objects as Function Arguments 233 • The Default Copy Constructor 238 • Returning Objects from Functions 240 •A Card-Game Example 243 •Structures and Classes 247 •Classes, Objects, and Memory 247 •Static Class Data 249 • const and Classes 252 • What Does It All Mean? 256 07 3087 CH06 11/29/01 2:15 PM Page 215Chapter 6216 And now, the topics you’ve all been waiting for: objects and classes. The preliminaries are out of the way. We’ve learned about structures, which provide a way to group data elements. We’ve examined functions, which organize program actions into named entities. In this chapter we’ll put these ideas together to create classes. We’ll introduce several classes, starting with simple ones and working toward more complicated examples. We’ll focus first on the details of classes and objects. At the end of the chapter we’ll take a wider view, discussing what is to be gained by using the OOP approach. As you read this chapter you may want to refer back to the concepts introduced in Chapter 1, “The Big Picture.” A Simple Class Our first program contains a class and two objects of that class. Although it’s simple, the program demonstrates the syntax and general features of classes in C++. Here’s the listing for the SMALLOBJ program: // smallobj.cpp // demonstrates a small, simple object #include using namespace std; //////////////////////////////////////////////////////////////// class smallobj //define a class { private: int somedata; //class data public: void setdata(int d) //member function to set data { somedata = d; } void showdata() //member function to display data { cout << “Data is “ << somedata << endl; } }; //////////////////////////////////////////////////////////////// int main() { smallobj s1, s2; //define two objects of class smallobj s1.setdata(1066); //call member function to set data s2.setdata(1776); s1.showdata(); //call member function to display data s2.showdata(); return 0; } 07 3087 CH06 11/29/01 2:15 PM Page 216The class smallobj defined in this program contains one data item and two member functions. The two member functions provide the only access to the data item from outside the class. The first member function sets the data item to a value, and the second displays the value. (This may sound like Greek, but we’ll see what these terms mean as we go along.) Placing data and functions together into a single entity is a central idea in object-oriented programming. This is shown in Figure 6.1. Objects and Classes 6 O BJECTS AND C LASSES 217 FIGURE 6.1 Classes contain data and functions. Classes and Objects Recall from Chapter 1 that an object has the same relationship to a class that a variable has to a data type. An object is said to be an instance of a class, in the same way my 1954 Chevrolet is an instance of a vehicle. In SMALLOBJ,the class—whose name is smallobj—is defined in the first part of the program. Later, in main(), we define two objects—s1 and s2—that are instances of that class. Each of the two objects is given a value, and each displays its value. Here’s the output of the program: Data is 1066 ← object s1 displayed this Data is 1776 ← object s2 displayed this 07 3087 CH06 11/29/01 2:15 PM Page 217We’ll begin by looking in detail at the first part of the program—the definition of the class smallobj. Later we’ll focus on what main() does with objects of this class. Defining the Class Here’s the definition (sometimes called a specifier) for the class smallobj, copied from the SMALLOBJ listing: class smallobj //define a class { private: int somedata; //class data public: void setdata(int d) //member function to set data { somedata = d; } void showdata() //member function to display data { cout << “\nData is “ << somedata; } }; The definition starts with the keyword class,followed by the class name—smallobj in this example. Like a structure, the body of the class is delimited by braces and terminated by a semicolon. (Don’t forget the semicolon. Remember, data constructs such as structures and classes end with a semicolon, while control constructs such as functions and loops do not.) private and public The body of the class contains two unfamiliar keywords: private and public. What is their purpose? A key feature of object-oriented programming is data hiding. This term does not refer to the activities of particularly paranoid programmers; rather it means that data is concealed within a class so that it cannot be accessed mistakenly by functions outside the class. The primary mechanism for hiding data is to put it in a class and make it private. Private data or functions can only be accessed from within the class. Public data or functions, on the other hand, are accessible from outside the class. This is shown in Figure 6.2. Hidden from Whom? Don’t confuse data hiding with the security techniques used to protect computer databases. To provide a security measure you might, for example, require a user to supply a password before granting access to a database. The password is meant to keep unauthorized or malevolent users from altering (or often even reading) the data. Data hiding, on the other hand, means hiding data from parts of the program that don’t need to access it. More specifically, one class’s data is hidden from other classes. Data hiding is designed to protect well-intentioned programmers from honest mistakes. Programmers who really want to can figure out a way to access private data, but they will find it hard to do so by accident. Chapter 6218 07 3087 CH06 11/29/01 2:15 PM Page 218FIGURE 6.2 Private and public. Class Data The smallobj class contains one data item: somedata, which is of type int. The data items within a class are called data members (or sometimes member data). There can be any number of data members in a class, just as there can be any number of data items in a structure. The data member somedata follows the keyword private,so it can be accessed from within the class, but not from outside. Member Functions Member functions are functions that are included within a class. (In some object-oriented languages, such as Smalltalk, member functions are called methods; some writers use this term in C++ as well.) There are two member functions in smallobj: setdata() and showdata(). The function bodies of these functions have been written on the same line as the braces that delimit them. You could also use the more traditional format for these function definitions: void setdata(int d) { somedata = d; } and Objects and Classes 6 O BJECTS AND C LASSES 219 07 3087 CH06 11/29/01 2:15 PM Page 219void showdata() { cout << “\nData is “ << somedata; } However, when member functions are small, it is common to compress their definitions this way to save space. Because setdata() and showdata() follow the keyword public,they can be accessed from outside the class. We’ll see how this is done in a moment. Figure 6.3 shows the syntax of a class definition. Functions Are Public, Data Is Private Usually the data within a class is private and the functions are public. This is a result of the way classes are used. The data is hidden so it will be safe from accidental manipulation, while the functions that operate on the data are public so they can be accessed from outside the class. However, there is no rule that says data must be private and functions public; in some circum- stances you may find you’ll need to use private functions and public data. Chapter 6220 Keyword Keyword public and colon Private functions and data Keyword private and colon Name of class Public functions and data Semicolon Braces class foo { private: int data; public: void memfunc (int d) { data = d; } }; FIGURE 6.3 Syntax of a class definition. Member Functions Within Class Definition The member functions in the smallobj class perform operations that are quite common in classes: setting and retrieving the data stored in the class. The setdata() function accepts a value as a parameter and sets the somedata variable to this value. The showdata() function displays the value stored in somedata. 07 3087 CH06 11/29/01 2:15 PM Page 220Note that the member functions setdata() and showdata() are definitions in that the actual code for the function is contained within the class definition. (The functions are not definitions in the sense that memory is set aside for the function code; this doesn’t happen until an object of the class is created.) Member functions defined inside a class this way are created as inline functions by default. (Inline functions were discussed in Chapter 5, “Functions.”) We’ll see later that it is also possible to declare a function within a class but to define it elsewhere. Functions defined outside the class are not normally inline. Using the Class Now that the class is defined, let’s see how main() makes use of it. We’ll see how objects are defined, and, once defined, how their member functions are accessed. Defining Objects The first statement in main() smallobj s1, s2; defines two objects, s1 and s2,of class smallobj. Remember that the definition of the class smallobj does not create any objects. It only describes how they will look when they are created, just as a structure definition describes how a structure will look but doesn’t create any structure variables. It is objects that participate in program operations. Defining an object is similar to defining a variable of any data type: Space is set aside for it in memory. Defining objects in this way means creating them. This is also called instantiating them. The term instantiating arises because an instance of the class is created. An object is an instance (that is, a specific example) of a class. Objects are sometimes called instance variables. Calling Member Functions The next two statements in main() call the member function setdata(): s1.setdata(1066); s2.setdata(1776); These statements don’t look like normal function calls. Why are the object names s1 and s2 connected to the function names with a period? This strange syntax is used to call a member function that is associated with a specific object. Because setdata() is a member function of the smallobj class, it must always be called in connection with an object of this class. It doesn’t make sense to say setdata(1066); Objects and Classes 6 O BJECTS AND C LASSES 221 07 3087 CH06 11/29/01 2:16 PM Page 221by itself, because a member function is always called to act on a specific object, not on the class in general. Attempting to access the class this way would be like trying to drive the blueprint of a car. Not only does this statement not make sense, but the compiler will issue an error message if you attempt it. Member functions of a class can be accessed only by an object of that class. To use a member function, the dot operator (the period) connects the object name and the member function. The syntax is similar to the way we refer to structure members, but the parentheses signal that we’re executing a member function rather than referring to a data item. (The dot operator is also called the class member access operator.) The first call to setdata() s1.setdata(1066); executes the setdata() member function of the s1 object. This function sets the variable somedata in object s1 to the value 1066. The second call s2.setdata(1776); causes the variable somedata in s2 to be set to 1776. Now we have two objects whose somedata variables have different values, as shown in Figure 6.4. Chapter 6222 FIGURE 6.4 Two objects of class smallobj. 07 3087 CH06 11/29/01 2:16 PM Page 222Similarly, the following two calls to the showdata() function will cause the two objects to display their values: s1.showdata(); s2.showdata(); Messages Some object-oriented languages refer to calls to member functions as messages. Thus the call s1.showdata(); can be thought of as sending a message to s1 telling it to show its data. The term message is not a formal term in C++, but it is a useful idea to keep in mind as we discuss member functions. Talking about messages emphasizes that objects are discrete entities and that we communicate with them by calling their member functions. Referring to the analogy with company organization in Chapter 1, it’s like sending a message to the secretary in the sales department asking for a list of products sold in the southwest distribution area. C++ Objects as Physical Objects In many programming situations, objects in programs represent physical objects: things that can be felt or seen. These situations provide vivid examples of the correspondence between the program and the real world. We’ll look at two such situations: widget parts and graphics circles. Widget Parts as Objects The smallobj class in the last example had only one data item. Let’s look at an example of a somewhat more ambitious class. (These are not the same ambitious classes discussed in political science courses.) We’ll create a class based on the structure for the widget parts inventory, last seen in such examples as PARTS in Chapter 4, “Structures.” Here’s the listing for OBJPART: // objpart.cpp // widget part as an object #include using namespace std; //////////////////////////////////////////////////////////////// class part //define class { private: int modelnumber; //ID number of widget int partnumber; //ID number of widget part float cost; //cost of part public: void setpart(int mn, int pn, float c) //set data { Objects and Classes 6 O BJECTS AND C LASSES 223 07 3087 CH06 11/29/01 2:16 PM Page 223modelnumber = mn; partnumber = pn; cost = c; } void showpart() //display data { cout << “Model “ << modelnumber; cout << “, part “ << partnumber; cout << “, costs $” << cost << endl; } }; //////////////////////////////////////////////////////////////// int main() { part part1; //define object // of class part part1.setpart(6244, 373, 217.55F); //call member function part1.showpart(); //call member function return 0; } This program features the class part. Instead of one data item, as SMALLOBJ had, this class has three: modelnumber, partnumber, and cost. A single member function, setpart(),supplies values to all three data items at once. Another function, showpart(),displays the values stored in all three items. In this example only one object of type part is created: part1. The member function setpart() sets the three data items in this part to the values 6244, 373, and 217.55. The mem- ber function showpart() then displays these values. Here’s the output: Model 6244, part 373, costs $217.55 This is a somewhat more realistic example than SMALLOBJ. If you were designing an inventory program you might actually want to create a class something like part. It’s an example of a C++ object representing a physical object in the real world—a widget part. Circles as Objects In our next example we’ll examine an object used to represent a circle: the kind of circle displayed on your computer screen. An image isn’t quite as tangible an object as a widget part, which you can presumably hold in your hand, but you can certainly see such a circle when your program runs. Our example is an object-oriented version of the CIRCSTRC program from Chapter 5. (As in that program, you’ll need to add the appropriate Console Graphics Lite files to your project. These Chapter 6224 07 3087 CH06 11/29/01 2:16 PM Page 224files can be downloaded from the publisher’s Web site as described in the Introduction. Appendix E, “Console Graphics Lite,” describes these files. See also the appendix for your particular compiler.) The program creates three circles with various characteristics and displays them. Here’s the listing for CIRCLES: // circles.cpp // circles as graphics objects #include “msoftcon.h” // for graphics functions //////////////////////////////////////////////////////////////// class circle //graphics circle { protected: int xCo, yCo; //coordinates of center int radius; color fillcolor; //color fstyle fillstyle; //fill pattern public: //sets circle attributes void set(int x, int y, int r, color fc, fstyle fs) { xCo = x; yCo = y; radius = r; fillcolor = fc; fillstyle = fs; } void draw() //draws the circle { set_color(fillcolor); //set color set_fill_style(fillstyle); //set fill draw_circle(xCo, yCo, radius); //draw solid circle } }; //////////////////////////////////////////////////////////////// int main() { init_graphics(); //initialize graphics system circle c1; //create circles circle c2; circle c3; //set circle attributes c1.set(15, 7, 5, cBLUE, X_FILL); c2.set(41, 12, 7, cRED, O_FILL); c3.set(65, 18, 4, cGREEN, MEDIUM_FILL); Objects and Classes 6 O BJECTS AND C LASSES 225 07 3087 CH06 11/29/01 2:16 PM Page 225c1.draw(); //draw circles c2.draw(); c3.draw(); set_cursor_pos(1, 25); //lower left corner return 0; } The output of this program is the same as that of the CIRCSTRC program in Chapter 5, shown in Figure 5.5 in that chapter. You may find it interesting to compare the two programs. In CIRCLES, each circle is represented as a C++ object rather than as a combination of a structure variable and an unrelated circ_draw() function, as it was in CIRCSTRC. Notice in CIRCLES how everything connected with a circle—attributes and functions—is brought together in the class definition. In CIRCLES, besides the draw() function, the circle class also requires the five-argument set() function to set its attributes. We’ll see later that it’s advantageous to dispense with this function and use a constructor instead. C++ Objects as Data Types Here’s another kind of entity C++ objects can represent: variables of a user-defined data type. We’ll use objects to represent distances measured in the English system, as discussed in Chapter 4. Here’s the listing for ENGLOBJ: // englobj.cpp // objects using English measurements #include using namespace std; //////////////////////////////////////////////////////////////// class Distance //English Distance class { private: int feet; float inches; public: void setdist(int ft, float in) //set Distance to args { feet = ft; inches = in; } void getdist() //get length from user { cout << “\nEnter feet: “; cin >> feet; cout << “Enter inches: “; cin >> inches; } void showdist() //display distance { cout << feet << “\’-” << inches << ‘\”’; } Chapter 6226 07 3087 CH06 11/29/01 2:16 PM Page 226}; //////////////////////////////////////////////////////////////// int main() { Distance dist1, dist2; //define two lengths dist1.setdist(11, 6.25); //set dist1 dist2.getdist(); //get dist2 from user //display lengths cout << “\ndist1 = “; dist1.showdist(); cout << “\ndist2 = “; dist2.showdist(); cout << endl; return 0; } In this program, the class Distance contains two data items, feet and inches. This is similar to the Distance structure seen in examples in Chapter 4, but here the class Distance also has three member functions: setdist(), which uses arguments to set feet and inches; getdist(), which gets values for feet and inches from the user at the keyboard; and showdist(), which displays the distance in feet-and-inches format. The value of an object of class Distance can thus be set in either of two ways. In main(),we define two objects of class Distance: dist1 and dist2. The first is given a value using the setdist() member function with the arguments 11 and 6.25, and the second is given a value that is supplied by the user. Here’s a sample interaction with the program: Enter feet: 10 Enter inches: 4.75 dist1 = 11’-6.25” ← provided by arguments dist2 = 10’-4.75” ← input by the user Constructors The ENGLOBJ example shows two ways that member functions can be used to give values to the data items in an object. Sometimes, however, it’s convenient if an object can initialize itself when it’s first created, without requiring a separate call to a member function. Automatic initial- ization is carried out using a special member function called a constructor. A constructor is a member function that is executed automatically whenever an object is created. (The term con- structor is sometimes abbreviated ctor,especially in comments in program listings.) Objects and Classes 6 O BJECTS AND C LASSES 227 07 3087 CH06 11/29/01 2:16 PM Page 227A Counter Example As an example, we’ll create a class of objects that might be useful as a general-purpose programming element. A counter is a variable that counts things. Maybe it counts file accesses, or the number of times the user presses the Enter key, or the number of customers entering a bank. Each time such an event takes place, the counter is incremented (1 is added to it). The counter can also be accessed to find the current count. Let’s assume that this counter is important in the program and must be accessed by many different functions. In procedural languages such as C, a counter would probably be implemented as a global variable. However, as we noted in Chapter 1, global variables complicate the program’s design and may be modified accidentally. This example, COUNTER,provides a counter variable that can be modified only through its member functions. // counter.cpp // object represents a counter variable #include using namespace std; //////////////////////////////////////////////////////////////// class Counter { private: unsigned int count; //count public: Counter() : count(0) //constructor { /*empty body*/ } void inc_count() //increment count { count++; } int get_count() //return count { return count; } }; //////////////////////////////////////////////////////////////// int main() { Counter c1, c2; //define and initialize cout << “\nc1=” << c1.get_count(); //display cout << “\nc2=” << c2.get_count(); c1.inc_count(); //increment c1 c2.inc_count(); //increment c2 c2.inc_count(); //increment c2 cout << “\nc1=” << c1.get_count(); //display again cout << “\nc2=” << c2.get_count(); Chapter 6228 07 3087 CH06 11/29/01 2:16 PM Page 228cout << endl; return 0; } The Counter class has one data member: count,of type unsigned int (since the count is always positive). It has three member functions: the constructor Counter(), which we’ll look at in a moment; inc_count(), which adds 1 to count; and get_count(), which returns the cur- rent value of count. Automatic Initialization When an object of type Counter is first created, we want its count to be initialized to 0. After all, most counts start at 0. We could provide a set_count() function to do this and call it with an argument of 0, or we could provide a zero_count() function, which would always set count to 0. However, such functions would need to be executed every time we created a Counter object. Counter c1; //every time we do this, c1.zero_count(); //we must do this too This is mistake prone, because the programmer may forget to initialize the object after creating it. It’s more reliable and convenient, especially when there are a great many objects of a given class, to cause each object to initialize itself when it’s created. In the Counter class, the constructor Counter() does this. This function is called automatically whenever a new object of type Counter is created. Thus in main() the statement Counter c1, c2; creates two objects of type Counter. As each is created, its constructor, Counter(), is executed. This function sets the count variable to 0. So the effect of this single statement is to not only create two objects, but also to initialize their count variables to 0. Same Name as the Class There are some unusual aspects of constructor functions. First, it is no accident that they have exactly the same name (Counter in this example) as the class of which they are members. This is one way the compiler knows they are constructors. Second, no return type is used for constructors. Why not? Since the constructor is called automatically by the system, there’s no program for it to return anything to; a return value wouldn’t make sense. This is the second way the compiler knows they are constructors. Initializer List One of the most common tasks a constructor carries out is initializing data members. In the Counter class the constructor must initialize the count member to 0. You might think that this would be done in the constructor’s function body, like this: Objects and Classes 6 O BJECTS AND C LASSES 229 07 3087 CH06 11/29/01 2:16 PM Page 229count() { count = 0; } However, this is not the preferred approach (although it does work). Here’s how you should initialize a data member: count() : count(0) { } The initialization takes place following the member function declarator but before the function body. It’s preceded by a colon. The value is placed in parentheses following the member data. If multiple members must be initialized, they’re separated by commas. The result is the initializer list (sometimes called by other names, such as the member-initialization list). someClass() : m1(7), m2(33), m2(4) ← initializer list { } Why not initialize members in the body of the constructor? The reasons are complex, but have to do with the fact that members initialized in the initializer list are given a value before the constructor even starts to execute. This is important in some situations. For example, the initializer list is the only way to initialize const member data and references. Actions more complicated than simple initialization must be carried out in the constructor body, as with ordinary functions. Counter Output The main() part of this program exercises the Counter class by creating two counters, c1 and c2. It causes the counters to display their initial values, which—as arranged by the constructor—are 0. It then increments c1 once and c2 twice, and again causes the counters to display themselves (non-criminal behavior in this context). Here’s the output: c1=0 c2=0 c1=1 c2=2 If this isn’t enough proof that the constructor is operating as advertised, we can rewrite the constructor to print a message when it executes. Counter() : count(0) { cout << “I’m the constructor\n”; } Now the program’s output looks like this: I’m the constructor I’m the constructor Chapter 6230 07 3087 CH06 11/29/01 2:16 PM Page 230c1=0 c2=0 c1=1 c2=2 As you can see, the constructor is executed twice—once for c1 and once for c2—when the statement Counter c1, c2; is executed in main(). Do-It-Yourself Data Constructors are pretty amazing when you think about it. Whoever writes language compilers (for C or VB or even C++) must execute the equivalent of a constructor when the user defines a variable. If you define an int,for example, somewhere there’s a constructor allocating four bytes of memory for it. If we can write our own constructors, we can start to take over some of the tasks of a compiler writer. This is one step on the path to creating our own data types, as we’ll see later. A Graphics Example Let’s rewrite our earlier CIRCLES example to use a constructor instead of a set() function. To handle the initialization of the five attributes of circles, this constructor will have five arguments and five items in its initialization list. Here’s the listing for CIRCTOR: // circtor.cpp // circles use constructor for initialization #include “msoftcon.h” // for graphics functions //////////////////////////////////////////////////////////////// class circle //graphics circle { protected: int xCo, yCo; //coordinates of center int radius; color fillcolor; //color fstyle fillstyle; //fill pattern public: //constructor circle(int x, int y, int r, color fc, fstyle fs) : xCo(x), yCo(y), radius(r), fillcolor(fc), fillstyle(fs) { } void draw() //draws the circle { set_color(fillcolor); //set color Objects and Classes 6 O BJECTS AND C LASSES 231 07 3087 CH06 11/29/01 2:16 PM Page 231set_fill_style(fillstyle); //set fill draw_circle(xCo, yCo, radius); //draw solid circle } }; //////////////////////////////////////////////////////////////// int main() { init_graphics(); //initialize graphics system //create circles circle c1(15, 7, 5, cBLUE, X_FILL); circle c2(41, 12, 7, cRED, O_FILL); circle c3(65, 18, 4, cGREEN, MEDIUM_FILL); c1.draw(); //draw circles c2.draw(); c3.draw(); set_cursor_pos(1, 25); //lower left corner return 0; } This program is similar to CIRCLES,except that set() has been replaced by the constructor. Note how this simplifies main(). Instead of two separate statements for each object, one to create it and one to set its attributes, now one statement both creates the object and sets its attributes at the same time. Destructors We’ve seen that a special member function—the constructor—is called automatically when an object is first created. You might guess that another function is called automatically when an object is destroyed. This is indeed the case. Such a function is called a destructor. A destructor has the same name as the constructor (which is the same as the class name) but is preceded by a tilde: class Foo { private: int data; public: Foo() : data(0) //constructor (same name as class) { } ~Foo() //destructor (same name with tilde) { } }; Chapter 6232 07 3087 CH06 11/29/01 2:16 PM Page 232Like constructors, destructors do not have a return value. They also take no arguments (the assumption being that there’s only one way to destroy an object). The most common use of destructors is to deallocate memory that was allocated for the object by the constructor. We’ll investigate these activities in Chapter 10, “Pointers.” Until then we won’t have much use for destructors. Objects as Function Arguments Our next program adds some embellishments to the ENGLOBJ example. It also demonstrates some new aspects of classes: constructor overloading, defining member functions outside the class, and—perhaps most importantly—objects as function arguments. Here’s the listing for ENGLCON: // englcon.cpp // constructors, adds objects using member function #include using namespace std; //////////////////////////////////////////////////////////////// class Distance //English Distance class { private: int feet; float inches; public: //constructor (no args) Distance() : feet(0), inches(0.0) { } //constructor (two args) Distance(int ft, float in) : feet(ft), inches(in) { } void getdist() //get length from user { cout << “\nEnter feet: “; cin >> feet; cout << “Enter inches: “; cin >> inches; } void showdist() //display distance { cout << feet << “\’-” << inches << ‘\”’; } void add_dist( Distance, Distance ); //declaration }; //-------------------------------------------------------------- //add lengths d2 and d3 void Distance::add_dist(Distance d2, Distance d3) Objects and Classes 6 O BJECTS AND C LASSES 233 07 3087 CH06 11/29/01 2:16 PM Page 233{ inches = d2.inches + d3.inches; //add the inches feet = 0; //(for possible carry) if(inches >= 12.0) //if total exceeds 12.0, { //then decrease inches inches -= 12.0; //by 12.0 and feet++; //increase feet } //by 1 feet += d2.feet + d3.feet; //add the feet } //////////////////////////////////////////////////////////////// int main() { Distance dist1, dist3; //define two lengths Distance dist2(11, 6.25); //define and initialize dist2 dist1.getdist(); //get dist1 from user dist3.add_dist(dist1, dist2); //dist3 = dist1 + dist2 //display all lengths cout << “\ndist1 = “; dist1.showdist(); cout << “\ndist2 = “; dist2.showdist(); cout << “\ndist3 = “; dist3.showdist(); cout << endl; return 0; } This program starts with a distance dist2 set to an initial value and adds to it a distance dist1, whose value is supplied by the user, to obtain the sum of the distances. It then displays all three distances: Enter feet: 17 Enter inches: 5.75 dist1 = 17’-5.75” dist2 = 11’-6.25” dist3 = 29’-0” Let’s see how the new features in this program are implemented. Overloaded Constructors It’s convenient to be able to give variables of type Distance a value when they are first created. That is, we would like to use definitions like Distance width(5, 6.25); Chapter 6234 07 3087 CH06 11/29/01 2:16 PM Page 234which defines an object, width, and simultaneously initializes it to a value of 5 for feet and 6.25 for inches. To do this we write a constructor like this: Distance(int ft, float in) : feet(ft), inches(in) { } This sets the member data feet and inches to whatever values are passed as arguments to the constructor. So far so good. However, we also want to define variables of type Distance without initializing them, as we did in ENGLOBJ. Distance dist1, dist2; In that program there was no constructor, but our definitions worked just fine. How could they work without a constructor? Because an implicit no-argument constructor is built into the program automatically by the compiler, and it’s this constructor that created the objects, even though we didn’t define it in the class. This no-argument constructor is called the default constructor. If it weren’t created automatically by the constructor, you wouldn’t be able to create objects of a class for which no constructor was defined. Often we want to initialize data members in the default (no-argument) constructor as well. If we let the default constructor do it, we don’t really know what values the data members may be given. If we care what values they may be given, we need to explicitly define the constructor. In ENGLECON we show how this looks: Distance() : feet(0), inches(0.0) //default constructor { } //no function body, doesn’t do anything The data members are initialized to constant values, in this case the integer value 0 and the float value 0.0, for feet and inches respectively. Now we can use objects initialized with the no-argument constructor and be confident that they represent no distance (0 feet plus 0.0 inches) rather than some arbitrary value. Since there are now two explicit constructors with the same name, Distance(), we say the constructor is overloaded. Which of the two constructors is executed when an object is created depends on how many arguments are used in the definition: Distance length; // calls first constructor Distance width(11, 6.0); // calls second constructor Objects and Classes 6 O BJECTS AND C LASSES 235 07 3087 CH06 11/29/01 2:16 PM Page 235Member Functions Defined Outside the Class So far we’ve seen member functions that were defined inside the class definition. This need not always be the case. ENGLCON shows a member function, add_dist(),that is not defined within the Distance class definition. It is only declared inside the class, with the statement void add_dist( Distance, Distance ); This tells the compiler that this function is a member of the class but that it will be defined outside the class declaration, someplace else in the listing. In ENGLCON the add_dist() function is defined following the class definition. It is adapted from the ENGLSTRC program in Chapter 4: //add lengths d2 and d3 void Distance::add_dist(Distance d2, Distance d3) { inches = d2.inches + d3.inches; //add the inches feet = 0; //(for possible carry) if(inches >= 12.0) //if total exceeds 12.0, { //then decrease inches inches -= 12.0; //by 12.0 and feet++; //increase feet } //by 1 feet += d2.feet + d3.feet; //add the feet } The declarator in this definition contains some unfamiliar syntax. The function name, add_dist(), is preceded by the class name, Distance, and a new symbol—the double colon (::). This symbol is called the scope resolution operator. It is a way of specifying what class something is associated with. In this situation, Distance::add_dist() means “the add_dist() member function of the Distance class.” Figure 6.5 shows its usage. Chapter 6236 FIGURE 6.5 The scope resolution operator. 07 3087 CH06 11/29/01 2:16 PM Page 236Objects as Arguments Now we can see how ENGLCON works. The distances dist1 and dist3 are created using the default constructor (the one that takes no arguments). The distance dist2 is created with the constructor that takes two arguments, and is initialized to the values passed in these arguments. A value is obtained for dist1 by calling the member function getdist(), which obtains values from the user. Now we want to add dist1 and dist2 to obtain dist3. The function call in main() dist3.add_dist(dist1, dist2); does this. The two distances to be added, dist1 and dist2,are supplied as arguments to add_dist(). The syntax for arguments that are objects is the same as that for arguments that are simple data types such as int:The object name is supplied as the argument. Since add_dist() is a member function of the Distance class, it can access the private data in any object of class Distance supplied to it as an argument, using names like dist1.inches and dist2.feet. Close examination of add_dist() emphasizes some important truths about member functions. A member function is always given access to the object for which it was called: the object connected to it with the dot operator. But it may be able to access other objects. In the following statement in ENGLCON, what objects can add_dist() access? dist3.add_dist(dist1, dist2); Besides dist3,the object for which it was called, it can also access dist1 and dist2, because they are supplied as arguments. You might think of dist3 as a sort of phantom argument; the member function always has access to it, even though it is not supplied as an argument. That’s what this statement means: “Execute the add_dist() member function of dist3.” When the variables feet and inches are referred to within this function, they refer to dist3.feet and dist3.inches. Notice that the result is not returned by the function. The return type of add_dist() is void. The result is stored automatically in the dist3 object. Figure 6.6 shows the two distances dist1 and dist2 being added together, with the result stored in dist3. To summarize, every call to a member function is associated with a particular object (unless it’s a static function; we’ll get to that later). Using the member names alone (feet and inches), the function has direct access to all the members, whether private or public, of that object. It also has indirect access, using the object name and the member name, connected with the dot operator (dist1.inches or dist2.feet) to other objects of the same class that are passed as arguments. Objects and Classes 6 O BJECTS AND C LASSES 237 07 3087 CH06 11/29/01 2:16 PM Page 237FIGURE 6.6 Result in this object. The Default Copy Constructor We’ve seen two ways to initialize objects. A no-argument constructor can initialize data members to constant values, and a multi-argument constructor can initialize data members to values passed as arguments. Let’s mention another way to initialize an object: you can initialize it with another object of the same type. Surprisingly, you don’t need to create a special constructor for this; one is already built into all classes. It’s called the default copy constructor. It’s a one- argument constructor whose argument is an object of the same class as the constructor. The ECOPYCON program shows how this constructor is used. // ecopycon.cpp // initialize objects using default copy constructor #include using namespace std; //////////////////////////////////////////////////////////////// Chapter 6238 dist3 feet inches feet inches dist1 feet inches dist1.feet dist1.inches dist2.feet dist2.inches dist2 feet inches dist3.add_dist(dist1, dist2) Member functions of dist3 can refer to its data directly. Data in objects passed as arguments is referred to with the dot operator. 07 3087 CH06 11/29/01 2:16 PM Page 238class Distance //English Distance class { private: int feet; float inches; public: //constructor (no args) Distance() : feet(0), inches(0.0) { } //Note: no one-arg constructor //constructor (two args) Distance(int ft, float in) : feet(ft), inches(in) { } void getdist() //get length from user { cout << “\nEnter feet: “; cin >> feet; cout << “Enter inches: “; cin >> inches; } void showdist() //display distance { cout << feet << “\’-” << inches << ‘\”’; } }; //////////////////////////////////////////////////////////////// int main() { Distance dist1(11, 6.25); //two-arg constructor Distance dist2(dist1); //one-arg constructor Distance dist3 = dist1; //also one-arg constructor //display all lengths cout << “\ndist1 = “; dist1.showdist(); cout << “\ndist2 = “; dist2.showdist(); cout << “\ndist3 = “; dist3.showdist(); cout << endl; return 0; } We initialize dist1 to the value of 11’-6.25” using the two-argument constructor. Then we define two more objects of type Distance, dist2 and dist3,initializing both to the value of dist1. You might think this would require us to define a one-argument constructor, but initial- izing an object with another object of the same type is a special case. These definitions both use the default copy constructor. The object dist2 is initialized in the statement Distance dist2(dist1); Objects and Classes 6 O BJECTS AND C LASSES 239 07 3087 CH06 11/29/01 2:16 PM Page 239This causes the default copy constructor for the Distance class to perform a member-by-member copy of dist1 into dist2. Surprisingly, a different format has exactly the same effect, causing dist1 to be copied member-by-member into dist3: Distance dist3 = dist1; Although this looks like an assignment statement, it is not. Both formats invoke the default copy constructor, and can be used interchangeably. Here’s the output from the program: dist1 = 11’-6.25” dist2 = 11’-6.25” dist3 = 11’-6.25” This shows that the dist2 and dist3 objects have been initialized to the same value as dist1. In Chapter 11, “Virtual Functions,” we discuss how to create your own custom copy construc- tor by overloading the default. Returning Objects from Functions In the ENGLCON example, we saw objects being passed as arguments to functions. Now we’ll see an example of a function that returns an object. We’ll modify the ENGLCON program to produce ENGLRET: // englret.cpp // function returns value of type Distance #include using namespace std; //////////////////////////////////////////////////////////////// class Distance //English Distance class { private: int feet; float inches; public: //constructor (no args) Distance() : feet(0), inches(0.0) { } //constructor (two args) Distance(int ft, float in) : feet(ft), inches(in) { } void getdist() //get length from user { cout << “\nEnter feet: “; cin >> feet; cout << “Enter inches: “; cin >> inches; } void showdist() //display distance { cout << feet << “\’-” << inches << ‘\”’; } Chapter 6240 07 3087 CH06 11/29/01 2:16 PM Page 240Distance add_dist(Distance); //add }; //-------------------------------------------------------------- //add this distance to d2, return the sum Distance Distance::add_dist(Distance d2) { Distance temp; //temporary variable temp.inches = inches + d2.inches; //add the inches if(temp.inches >= 12.0) //if total exceeds 12.0, { //then decrease inches temp.inches -= 12.0; //by 12.0 and temp.feet = 1; //increase feet } //by 1 temp.feet += feet + d2.feet; //add the feet return temp; } //////////////////////////////////////////////////////////////// int main() { Distance dist1, dist3; //define two lengths Distance dist2(11, 6.25); //define, initialize dist2 dist1.getdist(); //get dist1 from user dist3 = dist1.add_dist(dist2); //dist3 = dist1 + dist2 //display all lengths cout << “\ndist1 = “; dist1.showdist(); cout << “\ndist2 = “; dist2.showdist(); cout << “\ndist3 = “; dist3.showdist(); cout << endl; return 0; } The ENGLRET program is very similar to ENGLCON,but the differences reveal important aspects of how functions work with objects. Arguments and Objects In ENGLCON,two distances were passed to add_dist() as arguments, and the result was stored in the object of which add_dist() was a member, namely dist3. In ENGLRET, one distance, dist2, is passed to add_dist() as an argument. It is added to the object, dist1,of which add_dist() is a member, and the result is returned from the function. In main(),the result is assigned to dist3 in the statement dist3 = dist1.add_dist(dist2); Objects and Classes 6 O BJECTS AND C LASSES 241 07 3087 CH06 11/29/01 2:16 PM Page 241The effect is the same as the corresponding statement in ENGLCON,but it is somewhat more natural looking, since the assignment operator, =, is used in a natural way. In Chapter 8, “Operator Overloading,” we’ll see how to use the arithmetic + operator to achieve the even more natural expression dist3 = dist1 + dist2; Here’s the add_dist() function from ENGLRET: //add this distance to d2, return the sum Distance Distance::add_dist(Distance d2) { Distance temp; //temporary variable temp.inches = inches + d2.inches; //add the inches if(temp.inches >= 12.0) //if total exceeds 12.0, { //then decrease inches temp.inches -= 12.0; //by 12.0 and temp.feet = 1; //increase feet } //by 1 temp.feet += feet + d2.feet; //add the feet return temp; } Compare this with the same function in ENGLCON. As you can see, there are some subtle differences. In the ENGLRET version, a temporary object of class Distance is created. This object holds the sum until it can be returned to the calling program. The sum is calculated by adding two distances. The first is the object of which add_dist() is a member, dist1. Its member data is accessed in the function as feet and inches. The second is the object passed as an argument, dist2. Its member data is accessed as d2.feet and d2.inches. The result is stored in temp and accessed as temp.feet and temp.inches. The temp object is then returned by the function using the statement return temp; and the statement in main() assigns it to dist3. Notice that dist1 is not modified; it simply supplies data to add_dist(). Figure 6.7 shows how this looks. Chapter 6242 07 3087 CH06 11/29/01 2:16 PM Page 242FIGURE 6.7 Result returned from the temporary object. A Card-Game Example As a larger example of objects modeling the real world, let’s look at a variation of the CARDS program from Chapter 4. This program, CARDOBJ, has been rewritten to use objects. It does not introduce any new concepts, but it does use almost all the programming ideas we’ve discussed up to this point. As the CARDS example did, CARDOBJ creates three cards with fixed values and switches them around in an attempt to confuse the user about their location. But in CARDOBJ each card is an object of class card. Here’s the listing: // cardobj.cpp // cards as objects #include using namespace std; Objects and Classes 6 O BJECTS AND C LASSES 243 07 3087 CH06 11/29/01 2:16 PM Page 243enum Suit { clubs, diamonds, hearts, spades }; const int jack = 11; //from 2 to 10 are const int queen = 12; //integers without names const int king = 13; const int ace = 14; //////////////////////////////////////////////////////////////// class card { private: int number; //2 to 10, jack, queen, king, ace Suit suit; //clubs, diamonds, hearts, spades public: card () //constructor (no args) { } //constructor (two args) card (int n, Suit s) : number(n), suit(s) { } void display(); //display card bool isEqual(card); //same as another card? }; //-------------------------------------------------------------- void card::display() //display the card { if( number >= 2 && number <= 10 ) cout << number << “ of “; else switch(number) { case jack: cout << “jack of “; break; case queen: cout << “queen of “; break; case king: cout << “king of “; break; case ace: cout << “ace of “; break; } switch(suit) { case clubs: cout << “clubs”; break; case diamonds: cout << “diamonds”; break; case hearts: cout << “hearts”; break; case spades: cout << “spades”; break; } } //-------------------------------------------------------------- bool card::isEqual(card c2) //return true if cards equal { return ( number==c2.number && suit==c2.suit ) ? true : false; } Chapter 6244 07 3087 CH06 11/29/01 2:16 PM Page 244//////////////////////////////////////////////////////////////// int main() { card temp, chosen, prize; //define various cards int position; card card1( 7, clubs ); //define & initialize card1 cout << “\nCard 1 is the “; card1.display(); //display card1 card card2( jack, hearts ); //define & initialize card2 cout << “\nCard 2 is the “; card2.display(); //display card2 card card3( ace, spades ); //define & initialize card3 cout << “\nCard 3 is the “; card3.display(); //display card3 prize = card3; //prize is the card to guess cout << “\nI’m swapping card 1 and card 3”; temp = card3; card3 = card1; card1 = temp; cout << “\nI’m swapping card 2 and card 3”; temp = card3; card3 = card2; card2 = temp; cout << “\nI’m swapping card 1 and card 2”; temp = card2; card2 = card1; card1 = temp; cout << “\nNow, where (1, 2, or 3) is the “; prize.display(); //display prize card cout << “? “; cin >> position; //get user’s guess of position switch (position) { //set chosen to user’s choice case 1: chosen = card1; break; case 2: chosen = card2; break; case 3: chosen = card3; break; } if( chosen.isEqual(prize) ) //is chosen card the prize? cout << “That’s right! You win!”; else cout << “Sorry. You lose.”; cout << “ You chose the “; chosen.display(); //display chosen card Objects and Classes 6 O BJECTS AND C LASSES 245 07 3087 CH06 11/29/01 2:16 PM Page 245cout << endl; return 0; } There are two constructors in class card. The first, which takes no arguments, is used in main() to create the cards temp, chosen, and prize, which are not initialized. The second constructor, which takes two arguments, is used to create card1, card2,and card3 and to initialize them to specific values. Besides the constructors, card has two other member functions, both defined outside the class. The display() function takes no arguments; it simply displays the card object of which it is a member, using the number and suit data items in the card. The statement in main() chosen.display(); displays the card chosen by the user. The isEqual() function checks whether the card is equal to a card supplied as an argument. It uses the conditional operator to compare the card of which it is a member with a card supplied as an argument. This function could also have been written with an if...else statement if( number==c2.number && suit==c2.suit ) return true; else return false; but the conditional operator is more compact. In isEqual() the argument is called c2 as a reminder that there are two cards in the comparison: The first card is the object of which isEqual() is a member. The expression if( chosen.isEqual(prize) ) in main() compares the card chosen with the card prize. Here’s the output when the user guesses an incorrect card: Card 1 is the 7 of clubs Card 2 is the jack of hearts Card 3 is the ace of spades I’m swapping card 1 and card 3 I’m swapping card 2 and card 3 I’m swapping card 1 and card 2 Now, where (1, 2, or 3) is the ace of spades? 1 Sorry, you lose. You chose the 7 of clubs Chapter 6246 07 3087 CH06 11/29/01 2:16 PM Page 246Structures and Classes The examples so far in this book have portrayed structures as a way to group data and classes as a way to group both data and functions. In fact, you can use structures in almost exactly the same way that you use classes. The only formal difference between class and struct is that in a class the members are private by default, while in a structure they are public by default. Here’s the format we’ve been using for classes: class foo { private: int data1; public: void func(); }; Because private is the default in classes, this keyword is unnecessary. You can just as well write class foo { int data1; public: void func(); }; and the data1 will still be private. Many programmers prefer this style. We like to include the private keyword because it offers an increase in clarity. If you want to use a structure to accomplish the same thing as this class, you can dispense with the keyword public,provided you put the public members before the private ones struct foo { void func(); private: int data1; }; since public is the default. However, in most situations programmers don’t use a struct this way. They use structures to group only data, and classes to group both data and functions. Classes, Objects, and Memory We’ve probably given you the impression that each object created from a class contains separate copies of that class’s data and member functions. This is a good first approximation, since it Objects and Classes 6 O BJECTS AND C LASSES 247 07 3087 CH06 11/29/01 2:16 PM Page 247emphasizes that objects are complete, self-contained entities, designed using the class definition. The mental image here is of cars (objects) rolling off an assembly line, each one made according to a blueprint (the class definitions). Actually, things are not quite so simple. It’s true that each object has its own separate data items. On the other hand, contrary to what you may have been led to believe, all the objects in a given class use the same member functions. The member functions are created and placed in memory only once—when they are defined in the class definition. This makes sense; there’s really no point in duplicating all the member functions in a class every time you create another object of that class, since the functions for each object are identical. The data items, however, will hold different values, so there must be a separate instance of each data item for each object. Data is therefore placed in memory when each object is defined, so there is a separate set of data for each object. Figure 6.8 shows how this looks. Chapter 6248 FIGURE 6.8 Objects, data, functions, and memory. 07 3087 CH06 11/29/01 2:16 PM Page 248In the SMALLOBJ example at the beginning of this chapter there are two objects of type smallobj, so there are two instances of somedata in memory. However, there is only one instance of the functions setdata() and showdata(). These functions are shared by all the objects of the class. There is no conflict because (at least in a single-threaded system) only one function is executed at a time. In most situations you don’t need to know that there is only one member function for an entire class. It’s simpler to visualize each object as containing both its own data and its own member functions. But in some situations, such as in estimating the size of an executing program, it’s helpful to know what’s happening behind the scenes. Static Class Data Having said that each object contains its own separate data, we must now amend that slightly. If a data item in a class is declared as static, only one such item is created for the entire class, no matter how many objects there are. A static data item is useful when all objects of the same class must share a common item of information. A member variable defined as static has characteristics similar to a normal static variable: It is visible only within the class, but its lifetime is the entire program. It continues to exist even if there are no objects of the class. (See Chapter 5 for a discussion of static variables.) However, while a normal static variable is used to retain information between calls to a function, static class member data is used to share information among the objects of a class. Uses of Static Class Data Why would you want to use static member data? As an example, suppose an object needed to know how many other objects of its class were in the program. In a road-racing game, for example, a race car might want to know how many other cars are still in the race. In this case a static variable count could be included as a member of the class. All the objects would have access to this variable. It would be the same variable for all of them; they would all see the same count. An Example of Static Class Data Here’s an example, STATDATA,that demonstrates a simple static data member: // statdata.cpp // static class data #include using namespace std; //////////////////////////////////////////////////////////////// class foo { Objects and Classes 6 O BJECTS AND C LASSES 249 07 3087 CH06 11/29/01 2:16 PM Page 249private: static int count; //only one data item for all objects //note: “declaration” only! public: foo() //increments count when object created { count++; } int getcount() //returns count { return count; } }; //-------------------------------------------------------------- int foo::count = 0; //*definition* of count //////////////////////////////////////////////////////////////// int main() { foo f1, f2, f3; //create three objects cout << “count is “ << f1.getcount() << endl; //each object cout << “count is “ << f2.getcount() << endl; //sees the cout << “count is “ << f3.getcount() << endl; //same value return 0; } The class foo in this example has one data item, count, which is type static int. The constructor for this class causes count to be incremented. In main() we define three objects of class foo. Since the constructor is called three times, count is incremented three times. Another member function, getcount(),returns the value in count. We call this function from all three objects, and—as we expected—each prints the same value. Here’s the output: count is 3 ← static data count is 3 count is 3 If we had used an ordinary automatic variable—as opposed to a static variable—for count, each constructor would have incremented its own private copy of count once, and the output would have been count is 1 ← automatic data count is 1 count is 1 Static class variables are not used as often as ordinary non-static variables, but they are important in many situations. Figure 6.9 shows how static variables compare with automatic variables. Chapter 6250 07 3087 CH06 11/29/01 2:16 PM Page 250FIGURE 6.9 Static versus automatic member variables. Separate Declaration and Definition Static member data requires an unusual format. Ordinary variables are usually declared (the compiler is told about their name and type) and defined (the compiler sets aside memory to hold the variable) in the same statement. Static member data, on the other hand, requires two separate statements. The variable’s declaration appears in the class definition, but the variable is actually defined outside the class, in much the same way as a global variable. Why is this two-part approach used? If static member data were defined inside the class (as it actually was in early versions of C++), it would violate the idea that a class definition is only a blueprint and does not set aside any memory. Putting the definition of static member data outside the class also serves to emphasize that the memory space for such data is allocated only once, Objects and Classes 6 O BJECTS AND C LASSES 251 07 3087 CH06 11/29/01 2:16 PM Page 251before the program starts to execute, and that one static member variable is accessed by an entire class; each object does not have its own version of the variable, as it would with ordinary member data. In this way a static member variable is more like a global variable. It’s easy to handle static data incorrectly, and the compiler is not helpful about such errors. If you include the declaration of a static variable but forget its definition, there will be no warning from the compiler. Everything looks fine until you get to the linker, which will tell you that you’re trying to reference an undeclared global variable. This happens even if you include the definition but forget the class name (the foo:: in the STATDATA example). const and Classes We’ve seen several examples of const used on normal variables to prevent them from being modified, and in Chapter 5 we saw that const can be used with function arguments to keep a function from modifying a variable passed to it by reference. Now that we know about classes, we can introduce some other uses of const: on member functions, on member function arguments, and on objects. These concepts work together to provide some surprising benefits. const Member Functions A const member function guarantees that it will never modify any of its class’s member data. The CONSTFU program shows how this works. //constfu.cpp //demonstrates const member functions / class aClass { private: int alpha; public: void nonFunc() //non-const member function { alpha = 99; } //OK void conFunc() const //const member function { alpha = 99; } //ERROR: can’t modify a member }; The non-const function nonFunc() can modify member data alpha, but the constant function conFunc() can’t. If it tries to, a compiler error results. A function is made into a constant function by placing the keyword const after the declarator but before the function body. If there is a separate function declaration, const must be used in both declaration and definition. Member functions that do nothing but acquire data from an object are obvious candidates for being made const, because they don’t need to modify any data. Chapter 6252 07 3087 CH06 11/29/01 2:16 PM Page 252Making a function const helps the compiler flag errors, and tells anyone looking at the listing that you intended the function not to modify anything in its object. It also makes possible the creation and use of const objects, which we’ll discuss soon. A Distance Example To avoid raising too many subjects at once we have, up to now, avoided using const member functions in the example programs. However, there are many places where const member functions should be used. For example, in the Distance class, shown in several programs, the showdist() member function could be made const because it doesn’t (or certainly shouldn’t!) modify any of the data in the object for which it was called. It should simply display the data. Also, in ENGLRET,the add_dist() function should not modify any of the data in the object for which it was called. This object should simply be added to the object passed as an argument, and the resulting sum should be returned. We’ve modified the ENGLRET program to show how these two constant functions look. Note that const is used in both the declaration and the defi- nition of add_dist(). Here’s the listing for ENGCONST: // engConst.cpp // const member functions and const arguments to member functions #include using namespace std; //////////////////////////////////////////////////////////////// class Distance //English Distance class { private: int feet; float inches; public: //constructor (no args) Distance() : feet(0), inches(0.0) { } //constructor (two args) Distance(int ft, float in) : feet(ft), inches(in) { } void getdist() //get length from user { cout << “\nEnter feet: “; cin >> feet; cout << “Enter inches: “; cin >> inches; } void showdist() const //display distance { cout << feet << “\’-” << inches << ‘\”’; } Distance add_dist(const Distance&) const; //add }; //-------------------------------------------------------------- //add this distance to d2, return the sum Objects and Classes 6 O BJECTS AND C LASSES 253 07 3087 CH06 11/29/01 2:16 PM Page 253Distance Distance::add_dist(const Distance& d2) const { Distance temp; //temporary variable // feet = 0; //ERROR: can’t modify this // d2.feet = 0; //ERROR: can’t modify d2 temp.inches = inches + d2.inches; //add the inches if(temp.inches >= 12.0) //if total exceeds 12.0, { //then decrease inches temp.inches -= 12.0; //by 12.0 and temp.feet = 1; //increase feet } //by 1 temp.feet += feet + d2.feet; //add the feet return temp; } //////////////////////////////////////////////////////////////// int main() { Distance dist1, dist3; //define two lengths Distance dist2(11, 6.25); //define, initialize dist2 dist1.getdist(); //get dist1 from user dist3 = dist1.add_dist(dist2); //dist3 = dist1 + dist2 //display all lengths cout << “\ndist1 = “; dist1.showdist(); cout << “\ndist2 = “; dist2.showdist(); cout << “\ndist3 = “; dist3.showdist(); cout << endl; return 0; } Here, showdist() and add_dist() are both constant member functions. In add_dist() we show in the first commented statement, feet = 0,that a compiler error is generated if you attempt to modify any of the data in the object for which this constant function was called. const Member Function Arguments We mentioned in Chapter 5 that if an argument is passed to an ordinary function by reference, and you don’t want the function to modify it, the argument should be made const in the func- tion declaration (and definition). This is true of member functions as well. In ENGCONST the argument to add_dist() is passed by reference, and we want to make sure that ENGCONST won’t modify this variable, which is dist2 in main(). Therefore we make the argument d2 to add_dist() const in both declaration and definition. The second commented statement shows that the compiler will flag as an error any attempt by add_dist() to modify any member data of its argument dist2. Chapter 6254 07 3087 CH06 11/29/01 2:16 PM Page 254const Objects In several example programs, we’ve seen that we can apply const to variables of basic types such as int to keep them from being modified. In a similar way, we can apply const to objects of classes. When an object is declared as const, you can’t modify it. It follows that you can use only const member functions with it, because they’re the only ones that guarantee not to modify it. The CONSTOBJ program shows an example. // constObj.cpp // constant Distance objects #include using namespace std; //////////////////////////////////////////////////////////////// class Distance //English Distance class { private: int feet; float inches; public: //2-arg constructor Distance(int ft, float in) : feet(ft), inches(in) { } void getdist() //user input; non-const func { cout << “\nEnter feet: “; cin >> feet; cout << “Enter inches: “; cin >> inches; } void showdist() const //display distance; const func { cout << feet << “\’-” << inches << ‘\”’; } }; //////////////////////////////////////////////////////////////// int main() { const Distance football(300, 0); // football.getdist(); //ERROR: getdist() not const cout << “football = “; football.showdist(); //OK cout << endl; return 0; } A football field (for American-style football) is exactly 300 feet long. If we were to use the length of a football field in a program, it would make sense to make it const, because changing it would represent the end of the world for football fans. The CONSTOBJ program makes football a const variable. Now only const functions, such as showdist(), can be called for this object. Non-const functions, such as getdist(), which gives the object a new value obtained from the user, are illegal. In this way the compiler enforces the const value of football. Objects and Classes 6 O BJECTS AND C LASSES 255 07 3087 CH06 11/29/01 2:16 PM Page 255When you’re designing classes it’s a good idea to make const any function that does not modify any of the data in its object. This allows the user of the class to create const objects. These objects can use any const function, but cannot use any non-const function. Remember, using const helps the compiler to help you. What Does It All Mean? Now that you’ve been introduced to classes and objects, you may wonder what benefit they really offer. After all, as you can see by comparing several of the programs in this chapter with those in Chapter 4, it’s possible to do the same sorts of things with a procedural approach as it is with objects. One benefit of OOP that you may have glimpsed already is the close correspondence between the real-world things being modeled by the program and the C++ objects in the program. A widget part object in a program represents a widget part in the real world, a card object represents a card, a circle object represents a graphics circle, and so on. In C++, everything about a wid- get part is included in its class description—the part number and other data items, and the func- tions necessary to access and operate on this data. This makes it easy to conceptualize a programming problem. You figure out what parts of the problem can be most usefully repre- sented as objects, and then put all the data and functions connected with that object into the class. If you’re using a C++ class to represent a playing card, you put into this class the data items that represent the value of the card, and also the functions to set value, retrieve it, display it, compare it, and so on. In a procedural program, by contrast, the global variables and functions connected with a real- world object are distributed all over the listing; they don’t form a single, easily grasped unit. In some situations it may not be obvious what parts of a real-life situation should be made into objects. If you’re writing a program that plays chess, for instance, what are the objects? The chessmen, the squares on the board, or possibly entire board positions? In small programs, such as many of the ones in this book, you can often proceed by trial and error. You break a problem into objects in one way and write trial class definitions for these objects. If the classes seem to match reality in a useful way, you continue. If they don’t, you may need to start over, selecting different entities to be classes. The more experience you have with OOP, the easier it will be to break a programming problem into classes. Larger programs may prove too complex for this trial-and-error approach. A new field, object- oriented design (OOD) is increasingly applied to analyzing a programming problem and figuring out what classes and objects should be used to represent the real-world situation (which is often called the problem domain). We’ll discuss this methodology in detail in Chapter 16, “Object-Oriented Software Development.” Chapter 6256 07 3087 CH06 11/29/01 2:16 PM Page 256Some of the benefits of object-oriented programming are probably not apparent at this point. Remember that OOP was devised to cope with the complexity of large programs. Smaller programs, such as the examples in this chapter, have less need for the organizational power that OOP provides. The larger the program, the greater the benefit. But even for small programs, once you start thinking in object-oriented terms, the OO design approach becomes natural and surprisingly helpful. One advantage is that in an OO program the compiler can find many more conceptual errors than in a procedural program. Summary A class is a specification or blueprint for a number of objects. Objects consist of both data and functions that operate on that data. In a class definition, the members—whether data or functions— can be private, meaning they can be accessed only by member functions of that class, or public, meaning they can be accessed by any function in the program. A member function is a function that is a member of a class. Member functions have access to an object’s private data, while non-member functions do not. A constructor is a member function, with the same name as its class, that is executed every time an object of the class is created. A constructor has no return type but can take arguments. It is often used to give initial values to object data members. Constructors can be overloaded, so an object can be initialized in different ways. A destructor is a member function with the same name as its class but preceded by a tilde (~). It is called when an object is destroyed. A destructor takes no arguments and has no return value. In the computer’s memory there is a separate copy of the data members for each object that is created from a class, but there is only one copy of a class’s member functions. You can restrict a data item to a single instance for all objects of a class by making it static. One reason to use OOP is the close correspondence between real-world objects and OOP classes. Deciding what objects and classes to use in a program can be complicated. For small programs, trial and error may be sufficient. For large programs, a more systematic approach is usually needed. Questions Answers to these questions can be found in Appendix G. 1. What is the purpose of a class definition? 2. A ________ has the same relation to an ________ that a basic data type has to a variable of that type. Objects and Classes 6 O BJECTS AND C LASSES 257 07 3087 CH06 11/29/01 2:16 PM Page 2573. In a class definition, data or functions designated private are accessible a. to any function in the program. b. only if you know the password. c. to member functions of that class. d. only to public members of the class. 4. Write a class definition that creates a class called leverage with one private data member, crowbar,of type int and one public function whose declaration is void pry(). 5. True or false: Data items in a class must be private. 6. Write a statement that defines an object called lever1 of the leverage class described in Question 4. 7. The dot operator (or class member access operator) connects the following two entities (reading from left to right): a. A class member and a class object b. A class object and a class c. A class and a member of that class d. A class object and a member of that class 8. Write a statement that executes the pry() function in the lever1 object, as described in Questions 4 and 6. 9. Member functions defined inside a class definition are ________ by default. 10. Write a member function called getcrow() for the leverage class described in Question 4. This function should return the value of the crowbar data. Assume the function is defined within the class definition. 11. A constructor is executed automatically when an object is ________. 12. A constructor’s name is the same as _________. 13. Write a constructor that initializes to 0 the crowbar data, a member of the leverage class described in Question 4. Assume that the constructor is defined within the class defini- tion. 14. True or false: In a class you can have more than one constructor with the same name. 15. A member function can always access the data a. in the object of which it is a member. b. in the class of which it is a member. c. in any object of the class of which it is a member. d. in the public part of its class. 16. Assume that the member function getcrow() described in Question 10 is defined outside the class definition. Write the declaration that goes inside the class definition. Chapter 6258 07 3087 CH06 11/29/01 2:16 PM Page 25817. Write a revised version of the getcrow() member function from Question 10 that is defined outside the class definition. 18. The only technical difference between structures and classes in C++ is that _________. 19. If three objects of a class are defined, how many copies of that class’s data items are stored in memory? How many copies of its member functions? 20. Sending a message to an object is the same as _________. 21. Classes are useful because they a. are removed from memory when not in use. b. permit data to be hidden from other classes. c. bring together all aspects of an entity in one place. d. can closely model objects in the real world. 22. True or false: There is a simple but precise methodology for dividing a real-world programming problem into classes. 23. For the object for which it was called, a const member function a. can modify both const and non-const member data. b. can modify only const member data. c. can modify only non-const member data. d. can modify neither const nor non-const member data. 24. True or false: If you declare a const object, it can only be used with const member functions. 25. Write a declaration (not a definition) for a const void function called aFunc() that takes one const argument called jerry of type float. Exercises Answers to the starred exercises can be found in Appendix G. *1. Create a class that imitates part of the functionality of the basic data type int. Call the class Int (note different capitalization). The only data in this class is an int variable. Include member functions to initialize an Int to 0, to initialize it to an int value, to dis- play it (it looks just like an int), and to add two Int values. Write a program that exercises this class by creating one uninitialized and two initialized Int values, adding the two initialized values and placing the response in the uninitialized value, and then displaying this result. *2. Imagine a tollbooth at a bridge. Cars passing by the booth are expected to pay a 50 cent toll. Mostly they do, but sometimes a car goes by without paying. The tollbooth keeps track of the number of cars that have gone by, and of the total amount of money collected. Objects and Classes 6 O BJECTS AND C LASSES 259 07 3087 CH06 11/29/01 2:16 PM Page 259Model this tollbooth with a class called tollBooth. The two data items are a type unsigned int to hold the total number of cars, and a type double to hold the total amount of money collected. A constructor initializes both of these to 0. A member function called payingCar() increments the car total and adds 0.50 to the cash total. Another function, called nopayCar(),increments the car total but adds nothing to the cash total. Finally, a member function called display() displays the two totals. Make appropriate member functions const. Include a program to test this class. This program should allow the user to push one key to count a paying car, and another to count a nonpaying car. Pushing the Esc key should cause the program to print out the total cars and total cash and then exit. *3. Create a class called time that has separate int member data for hours, minutes, and seconds. One constructor should initialize this data to 0, and another should initialize it to fixed values. Another member function should display it, in 11:59:59 format. The final member function should add two objects of type time passed as arguments. A main() program should create two initialized time objects (should they be const?) and one that isn’t initialized. Then it should add the two initialized values together, leaving the result in the third time variable. Finally it should display the value of this third variable. Make appropriate member functions const. 4. Create an employee class, basing it on Exercise 4 of Chapter 4. The member data should comprise an int for storing the employee number and a float for storing the employee’s compensation. Member functions should allow the user to enter this data and display it. Write a main() that allows the user to enter data for three employees and display it. 5. Start with the date structure in Exercise 5 in Chapter 4 and transform it into a date class. Its member data should consist of three ints: month, day, and year. It should also have two member functions: getdate(), which allows the user to enter a date in 12/31/02 format, and showdate(), which displays the date. 6. Extend the employee class of Exercise 4 to include a date class (see Exercise 5) and an etype enum (see Exercise 6 in Chapter 4). An object of the date class should be used to hold the date of first employment; that is, the date when the employee was hired. The etype variable should hold the employee’s type: laborer, secretary, manager, and so on. These two items will be private member data in the employee definition, just like the employee number and salary. You’ll need to extend the getemploy() and putemploy() functions to obtain this new information from the user and display it. These functions will probably need switch statements to handle the etype variable. Write a main() program that allows the user to enter data for three employee variables and then displays this data. 7. In ocean navigation, locations are measured in degrees and minutes of latitude and longi- tude. Thus if you’re lying off the mouth of Papeete Harbor in Tahiti, your location is 149 degrees 34.8 minutes west longitude, and 17 degrees 31.5 minutes south latitude. This is Chapter 6260 07 3087 CH06 11/29/01 2:16 PM Page 260written as 149°34.8’ W, 17°31.5’ S. There are 60 minutes in a degree. (An older system also divided a minute into 60 seconds, but the modern approach is to use decimal minutes instead.) Longitude is measured from 0 to 180 degrees, east or west from Greenwich, England, to the international dateline in the Pacific. Latitude is measured from 0 to 90 degrees, north or south from the equator to the poles. Create a class angle that includes three member variables: an int for degrees, a float for minutes, and a char for the direction letter (N, S, E, or W). This class can hold either a latitude variable or a longitude variable. Write one member function to obtain an angle value (in degrees and minutes) and a direction from the user, and a second to display the angle value in 179°59.9’ E format. Also write a three-argument constructor. Write a main() program that displays an angle initialized with the constructor, and then, within a loop, allows the user to input any angle value, and then displays the value. You can use the hex character constant ‘\xF8’, which usually prints a degree (°) symbol. 8. Create a class that includes a data member that holds a “serial number” for each object created from the class. That is, the first object created will be numbered 1, the second 2, and so on. To do this, you’ll need another data member that records a count of how many objects have been created so far. (This member should apply to the class as a whole; not to individual objects. What keyword specifies this?) Then, as each object is created, its constructor can examine this count member variable to determine the appropriate serial number for the new object. Add a member function that permits an object to report its own serial number. Then write a main() program that creates three objects and queries each one about its serial number. They should respond I am object number 2,and so on. 9. Transform the fraction structure from Exercise 8 in Chapter 4 into a fraction class. Member data is the fraction’s numerator and denominator. Member functions should accept input from the user in the form 3/5, and output the fraction’s value in the same format. Another member function should add two fraction values. Write a main() program that allows the user to repeatedly input two fractions and then displays their sum. After each operation, ask whether the user wants to continue. 10. Create a class called ship that incorporates a ship’s number and location. Use the approach of Exercise 8 to number each ship object as it is created. Use two variables of the angle class from Exercise 7 to represent the ship’s latitude and longitude. A member function of the ship class should get a position from the user and store it in the object; another should report the serial number and position. Write a main() program that cre- ates three ships, asks the user to input the position of each, and then displays each ship’s number and position. Objects and Classes 6 O BJECTS AND C LASSES 261 07 3087 CH06 11/29/01 2:16 PM Page 26111. Modify the four-function fraction calculator of Exercise 12 in Chapter 5 to use a fraction class rather than a structure. There should be member functions for input and output, as well as for the four arithmetical operations. While you’re at it, you might as well install the capability to reduce fractions to lowest terms. Here’s a member function that will reduce the fraction object of which it is a member to lowest terms. It finds the greatest common divisor (gcd) of the fraction’s numerator and denominator, and uses this gcd to divide both numbers. void fraction::lowterms() // change ourself to lowest terms { long tnum, tden, temp, gcd; tnum = labs(num); // use non-negative copies tden = labs(den); // (needs cmath) if(tden==0 ) // check for n/0 { cout << “Illegal fraction: division by 0”; exit(1); } else if( tnum==0 ) // check for 0/n { num=0; den = 1; return; } // this ‘while’ loop finds the gcd of tnum and tden while(tnum != 0) { if(tnum < tden) // ensure numerator larger { temp=tnum; tnum=tden; tden=temp; } // swap them tnum = tnum - tden; // subtract them } gcd = tden; // this is greatest common divisor num = num / gcd; // divide both num and den by gcd den = den / gcd; // to reduce frac to lowest terms } You can call this function at the end of each arithmetic function, or just before you per- form output. You’ll also need the usual member functions: four arithmetic operations, input, and display. You may find a two-argument constructor useful. 12. Note that one advantage of the OOP approach is that an entire class can be used, without modification, in a different program. Use the fraction class from Exercise 11 in a pro- gram that generates a multiplication table for fractions. Let the user input a denominator, and then generate all combinations of two such fractions that are between 0 and 1, and multiply them together. Here’s an example of the output if the denominator is 6: 1/6 1/3 1/2 2/3 5/6 ----------------------------------------- 1/6 1/36 1/18 1/12 1/9 5/36 1/3 1/18 1/9 1/6 2/9 5/18 1/2 1/12 1/6 1/4 1/3 5/12 2/3 1/9 2/9 1/3 4/9 5/9 5/6 5/36 5/18 5/12 5/9 25/36 Chapter 6262 07 3087 CH06 11/29/01 2:16 PM Page 262CHAPTER 7 Arrays and Strings IN THIS CHAPTER •Array Fundamentals 264 •Arrays as Class Member Data 279 •Arrays of Objects 283 •C-Strings 290 • The Standard C++ string Class 302 08 3087 CH07 11/29/01 2:14 PM Page 263Chapter 7264 In everyday life we commonly group similar objects into units. We buy peas by the can and eggs by the carton. In computer languages we also need to group together data items of the same type. The most basic mechanism that accomplishes this in C++ is the array. Arrays can hold a few data items or tens of thousands. The data items grouped in an array can be simple types such as int or float,or they can be user-defined types such as structures and objects. Arrays are like structures in that they both group a number of items into a larger unit. But while a structure usually groups items of different types, an array groups items of the same type. More importantly, the items in a structure are accessed by name, while those in an array are accessed by an index number. Using an index number to specify an item allows easy access to a large number of items. Arrays exist in almost every computer language. Arrays in C++ are similar to those in other languages, and identical to those in C. In this chapter we’ll look first at arrays of basic data types such as int and char. Then we’ll examine arrays used as data members in classes, and arrays used to hold objects. Thus this chapter is intended not only to introduce arrays, but to increase your understanding of object- oriented programming. In Standard C++ the array is not the only way to group elements of the same type. A vector, which is part of the Standard Template library, is another approach. We’ll look at vectors in Chapter 15, “The Standard Template Library.” In this chapter we’ll also look at two different approaches to strings, which are used to store and manipulate text. The first kind of string is an array of type char, and the second is a member of the Standard C++ string class. Array Fundamentals A simple example program will serve to introduce arrays. This program, REPLAY,creates an array of four integers representing the ages of four people. It then asks the user to enter four values, which it places in the array. Finally, it displays all four values. // replay.cpp // gets four ages from user, displays them #include using namespace std; int main() { int age[4]; //array ‘age’ of 4 ints 08 3087 CH07 11/29/01 2:14 PM Page 264for(int j=0; j<4; j++) //get 4 ages { cout << “Enter an age: “; cin >> age[j]; //access array element } for(j=0; j<4; j++) //display 4 ages cout << “You entered “ << age[j] << endl; return 0; } Here’s a sample interaction with the program: Enter an age: 44 Enter an age: 16 Enter an age: 23 Enter an age: 68 You entered 44 You entered 16 You entered 23 You entered 68 The first for loop gets the ages from the user and places them in the array, while the second reads them from the array and displays them. Defining Arrays Like other variables in C++, an array must be defined before it can be used to store information. And, like other definitions, an array definition specifies a variable type and a name. But it includes another feature: a size. The size specifies how many data items the array will contain. It immediately follows the name, and is surrounded by square brackets. Figure 7.1 shows the syntax of an array definition. In the REPLAY example, the array is type int. The name of the array comes next, followed immediately by an opening bracket, the array size, and a closing bracket. The number in brackets must be a constant or an expression that evaluates to a constant, and should also be an integer. In the example we use the value 4. Array Elements The items in an array are called elements (in contrast to the items in a structure, which are called members). As we noted, all the elements in an array are of the same type; only the values vary. Figure 7.2 shows the elements of the array age. Arrays and Strings 7 A RRAYS AND S TRINGS 265 08 3087 CH07 11/29/01 2:14 PM Page 265FIGURE 7.1 Syntax of array definition. Chapter 7266 age[0] age[1] age[2] age[3] 44 16 23 68 Memory FIGURE 7.2 Array elements. Following the conventional (although in some ways backward) approach, memory grows downward in the figure. That is, the first array elements are on the top of the page; later elements extend downward. As specified in the definition, the array has exactly four elements. 08 3087 CH07 11/29/01 2:14 PM Page 266Notice that the first array element is numbered 0. Thus, since there are four elements, the last one is number 3. This is a potentially confusing situation; you might think the last element in a four-element array would be number 4, but it’s not. Accessing Array Elements In the REPLAY example we access each array element twice. The first time, we insert a value into the array, with the line cin >> age[j]; The second time, we read it out with the line cout << “\nYou entered “ << age[j]; In both cases the expression for the array element is age[j] This consists of the name of the array, followed by brackets delimiting a variable j. Which of the four array elements is specified by this expression depends on the value of j; age[0] refers to the first element, age[1] to the second, age[2] to the third, and age[3] to the fourth. The variable (or constant) in the brackets is called the array index. Since j is the loop variable in both for loops, it starts at 0 and is incremented until it reaches 3, thereby accessing each of the array elements in turn. Averaging Array Elements Here’s another example of an array at work. This one, SALES,invites the user to enter a series of six values representing widget sales for each day of the week (excluding Sunday), and then calculates the average of these values. We use an array of type double so that monetary values can be entered. // sales.cpp // averages a weeks’s widget sales (6 days) #include using namespace std; int main() { const int SIZE = 6; //size of array double sales[SIZE]; //array of 6 variables cout << “Enter widget sales for 6 days\n”; for(int j=0; j> sales[j]; Arrays and Strings 7 A RRAYS AND S TRINGS 267 08 3087 CH07 11/29/01 2:14 PM Page 267double total = 0; for(j=0; j using namespace std; int main() { int month, day, total_days; int days_per_month[12] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }; cout << “\nEnter month (1 to 12): “; //get date cin >> month; cout << “Enter day (1 to 31): “; Chapter 7268 08 3087 CH07 11/29/01 2:14 PM Page 268cin >> day; total_days = day; //separate days for(int j=0; j #include //for setprecision, etc. using namespace std; const int DISTRICTS = 4; //array dimensions const int MONTHS = 3; int main() { int d, m; double sales[DISTRICTS][MONTHS]; //two-dimensional array //definition cout << endl; for(d=0; d> sales[d][m]; //put number in array } cout << “\n\n”; cout << “ Month\n”; cout << “ 1 2 3”; for(d=0; d #include //for setprecision, etc. using namespace std; const int DISTRICTS = 4; //array dimensions const int MONTHS = 3; int main() { int d, m; //initialize array elements double sales[DISTRICTS][MONTHS] Arrays and Strings 7 A RRAYS AND S TRINGS 273 08 3087 CH07 11/29/01 2:14 PM Page 273= { { 1432.07, 234.50, 654.01 }, { 322.00, 13838.32, 17589.88 }, { 9328.34, 934.00, 4492.30 }, { 12838.29, 2332.63, 32.93 } }; cout << “\n\n”; cout << “ Month\n”; cout << “ 1 2 3”; for(d=0; d #include //for setprecision, etc. using namespace std; const int DISTRICTS = 4; //array dimensions const int MONTHS = 3; void display( double[DISTRICTS][MONTHS] ); //declaration //-------------------------------------------------------------- int main() { //initialize two-dimensional array double sales[DISTRICTS][MONTHS] Chapter 7274 08 3087 CH07 11/29/01 2:14 PM Page 274= { { 1432.07, 234.50, 654.01 }, { 322.00, 13838.32, 17589.88 }, { 9328.34, 934.00, 4492.30 }, { 12838.29, 2332.63, 32.93 } }; display(sales); //call function; array as argument cout << endl; return 0; } //end main //-------------------------------------------------------------- //display() //function to display 2-d array passed as argument void display( double funsales[DISTRICTS][MONTHS] ) { int d, m; cout << “\n\n”; cout << “ Month\n”; cout << “ 1 2 3”; for(d=0; d using namespace std; const int SIZE = 4; //number of parts in array //////////////////////////////////////////////////////////////// struct part //specify a structure { int modelnumber; //ID number of widget int partnumber; //ID number of widget part float cost; //cost of part }; //////////////////////////////////////////////////////////////// int main() { int n; part apart[SIZE]; //define array of structures for(n=0; n> apart[n].modelnumber; //get model number cout << “Enter part number: “; cin >> apart[n].partnumber; //get part number cout << “Enter cost: “; cin >> apart[n].cost; //get cost } cout << endl; for(n=0; n using namespace std; //////////////////////////////////////////////////////////////// class Stack { private: enum { MAX = 10 }; //(non-standard syntax) int st[MAX]; //stack: array of integers int top; //number of top of stack public: Stack() //constructor { top = 0; } void push(int var) //put number on stack { st[++top] = var; } int pop() //take number off stack { return st[top--]; } }; //////////////////////////////////////////////////////////////// int main() { Stack s1; s1.push(11); s1.push(22); cout << “1: “ << s1.pop() << endl; //22 cout << “2: “ << s1.pop() << endl; //11 s1.push(33); s1.push(44); s1.push(55); s1.push(66); cout << “3: “ << s1.pop() << endl; //66 cout << “4: “ << s1.pop() << endl; //55 cout << “5: “ << s1.pop() << endl; //44 cout << “6: “ << s1.pop() << endl; //33 return 0; } The important member of the stack is the array st. An int variable, top,indicates the index of the last item placed on the stack; the location of this item is the top of the stack. The size of the array used for the stack is specified by MAX,in the statement enum { MAX = 10 }; Chapter 7280 08 3087 CH07 11/29/01 2:14 PM Page 280This definition of MAX is unusual. In keeping with the philosophy of encapsulation, it’s preferable to define constants that will be used entirely within a class, as MAX is here, within the class. Thus the use of global const variables for this purpose is nonoptimal. Standard C++ mandates that we should be able to declare MAX within the class as static const int MAX = 10; This means that MAX is constant and applies to all objects in the class. Unfortunately, some compilers, including the current version of Microsoft Visual C++, do not allow this newly- approved construction. As a workaround we can define such constants to be enumerators (described in Chapter 4). We don’t need to name the enumeration, and we need only the one enumerator: enum { MAX = 10 }; This defines MAX as an integer with the value 10, and the definition is contained entirely within the class. This approach works, but it’s awkward. If your compiler supports the static const approach, you should use it instead to define constants within the class. Figure 7.6 shows a stack. Since memory grows downward in the figure, the top of the stack is at the bottom in the figure. When an item is added to the stack, the index in top is incremented to point to the new top of the stack. When an item is removed, the index in top is decremented. (We don’t need to erase the old value left in memory when an item is removed; it just becomes irrelevant.) To place an item on the stack—a process called pushing the item—you call the push() mem- ber function with the value to be stored as an argument. To retrieve (or pop) an item from the stack, you use the pop() member function, which returns the value of the item. The main() program in STAKARAY exercises the stack class by creating an object, s1,of the class. It pushes two items onto the stack, and pops them off and displays them. Then it pushes four more items onto the stack, and pops them off and displays them. Here’s the output: 1: 22 2: 11 3: 66 4: 55 5: 44 6: 33 Arrays and Strings 7 A RRAYS AND S TRINGS 281 08 3087 CH07 11/29/01 2:14 PM Page 281FIGURE 7.6 A stack. As you can see, items are popped off the stack in reverse order; the last thing pushed is the first thing popped. Notice the subtle use of prefix and postfix notation in the increment and decrement operators. The statement st[++top] = var; in the push() member function first increments top so that it points to the next available array element—one past the last element. It then assigns var to this element, which becomes the new top of the stack. The statement return st[top--]; first returns the value it finds at the top of the stack, then decrements top so that it points to the preceding element. Chapter 7282 08 3087 CH07 11/29/01 2:14 PM Page 282The stack class is an example of an important feature of object-oriented programming: using a class to implement a container or data-storage mechanism. In Chapter 15 we’ll see that a stack is only one of a number of ways to store data. There are also queues, sets, linked lists, and so on. A data-storage scheme is chosen that matches the specific requirements of the program. Using a preexisting class to provide data storage means that the programmer does not need to waste time duplicating the details of the data-storage mechanism. Arrays of Objects We’ve seen how an object can contain an array. We can also reverse that situation and create an array of objects. We’ll look at two situations: an array of English distances and a deck of cards. Arrays of English Distances In Chapter 6, “Objects and Classes,” we showed several examples of an English Distance class that incorporated feet and inches into an object representing a new data type. The next program, ENGLARAY, demonstrates an array of such objects. // englaray.cpp // objects using English measurements #include using namespace std; //////////////////////////////////////////////////////////////// class Distance //English Distance class { private: int feet; float inches; public: void getdist() //get length from user { cout << “\n Enter feet: “; cin >> feet; cout << “ Enter inches: “; cin >> inches; } void showdist() const //display distance { cout << feet << “\’-” << inches << ‘\”’; } }; //////////////////////////////////////////////////////////////// int main() { Distance dist[100]; //array of distances int n=0; //count the entries char ans; //user response (‘y’ or ‘n’) Arrays and Strings 7 A RRAYS AND S TRINGS 283 08 3087 CH07 11/29/01 2:14 PM Page 283cout << endl; do { //get distances from user cout << “Enter distance number “ << n+1; dist[n++].getdist(); //store distance in array cout << “Enter another (y/n)?: “; cin >> ans; } while( ans != ‘n’ ); //quit if user types ‘n’ for(int j=0; j= MAX ) { cout << “\nThe array is full!!!”; break; } This causes a break out of the loop and prevents the array from overflowing. Accessing Objects in an Array The declaration of the Distance class in this program is similar to that used in previous programs. However, in the main() program we define an array of such objects: Distance dist[MAX]; Here the data type of the dist array is Distance, and it has MAX elements. Figure 7.7 shows what this looks like. A class member function that is an array element is accessed similarly to a structure member that is an array element, as in the PARTARAY example. Here’s how the showdist() member function of the jth element of the array dist is invoked: dist[j].showdist(); As you can see, a member function of an object that is an array element is accessed using the dot operator: The array name followed by the index in brackets is joined, using the dot opera- tor, to the member function name followed by parentheses. This is similar to accessing a struc- ture (or class) data member, except that the function name and parentheses are used instead of the data name. Notice that when we call the getdist() member function to put a distance into the array, we take the opportunity to increment the array index n: dist[n++].getdist(); Arrays and Strings 7 A RRAYS AND S TRINGS 285 08 3087 CH07 11/29/01 2:14 PM Page 285FIGURE 7.7 Array of objects. This way the next group of data obtained from the user will be placed in the structure in the next array element in dist. The n variable must be incremented manually like this because we use a do loop instead of a for loop. In the for loop, the loop variable—which is incremented automatically—can serve as the array index. Arrays of Cards Here’s another, somewhat longer, example of an array of objects. You will no doubt remember the CARDOBJ example from Chapter 6. We’ll borrow the card class from that example, and group an array of 52 such objects together in an array, thus creating a deck of cards. Here’s the listing for CARDARAY: // cardaray.cpp // cards as objects #include Chapter 7286 08 3087 CH07 11/29/01 2:14 PM Page 286#include //for srand(), rand() #include //for time for srand() using namespace std; enum Suit { clubs, diamonds, hearts, spades }; //from 2 to 10 are integers without names const int jack = 11; const int queen = 12; const int king = 13; const int ace = 14; //////////////////////////////////////////////////////////////// class card { private: int number; //2 to 10, jack, queen, king, ace Suit suit; //clubs, diamonds, hearts, spades public: card() //constructor { } void set(int n, Suit s) //set card { suit = s; number = n; } void display(); //display card }; //-------------------------------------------------------------- void card::display() //display the card { if( number >= 2 && number <= 10 ) cout << number; else switch(number) { case jack: cout << “J”; break; case queen: cout << “Q”; break; case king: cout << “K”; break; case ace: cout << “A”; break; } switch(suit) { case clubs: cout << static_cast(5); break; case diamonds: cout << static_cast(4); break; case hearts: cout << static_cast(3); break; case spades: cout << static_cast(6); break; } } //////////////////////////////////////////////////////////////// Arrays and Strings 7 A RRAYS AND S TRINGS 287 08 3087 CH07 11/29/01 2:14 PM Page 287int main() { card deck[52]; int j; cout << endl; for(j=0; j<52; j++) //make an ordered deck { int num = (j % 13) + 2; //cycles through 2 to 14, 4 times Suit su = Suit(j / 13); //cycles through 0 to 3, 13 times deck[j].set(num, su); //set card } cout << “\nOrdered deck:\n”; for(j=0; j<52; j++) //display ordered deck { deck[j].display(); cout << “ “; if( !( (j+1) % 13) ) //newline every 13 cards cout << endl; } srand( time(NULL) ); //seed random numbers with time for(j=0; j<52; j++) //for each card in the deck, { int k = rand() % 52; //pick another card at random card temp = deck[j]; //and swap them deck[j] = deck[k]; deck[k] = temp; } cout << “\nShuffled deck:\n”; for(j=0; j<52; j++) //display shuffled deck { deck[j].display(); cout << “, “; if( !( (j+1) % 13) ) //newline every 13 cards cout << endl; } return 0; } //end main Once we’ve created a deck, it’s hard to resist the temptation to shuffle it. We display the cards in the deck, shuffle it, and then display it again. To conserve space we use graphics characters for the club, diamond, heart, and spade. Figure 7.8 shows the output from the program. This program incorporates several new ideas, so let’s look at them in turn. Chapter 7288 08 3087 CH07 11/29/01 2:14 PM Page 288FIGURE 7.8 Output of the CARDARAY program. Graphics Characters There are several special graphics characters in the range below ASCII code 32. (See Appendix A, “ASCII Table,” for a list of ASCII codes.) In the display() member function of card we use codes 5, 4, 3, and 6 to access the characters for a club, a diamond, a heart, and a spade, respectively. Casting these numbers to type char,as in static_cast(5) causes the << operator to print them as characters rather than as numbers. The Card Deck The array of structures that constitutes the deck of cards is defined in the statement card deck[52]; which creates an array called deck, consisting of 52 objects of type card. To display the jth card in the deck, we call the display() member function: deck[j].display(); Random Numbers It’s always fun and sometimes even useful to generate random numbers. In this program we use them to shuffle the deck. Two steps are necessary to obtain random numbers. First the random-number generator must be seeded,or initialized. To do this we call the srand() library function. This function uses the system time as the seed, so it requires two header files, CSTDLIB and CTIME. To actually generate a random number we call the rand() library function. This function returns a random integer. To get a number in the range from 0 to 51, we apply the remainder operator and 52 to the result of rand(). int k = rand() % 52; Arrays and Strings 7 A RRAYS AND S TRINGS 289 08 3087 CH07 11/29/01 2:14 PM Page 289The resulting random number k is then used as an index to swap two cards. We go through the for loop, swapping one card, whose index points to each card in 0-to-51 order, with another card, whose index is the random number. When all 52 cards have been exchanged with a random card, the deck is considered to be shuffled. This program could form the basis for a card-playing program, but we’ll leave these details for you. Arrays of objects are widely used in C++ programming. We’ll see other examples as we go along. C-Strings We noted at the beginning of this chapter that two kinds of strings are commonly used in C++: C-strings and strings that are objects of the string class. In this section we’ll describe the first kind, which fits the theme of the chapter in that C-strings are arrays of type char. We call these strings C-strings,or C-style strings, because they were the only kind of strings available in the C language (and in the early days of C++, for that matter). They may also be called char* strings, because they can be represented as pointers to type char. (The * indicates a pointer, as we’ll learn in Chapter 10.) Although strings created with the string class, which we’ll examine in the next section, have superseded C-strings in many situations, C-strings are still important for a variety of reasons. First, they are used in many C library functions. Second, they will continue to appear in legacy code for years to come. And third, for students of C++, C-strings are more primitive and therefore easier to understand on a fundamental level. C-String Variables As with other data types, strings can be variables or constants. We’ll look at these two entities before going on to examine more complex string operations. Here’s an example that defines a single string variable. (In this section we’ll assume the word string refers to a C-string.) It asks the user to enter a string, and places this string in the string variable. Then it displays the string. Here’s the listing for STRINGIN: // stringin.cpp // simple string variable #include using namespace std; int main() { const int MAX = 80; //max characters in string char str[MAX]; //string variable str Chapter 7290 08 3087 CH07 11/29/01 2:14 PM Page 290cout << “Enter a string: “; cin >> str; //put string in str //display string from str cout << “You entered: “ << str << endl; return 0; } The definition of the string variable str looks like (and is) the definition of an array of type char: char str[MAX]; We use the extraction operator >> to read a string from the keyboard and place it in the string variable str. This operator knows how to deal with strings; it understands that they are arrays of characters. If the user enters the string “Amanuensis” (one employed to copy manuscripts) in this program, the array str will look something like Figure 7.9. Arrays and Strings 7 A RRAYS AND S TRINGS 291 FIGURE 7.9 String stored in string variable. Each character occupies 1 byte of memory. An important aspect of C-strings is that they must terminate with a byte containing 0. This is often represented by the character constant ‘\0’, which is a character with an ASCII value of 0. This terminating zero is called the null character. When the << operator displays the string, it displays characters until it encounters the null character. 08 3087 CH07 11/29/01 2:14 PM Page 291Avoiding Buffer Overflow The STRINGIN program invites the user to type in a string. What happens if the user enters a string that is longer than the array used to hold it? As we mentioned earlier, there is no built-in mechanism in C++ to keep a program from inserting array elements outside an array. So an overly enthusiastic typist could end up crashing the system. However, it is possible to tell the >> operator to limit the number of characters it places in an array. The SAFETYIN program demonstrates this approach. // safetyin.cpp // avoids buffer overflow with cin.width #include #include //for setw using namespace std; int main() { const int MAX = 20; //max characters in string char str[MAX]; //string variable str cout << “\nEnter a string: “; cin >> setw(MAX) >> str; //put string in str, // no more than MAX chars cout << “You entered: “ << str << endl; return 0; } This program uses the setw manipulator to specify the maximum number of characters the input buffer can accept. The user may type more characters, but the >> operator won’t insert them into the array. Actually, one character fewer than the number specified is inserted, so there is room in the buffer for the terminating null character. Thus, in SAFETYIN,a maximum of 19 characters are inserted. String Constants You can initialize a string to a constant value when you define it. Here’s an example, STRINIT, that does just that (with the first line of a Shakespearean sonnet): // strinit.cpp // initialized string #include using namespace std; int main() { char str[] = “Farewell! thou art too dear for my possessing.”; Chapter 7292 08 3087 CH07 11/29/01 2:14 PM Page 292cout << str << endl; return 0; } Here the string constant is written as a normal English phrase, delimited by quotes. This may seem surprising, since a string is an array of type char. In past examples you’ve seen arrays initialized to a series of values delimited by braces and separated by commas. Why isn’t str initialized the same way? In fact you could use such a sequence of character constants: char str[] = { ‘F’, ‘a’, ‘r’, ‘e’, ‘w’, ‘e’, ‘l’, ‘l’, ‘!’,’ ‘, ‘t’, ‘h’, and so on. Fortunately, the designers of C++ (and C) took pity on us and provided the shortcut approach shown in STRINIT. The effect is the same: The characters are placed one after the other in the array. As with all C-strings, the last character is a null (zero). Reading Embedded Blanks If you tried the STRINGIN program with strings that contained more than one word, you may have had an unpleasant surprise. Here’s an example: Enter a string: Law is a bottomless pit. You entered: Law Where did the rest of the phrase (a quotation from the Scottish writer John Arbuthnot, 1667– 1735) go? It turns out that the extraction operator >> considers a space to be a terminating character. Thus it will read strings consisting of a single word, but anything typed after a space is thrown away. To read text containing blanks we use another function, cin.get(). This syntax means a mem- ber function get() of the stream class of which cin is an object. The following example, BLANKSIN,shows how it’s used. // blanksin.cpp // reads string with embedded blanks #include using namespace std; int main() { const int MAX = 80; //max characters in string char str[MAX]; //string variable str cout << “\nEnter a string: “; cin.get(str, MAX); //put string in str cout << “You entered: “ << str << endl; return 0; } Arrays and Strings 7 A RRAYS AND S TRINGS 293 08 3087 CH07 11/29/01 2:14 PM Page 293The first argument to cin::get() is the array address where the string being input will be placed. The second argument specifies the maximum size of the array, thus automatically avoiding buffer overrun. Using this function, the input string is now stored in its entirety. Enter a string: Law is a bottomless pit. You entered: Law is a bottomless pit. There’s a potential problem when you mix cin.get() with cin and the extraction operator (>>). We’ll discuss the use of the ignore() member function of cin to solve this problem in Chapter 12, “Streams and Files.” Reading Multiple Lines We may have solved the problem of reading strings with embedded blanks, but what about strings with multiple lines? It turns out that the cin::get() function can take a third argument to help out in this situation. This argument specifies the character that tells the function to stop reading. The default value for this argument is the newline (‘\n’) character, but if you call the function with some other character for this argument, the default will be overridden by the specified character. In the next example, LINESIN, we call the function with a dollar sign (‘$’) as the third argument: // linesin.cpp // reads multiple lines, terminates on ‘$’ character #include using namespace std; const int MAX = 2000; //max characters in string char str[MAX]; //string variable str int main() { cout << “\nEnter a string:\n”; cin.get(str, MAX, ‘$’); //terminate with $ cout << “You entered:\n” << str << endl; return 0; } Now you can type as many lines of input as you want. The function will continue to accept characters until you enter the terminating character (or until you exceed the size of the array). Remember, you must still press Enter after typing the ‘$’ character. Here’s a sample interac- tion with a poem from Thomas Carew (1595–1639): Chapter 7294 08 3087 CH07 11/29/01 2:14 PM Page 294Enter a string: Ask me no more where Jove bestows When June is past, the fading rose; For in your beauty’s orient deep These flowers, as in their causes, sleep. $ You entered: Ask me no more where Jove bestows When June is past, the fading rose; For in your beauty’s orient deep These flowers, as in their causes, sleep. We terminate each line with Enter, but the program continues to accept input until we enter ‘$’. Copying a String the Hard Way The best way to understand the true nature of strings is to deal with them character by character. The following program does this. // strcopy1.cpp // copies a string using a for loop #include #include //for strlen() using namespace std; int main() { //initialized string char str1[] = “Oh, Captain, my Captain! “ “our fearful trip is done”; const int MAX = 80; //size of str2 buffer char str2[MAX]; //empty string for(int j=0; j #include //for strcpy() using namespace std; int main() { char str1[] = “Tiger, tiger, burning bright\n” “In the forests of the night”; const int MAX = 80; //size of str2 buffer char str2[MAX]; //empty string strcpy(str2, str1); //copy str1 to str2 cout << str2 << endl; //display str2 return 0; } Note that you call this function with the destination first: strcpy(destination, source) The right-to-left order is reminiscent of the format of normal assignment statements: The variable on the right is copied to the variable on the left. Chapter 7296 08 3087 CH07 11/29/01 2:14 PM Page 296Arrays of Strings If there are arrays of arrays, of course there can be arrays of strings. This is actually quite a useful construction. Here’s an example, STRARAY,that puts the names of the days of the week in an array: // straray.cpp // array of strings #include using namespace std; int main() { const int DAYS = 7; //number of strings in array const int MAX = 10; //maximum size of each string //array of strings char star[DAYS][MAX] = { “Sunday”, “Monday”, “Tuesday”, “Wednesday”, “Thursday”, “Friday”, “Saturday” }; for(int j=0; j #include //for strcpy() using namespace std; //////////////////////////////////////////////////////////////// class part { private: char partname[30]; //name of widget part int partnumber; //ID number of widget part double cost; //cost of part Chapter 7298 08 3087 CH07 11/29/01 2:14 PM Page 298public: void setpart(char pname[], int pn, double c) { strcpy(partname, pname); partnumber = pn; cost = c; } void showpart() //display data { cout << “\nName=” << partname; cout << “, number=” << partnumber; cout << “, cost=$” << cost; } }; //////////////////////////////////////////////////////////////// int main() { part part1, part2; part1.setpart(“handle bolt”, 4473, 217.55); //set parts part2.setpart(“start lever”, 9924, 419.25); cout << “\nFirst part: “; part1.showpart(); //show parts cout << “\nSecond part: “; part2.showpart(); cout << endl; return 0; } This program defines two objects of class part and gives them values with the setpart() member function. Then it displays them with the showpart() member function. Here’s the output: First part: Name=handle bolt, number=4473, cost=$217.55 Second part: Name=start lever, number=9924, cost=$419.25 To reduce the size of the program we’ve dropped the model number from the class members. In the setpart() member function, we use the strcpy() string library function to copy the string from the argument pname to the class data member partname. Thus this function serves the same purpose with string variables that an assignment statement does with simple vari- ables. (A similar function, strncpy(),takes a third argument, which is the maximum number of characters it will copy. This can help prevent overrunning the array.) Besides those we’ve seen, there are library functions to add a string to another, compare strings, search for specific characters in strings, and perform many other actions. Descriptions of these functions can be found in your compiler’s documentation. Arrays and Strings 7 A RRAYS AND S TRINGS 299 08 3087 CH07 11/29/01 2:14 PM Page 299A User-Defined String Type There are some problems with C-strings as they are normally used in C++. For one thing, you can’t use the perfectly reasonable expression strDest = strSrc; to set one string equal to another. (In some languages, like BASIC, this is perfectly all right.) The Standard C++ string class we’ll examine in the next section will take care of this problem, but for the moment let’s see if we can use object-oriented technology to solve the problem ourselves. Creating our own string class will give us an insight into representing strings as objects of a class, which will illuminate the operation of the Standard C++ string class. If we define our own string type, using a C++ class, we can use assignment statements. (Many other C-string operations, such as concatenation, can be simplified this way as well, but we’ll have to wait until Chapter 8, “Operator Overloading,” to see how this is done.) The STROBJ program creates a class called String. (Don’t confuse this homemade class String with the Standard C++ built-in class string, which has a lowercase ‘s’.) Here’s the listing: // strobj.cpp // a string as a class #include #include // for strcpy(), strcat() using namespace std; //////////////////////////////////////////////////////////////// class String { private: enum { SZ = 80; }; //max size of Strings char str[SZ]; //array public: String() //constructor, no args { str[0] = ‘\0’; } String( char s[] ) //constructor, one arg { strcpy(str, s); } void display() //display string { cout << str; } void concat(String s2) //add arg string to { //this string if( strlen(str)+strlen(s2.str) < SZ ) strcat(str, s2.str); else cout << “\nString too long”; } }; Chapter 7300 08 3087 CH07 11/29/01 2:14 PM Page 300//////////////////////////////////////////////////////////////// int main() { String s1(“Merry Christmas! “); //uses constructor 2 String s2 = “Season’s Greetings!”; //alternate form of 2 String s3; //uses constructor 1 cout << “\ns1=”; s1.display(); //display them all cout << “\ns2=”; s2.display(); cout << “\ns3=”; s3.display(); s3 = s1; //assignment cout << “\ns3=”; s3.display(); //display s3 s3.concat(s2); //concatenation cout << “\ns3=”; s3.display(); //display s3 cout << endl; return 0; } The String class contains an array of type char. It may seem that our newly defined class is just the same as the original definition of a string: an array of type char. But, by wrapping the array in a class, we have achieved some interesting benefits. Since an object can be assigned the value of another object of the same class using the = operator, we can use statements like s3 = s1; as we do in main(),to set one String object equal to another. We can also define our own member functions to deal with Strings (objects of class String). In the STROBJ program, all Strings have the same length: SZ characters (which we set to 80). There are two constructors. The first sets the first character in str to the null character, ‘\0’, so the string has a length of 0. This constructor is called with statements like String s3; The second constructor sets the String object to a “normal” (that is, a C-string) string constant. It uses the strcpy() library function to copy the string constant into the object’s data. It’s called with statements like String s1(“Merry Christmas! “); The alternative format for calling this constructor, which works with any one-argument constructor, is String s1 = “Merry Christmas! “; Whichever format is used, this constructor effectively converts a C-string to a String—that is, a normal string constant to an object of class String. A member function, display(),displays the String. Arrays and Strings 7 A RRAYS AND S TRINGS 301 08 3087 CH07 11/29/01 2:14 PM Page 301Another member function of our String class, concat(), concatenates (adds) one String to another. The original String is the object of which concat() is a member. To this String will be added the String passed as an argument. Thus the statement in main() s3.concat(s2); causes s2 to be added to the existing s3. Since s2 has been initialized to “Season’s Greetings!” and s3 has been assigned the value of s1, which was “Merry Christmas!” the resulting value of s3 is “Merry Christmas! Season’s Greetings!” The concat() function uses the strcat() C library function to do the concatenation. This library function adds the string specified in the second argument to the string specified in the first argument. The output from the program is s1=Merry Christmas! s2=Season’s Greetings! s3= ← nothing here yet s3=Merry Christmas! ← set equal to s1 s3=Merry Christmas! Season’s Greetings! ← s2 concatenated If the two Strings given to the concat() function together exceed the maximum String length, then the concatenation is not carried out, and a message is sent to the user. We’ve just examined a simple string class. Now we’ll see a far more sophisticated version of the same approach. The Standard C++ string Class Standard C++ includes a new class called string. This class improves on the traditional C- string in many ways. For one thing, you no longer need to worry about creating an array of the right size to hold string variables. The string class assumes all the responsibility for memory management. Also, the string class allows the use of overloaded operators, so you can concatenate string objects with the + operator: s3 = s1 + s2 There are other benefits as well. This new class is more efficient and safer to use than C-strings were. In most situations it is the preferred approach. (However, as we noted earlier, there are still many situations in which C-strings must be used.) In this section we’ll examine the string class and its various member functions and operators. Defining and Assigning string Objects You can define a string object in several ways. You can use a constructor with no arguments, creating an empty string. You can also use a one-argument constructor, where the argument is a Chapter 7302 08 3087 CH07 11/29/01 2:14 PM Page 302C-string constant; that is, characters delimited by double quotes. As in our homemade String class, objects of class string can be assigned to one another with a simple assignment operator. The SSTRASS example shows how this looks. //sstrass.cpp //defining and assigning string objects #include #include using namespace std; int main() { string s1(“Man”); //initialize string s2 = “Beast”; //initialize string s3; s3 = s1; //assign cout << “s3 = “ << s3 << endl; s3 = “Neither “ + s1 + “ nor “; //concatenate s3 += s2; //concatenate cout << “s3 = “ << s3 << endl; s1.swap(s2); //swap s1 and s2 cout << s1 << “ nor “ << s2 << endl; return 0; } Here, the first three lines of code show three ways to define string objects. The first two initialize strings, and the second creates an empty string variable. The next line shows simple assignment with the = operator. The string class uses a number of overloaded operators. We won’t learn about the inner workings of operator overloading until the next chapter, but you can use these operators without knowing how they’re constructed. The overloaded + operator concatenates one string object with another. The statement s3 = “Neither “ + s1 + “ nor “; places the string “Neither Man nor “ in the variable s3. You can also use the += operator to append a string to the end of an existing string. The statement s3 += s2; appends s2, which is “Beast”,to the end of s3,producing the string “Neither Man nor Beast” and assigning it to s3. Arrays and Strings 7 A RRAYS AND S TRINGS 303 08 3087 CH07 11/29/01 2:14 PM Page 303This example also introduces our first string class member function: swap(), which exchanges the values of two string objects. It’s called for one object with the other as an argument. We apply it to s1 (“Man”) and s2 (“Beast”), and then display their values to show that s1 is now “Beast” and s2 is now “Man”. Here’s the output of SSTRASS: s3 = Man s3 = Neither Man nor Beast Beast nor Man Input/Output with string Objects Input and output are handled in a similar way to that of C-strings. The << and >> operators are overloaded to handle string objects, and a function getline() handles input that contains embedded blanks or multiple lines. The SSTRIO example shows how this looks. // sstrio.cpp // string class input/output #include #include //for string class using namespace std; int main() { //objects of string class string full_name, nickname, address; string greeting(“Hello, “); cout << “Enter your full name: “; getline(cin, full_name); //reads embedded blanks cout << “Your full name is: “ << full_name << endl; cout << “Enter your nickname: “; cin >> nickname; //input to string object greeting += nickname; //append name to greeting cout << greeting << endl; //output: “Hello, Jim” cout << “Enter your address on separate lines\n”; cout << “Terminate with ‘$’\n”; getline(cin, address, ‘$’); //reads multiple lines cout << “Your address is: “ << address << endl; return 0; } Chapter 7304 08 3087 CH07 11/29/01 2:14 PM Page 304The program reads the user’s name, which presumably contains embedded blanks, using getline(). This function is similar to the get() function used with C-strings, but is not a member function. Instead, its first argument is the stream object from which the input will come (here it’s cin), and the second is the string object where the text will be placed, full_name. This variable is then displayed using the cout and <<. The program then reads the user’s nickname, which is assumed to be one word, using cin and the >> operator. Finally the program uses a variation of getline(),with three arguments, to read the user’s address, which may require multiple lines. The third argument specifies the character to be used to terminate the input. In the program we use the ‘$’ character, which the user must input as the last character before pressing the Enter key. If no third argument is sup- plied to getline(),the delimiter is assumed to be ‘\n’, which represents the Enter key. Here’s some interaction with SSTRIO: Enter your full name: F. Scott Fitzgerald Your full name is: F. Scott Fitzgerald Enter your nickname: Scotty Hello, Scotty Enter your address on separate lines: Terminate with ‘$’ 1922 Zelda Lane East Egg, New York$ Your address is: 1922 Zelda Lane East Egg, New York Finding string Objects The string class includes a variety of member functions for finding strings and substrings in string objects. The SSTRFIND example shows some of them. //sstrfind.cpp //finding substrings in string objects #include #include using namespace std; int main() { string s1 = “In Xanadu did Kubla Kahn a stately pleasure dome decree”; int n; n = s1.find(“Kubla”); cout << “Found Kubla at “ << n << endl; Arrays and Strings 7 A RRAYS AND S TRINGS 305 08 3087 CH07 11/29/01 2:14 PM Page 305n = s1.find_first_of(“spde”); cout << “First of spde at “ << n << endl; n = s1.find_first_not_of(“aeiouAEIOU”); cout << “First consonant at “ << n << endl; return 0; } The find() function looks for the string used as its argument in the string for which it was called. Here it finds “Kubla” in s1, which holds the first line of the poem Kubla Kahn by Samuel Taylor Coleridge. It finds it at position 14. As with C-strings, the leftmost character position is numbered 0. The find_first_of() function looks for any of a group of characters, and returns the position of the first one it finds. Here it looks for any of the group ‘s’, ‘p’, ‘d’,or ‘e’. The first of these it finds is the ‘d’ in Xanadu,at position 7. A similar function find_first_not_of() finds the first character in its string that is not one of a specified group. Here the group consists of all the vowels, both upper- and lowercase, so the function finds the first consonant, which is the second letter. The output of SSTRFIND is Found Kubla at 14 First of spde at 7 First consonent at 1 There are variations on many of these functions that we don’t demonstrate here, such as rfind(), which scans its string backward; find_last_of(), which finds the last character matching one of a group of characters, and find_last_not_of(). All these functions return –1 if the target is not found. Modifying string Objects There are various ways to modify string objects. Our next example shows the member functions erase(), replace(), and insert() at work. //sstrchng.cpp //changing parts of string objects #include #include using namespace std; int main() { string s1(“Quick! Send for Count Graystone.”); string s2(“Lord”); string s3(“Don’t “); Chapter 7306 08 3087 CH07 11/29/01 2:14 PM Page 306s1.erase(0, 7); //remove “Quick! “ s1.replace(9, 5, s2); //replace “Count” with “Lord” s1.replace(0, 1, “s”); //replace ‘S’ with ‘s’ s1.insert(0, s3); //insert “Don’t “ at beginning s1.erase(s1.size()-1, 1); //remove ‘.’ s1.append(3, ‘!’); //append “!!!” int x = s1.find(‘ ‘); //find a space while( x < s1.size() ) //loop while spaces remain { s1.replace(x, 1, “/”); //replace with slash x = s1.find(‘ ‘); //find next space } cout << “s1: “ << s1 << endl; return 0; } The erase() function removes a substring from a string. Its first argument is the position of the first character in the substring, and the second is the length of the substring. In the example it removes “Quick “ from the beginning of the string. The replace() function replaces part of the string with another string. The first argument is the position where the replacement should begin, the second is the number of characters in the original string to be replaced, and the third is the replacement string. Here “Count” is replaced by “Lord”. The insert() function inserts the string specified by its second argument at the location specified by its first argument. Here it inserts “Don’t “ at the beginning of s1. The second use of erase() employs the size() member function, which returns the number of characters in the string object. The expression size()-1 is the position of the last character, the period, which is erased. The append() function installs three exclamation points at the end of the sentence. In this version of the function the first argument is the number of characters to append, and the second is the character to be appended. At the end of the program we show an idiom you can use to replace multiple instances of a substring with another string. Here, in a while loop, we look for the space character ‘ ‘ using find(), and replace each one with a slash using replace(). We start with s1 containing the string “Quick! Send for Count Graystone.” After these changes, the output of SSTRCHNG is s1: Don’t/send/for/Lord/Graystone!!! Comparing string Objects You can use overloaded operators or the compare() function to compare string objects. These discover whether strings are the same, or whether they precede or follow one another alphabet- ically. The SSTRCOM program shows some of the possibilities. Arrays and Strings 7 A RRAYS AND S TRINGS 307 08 3087 CH07 11/29/01 2:14 PM Page 307//sstrcom.cpp //comparing string objects #include #include using namespace std; int main() { string aName = “George”; string userName; cout << “Enter your first name: “; cin >> userName; if(userName==aName) //operator == cout << “Greetings, George\n”; else if(userName < aName) //operator < cout << “You come before George\n”; else cout << “You come after George\n”; //compare() function int n = userName.compare(0, 2, aName, 0, 2); cout << “The first two letters of your name “; if(n==0) cout << “match “; else if(n < 0) cout << “come before “; else cout << “come after “; cout << aName.substr(0, 2) << endl; return 0; } In the first part of the program the == and < operators are used to determine whether a name typed by the user is equal to, or precedes or follows alphabetically, the name George. In the second part of the program the compare() function compares only the first two letters of “George” with the first two letters of the name typed by the user (userName). The arguments to this version of compare() are the starting position in userName and the number of characters to compare, the string used for comparison (aName), and the starting position and number of characters in aName. Here’s some interaction with SSTRCOM: Enter your first name: Alfred You come before George The first two letters of your name come before Ge The first two letters of “George” are obtained using the substr() member function. It returns a substring of the string for which it was called. Its first argument is the position of the substring, and the second is the number of characters. Chapter 7308 08 3087 CH07 11/29/01 2:14 PM Page 308Accessing Characters in string Objects You can access individual characters within a string object in several ways. In our next example we’ll show access using the at() member function. You can also use the overloaded [] opera- tor, which makes the string object look like an array. However, the [] operator doesn’t warn you if you attempt to access a character that’s out of bounds (beyond the end of the string, for example). The [] operator behaves this way with real arrays, and it’s more efficient. However, it can lead to hard-to-diagnose program bugs. It’s safer to use the at() function, which causes the program to stop if you use an out-of-bounds index. (It actually throws an exception; we’ll discuss exceptions in Chapter 14, “Templates and Exceptions.”) //sstrchar.cpp //accessing characters in string objects #include #include using namespace std; int main() { char charray[80]; string word; cout << “Enter a word: “; cin >> word; int wlen = word.length(); //length of string object cout << “One character at a time: “; for(int j=0; j>) stops reading a string when it encounters a space. 24. You can read input that consists of multiple lines of text using a. the normal cout << combination. b. the cin.get() function with one argument. c. the cin.get() function with two arguments. d. the cin.get() function with three arguments. 25. Write a statement that uses a string library function to copy the string name to the string blank. 26. Write the declaration for a class called dog that contains two data members: a string called breed and an int called age. (Don’t include any member functions.) 27. True or false: You should prefer C-strings to the Standard C++ string class in new programs. 28. Objects of the string class a. are zero-terminated. b. can be copied with the assignment operator. c. do not require memory management. d. have no member functions. 29. Write a statement that finds where the string “cat” occurs in the string s1. 30. Write a statement that inserts the string “cat” into string s1 at position 12. Exercises Answers to the starred exercises can be found in Appendix G. *1. Write a function called reversit() that reverses a C-string (an array of char). Use a for loop that swaps the first and last characters, then the second and next-to-last characters, and so on. The string should be passed to reversit() as an argument. Write a program to exercise reversit(). The program should get a string from the user, call reversit(), and print out the result. Use an input method that allows embedded blanks. Test the program with Napoleon’s famous phrase, “Able was I ere I saw Elba.” Arrays and Strings 7 A RRAYS AND S TRINGS 313 08 3087 CH07 11/29/01 2:14 PM Page 313*2. Create a class called employee that contains a name (an object of class string) and an employee number (type long). Include a member function called getdata() to get data from the user for insertion into the object, and another function called putdata() to display the data. Assume the name has no embedded blanks. Write a main() program to exercise this class. It should create an array of type employee, and then invite the user to input data for up to 100 employees. Finally, it should print out the data for all the employees. *3. Write a program that calculates the average of up to 100 English distances input by the user. Create an array of objects of the Distance class, as in the ENGLARAY example in this chapter. To calculate the average, you can borrow the add_dist() member function from the ENGLCON example in Chapter 6. You’ll also need a member function that divides a Distance value by an integer. Here’s one possibility: void Distance::div_dist(Distance d2, int divisor) { float fltfeet = d2.feet + d2.inches/12.0; fltfeet /= divisor; feet = int(fltfeet); inches = (fltfeet-feet) * 12.0; } 4. Start with a program that allows the user to input a number of integers, and then stores them in an int array. Write a function called maxint() that goes through the array, element by element, looking for the largest one. The function should take as arguments the address of the array and the number of elements in it, and return the index number of the largest element. The program should call this function and then display the largest element and its index number. (See the SALES program in this chapter.) 5. Start with the fraction class from Exercises 11 and 12 in Chapter 6. Write a main() program that obtains an arbitrary number of fractions from the user, stores them in an array of type fraction,averages them, and displays the result. 6. In the game of contract bridge, each of four players is dealt 13 cards, thus exhausting the entire deck. Modify the CARDARAY program in this chapter so that, after shuffling the deck, it deals four hands of 13 cards each. Each of the four players’ hands should then be displayed. 7. One of the weaknesses of C++ for writing business programs is that it does not contain a built-in type for monetary values such as $173,698,001.32. Such a money type should be able to store a number with a fixed decimal point and about 17 digits of precision, which is enough to handle the national debt in dollars and cents. Fortunately, the built-in C++ type long double has 19 digits of precision, so we can use it as the basis of a money class, even though it uses a floating decimal. However, we’ll need to add the capability to input and output money amounts preceded by a dollar sign and divided by commas into Chapter 7314 08 3087 CH07 11/29/01 2:14 PM Page 314groups of three digits; this makes it much easier to read large numbers. As a first step toward developing such a class, write a function called mstold() that takes a money string,a string representing a money amount like “$1,234,567,890,123.99” as an argument, and returns the equivalent long double. You’ll need to treat the money string as an array of characters, and go through it character by character, copying only digits (1–9) and the decimal point into another string. Ignore everything else, including the dollar sign and the commas. You can then use the _atold() library function (note the initial underscore—header file STDLIB.H or MATH.H) to convert the resulting pure string to a long double. Assume that money values will never be negative. Write a main() program to test mstold() by repeatedly obtaining a money string from the user and displaying the corresponding long double. 8. Another weakness of C++ is that it does not automatically check array indexes to see whether they are in bounds. (This makes array operations faster but less safe.) We can use a class to create a safe array that checks the index of all array accesses. Write a class called safearay that uses an int array of fixed size (call it LIMIT) as its only data member. There will be two member functions. The first, putel(),takes an index number and an int value as arguments and inserts the int value into the array at the index. The second, getel(),takes an index number as an argument and returns the int value of the element with that index. safearay sa1; // define a safearay object int temp = 12345; // define an int value sa1.putel(7, temp); // insert value of temp into array at index 7 temp = sa1.getel(7); // obtain value from array at index 7 Both functions should check the index argument to make sure it is not less than 0 or greater than LIMIT-1. You can use this array without fear of writing over other parts of memory. Using functions to access array elements doesn’t look as eloquent as using the [] operator. In Chapter 8 we’ll see how to overload this operator to make our safearay class work more like built-in arrays. 9. A queue is a data storage device much like a stack. The difference is that in a stack the last data item stored is the first one retrieved, while in a queue the first data item stored is the first one retrieved. That is, a stack uses a last-in-first-out (LIFO) approach, while a queue uses first-in-first-out (FIFO). A queue is like a line of customers in a bank: The first one to join the queue is the first one served. Rewrite the STAKARAY program from this chapter to incorporate a class called queue instead of a class called stack. Besides a constructor, it should have two functions: one called put() to put a data item on the queue, and one called get() to get data from the queue. These are equivalent to push() and pop() in the stack class. Arrays and Strings 7 A RRAYS AND S TRINGS 315 08 3087 CH07 11/29/01 2:14 PM Page 315Both a queue and a stack use an array to hold the data. However, instead of a single int variable called top,as the stack has, you’ll need two variables for a queue: one called head to point to the head of the queue, and one called tail to point to the tail. Items are placed on the queue at the tail (like the last customer getting in line at the bank) and removed from the queue at the head. The tail will follow the head along the array as items are added and removed from the queue. This results in an added complexity: When either the tail or the head gets to the end of the array, it must wrap around to the beginning. Thus you’ll need a statement like if(tail == MAX-1) tail = -1; to wrap the tail, and a similar one for the head. The array used in the queue is sometimes called a circular buffer, because the head and tail circle around it, with the data between them. 10. A matrix is a two-dimensional array. Create a class matrix that provides the same safety feature as the array class in Exercise 7; that is, it checks to be sure no array index is out of bounds. Make the member data in the matrix class a 10-by-10 array. A constructor should allow the programmer to specify the actual dimensions of the matrix (provided they’re less than 10 by 10). The member functions that access data in the matrix will now need two index numbers: one for each dimension of the array. Here’s what a fragment of a main() program that operates on such a class might look like: matrix m1(3, 4); // define a matrix object int temp = 12345; // define an int value m1.putel(7, 4, temp); // insert value of temp into matrix at 7,4 temp = m1.getel(7, 4); // obtain value from matrix at 7,4 11. Refer back to the discussion of money strings in Exercise 6. Write a function called ldtoms() to convert a number represented as type long double to the same value represented as a money string. First you should check that the value of the original long double is not too large. We suggest that you don’t try to convert any number greater than 9,999,999,999,999,990.00. Then convert the long double to a pure string (no dollar sign or commas) stored in memory, using an ostrstream object, as discussed earlier in this chapter. The resulting formatted string can go in a buffer called ustring. You’ll then need to start another string with a dollar sign; copy one digit from ustring at a time, starting from the left, and inserting a comma into the new string every three digits. Also, you’ll need to suppress leading zeros. You want to display $3,124.95, for example, not $0,000,000,000,003,124.95. Don’t forget to terminate the string with a ‘\0’ character. Write a main() program to exercise this function by having the user repeatedly input numbers in type long double format, and printing out the result as a money string. Chapter 7316 08 3087 CH07 11/29/01 2:14 PM Page 31612. Create a class called bMoney. It should store money amounts as long doubles. Use the function mstold() to convert a money string entered as input into a long double, and the function ldtoms() to convert the long double to a money string for display. (See Exercises 6 and 10.) You can call the input and output member functions getmoney() and putmoney(). Write another member function that adds two bMoney amounts; you can call it madd(). Adding bMoney objects is easy: Just add the long double member data amounts in two bMoney objects. Write a main() program that repeatedly asks the user to enter two money strings, and then displays the sum as a money string. Here’s how the class specifier might look: class bMoney { private: long double money; public: bMoney(); bMoney(char s[]); void madd(bMoney m1, bMoney m2); void getmoney(); void putmoney(); }; Arrays and Strings 7 A RRAYS AND S TRINGS 317 08 3087 CH07 11/29/01 2:14 PM Page 31708 3087 CH07 11/29/01 2:14 PM Page 318CHAPTER 8 Operator Overloading IN THIS CHAPTER • Overloading Unary Operators 320 • Overloading Binary Operators 328 • Data Conversion 344 • UML Class Diagrams 357 •Pitfalls of Operator Overloading and Conversion 358 • Keywords explicit and mutable 360 09 3087 CH08 11/29/01 2:18 PM Page 319Chapter 8320 Operator overloading is one of the most exciting features of object-oriented programming. It can transform complex, obscure program listings into intuitively obvious ones. For example, statements like d3.addobjects(d1, d2); or the similar but equally obscure d3 = d1.addobjects(d2); can be changed to the much more readable d3 = d1 + d2; The rather forbidding term operator overloading refers to giving the normal C++ operators, such as +, *, <=, and +=, additional meanings when they are applied to user-defined data types. Normally a = b + c; works only with basic types such as int and float, and attempting to apply it when a, b, and c are objects of a user-defined class will cause complaints from the compiler. However, using overloading, you can make this statement legal even when a, b, and c are user-defined types. In effect, operator overloading gives you the opportunity to redefine the C++ language. If you find yourself limited by the way the C++ operators work, you can change them to do whatever you want. By using classes to create new kinds of variables, and operator overloading to create new definitions for operators, you can extend C++ to be, in many ways, a new language of your own design. Another kind of operation, data type conversion, is closely connected with operator overloading. C++ handles the conversion of simple types, such as int and float, automatically; but conver- sions involving user-defined types require some work on the programmer’s part. We’ll look at data conversions in the second part of this chapter. Overloaded operators are not all beer and skittles. We’ll discuss some of the dangers of their use at the end of the chapter. Overloading Unary Operators Let’s start off by overloading a unary operator. As you may recall from Chapter 2, unary operators act on only one operand. (An operand is simply a variable acted on by an operator.) Examples of unary operators are the increment and decrement operators ++ and --, and the unary minus, as in -33. 09 3087 CH08 11/29/01 2:18 PM Page 320In the COUNTER example in Chapter 6, “Objects and Classes,” we created a class Counter to keep track of a count. Objects of that class were incremented by calling a member function: c1.inc_count(); That did the job, but the listing would have been more readable if we could have used the increment operator ++ instead: ++c1; All dyed-in-the-wool C++ (and C) programmers would guess immediately that this expression increments c1. Let’s rewrite COUNTER to make this possible. Here’s the listing for COUNTPP1: // countpp1.cpp // increment counter variable with ++ operator #include using namespace std; //////////////////////////////////////////////////////////////// class Counter { private: unsigned int count; //count public: Counter() : count(0) //constructor { } unsigned int get_count() //return count { return count; } void operator ++ () //increment (prefix) { ++count; } }; //////////////////////////////////////////////////////////////// int main() { Counter c1, c2; //define and initialize cout << “\nc1=” << c1.get_count(); //display cout << “\nc2=” << c2.get_count(); ++c1; //increment c1 ++c2; //increment c2 ++c2; //increment c2 Operator Overloading 8 O PERATOR O VERLOADING 321 09 3087 CH08 11/29/01 2:18 PM Page 321cout << “\nc1=” << c1.get_count(); //display again cout << “\nc2=” << c2.get_count() << endl; return 0; } In this program we create two objects of class Counter: c1 and c2. The counts in the objects are displayed; they are initially 0. Then, using the overloaded ++ operator, we increment c1 once and c2 twice, and display the resulting values. Here’s the program’s output: c1=0 ← counts are initially 0 c2=0 c1=1 ← incremented once c2=2 ← incremented twice The statements responsible for these operations are ++c1; ++c2; ++c2; The ++ operator is applied once to c1 and twice to c2. We use prefix notation in this example; we’ll explore postfix later. The operator Keyword How do we teach a normal C++ operator to act on a user-defined operand? The keyword operator is used to overload the ++ operator in this declarator: void operator ++ () The return type (void in this case) comes first, followed by the keyword operator,followed by the operator itself (++), and finally the argument list enclosed in parentheses (which are empty here). This declarator syntax tells the compiler to call this member function whenever the ++ operator is encountered, provided the operand (the variable operated on by the ++) is of type Counter. We saw in Chapter 5, “Functions,” that the only way the compiler can distinguish between overloaded functions is by looking at the data types and the number of their arguments. In the same way, the only way it can distinguish between overloaded operators is by looking at the data type of their operands. If the operand is a basic type such as an int,as in ++intvar; then the compiler will use its built-in routine to increment an int. But if the operand is a Counter variable, the compiler will know to use our user-written operator++() instead. Chapter 8322 09 3087 CH08 11/29/01 2:18 PM Page 322Operator Arguments In main() the ++ operator is applied to a specific object, as in the expression ++c1. Yet operator++() takes no arguments. What does this operator increment? It increments the count data in the object of which it is a member. Since member functions can always access the particular object for which they’ve been invoked, this operator requires no arguments. This is shown in Figure 8.1. Operator Overloading 8 O PERATOR O VERLOADING 323 ++c1; count void operator++() This statement causes this function to increment this count.c1 object No arguments { ++count; } FIGURE 8.1 Overloaded unary operator: no arguments. Operator Return Values The operator++() function in the COUNTPP1program has a subtle defect. You will discover it if you use a statement like this in main(): c1 = ++c2; The compiler will complain. Why? Because we have defined the ++ operator to have a return type of void in the operator++() function, while in the assignment statement it is being asked to return a variable of type Counter. That is, the compiler is being asked to return whatever value c2 has after being operated on by the ++ operator, and assign this value to c1. So as defined in COUNTPP1, we can’t use ++ to increment Counter objects in assignments; it must always stand alone with its operand. Of course the normal ++ operator, applied to basic data types such as int,would not have this problem. To make it possible to use our homemade operator++() in assignment expressions, we must provide a way for it to return a value. The next program, COUNTPP2, does just that. 09 3087 CH08 11/29/01 2:18 PM Page 323// countpp2.cpp // increment counter variable with ++ operator, return value #include using namespace std; //////////////////////////////////////////////////////////////// class Counter { private: unsigned int count; //count public: Counter() : count(0) //constructor { } unsigned int get_count() //return count { return count; } Counter operator ++ () //increment count { ++count; //increment count Counter temp; //make a temporary Counter temp.count = count; //give it same value as this obj return temp; //return the copy } }; //////////////////////////////////////////////////////////////// int main() { Counter c1, c2; //c1=0, c2=0 cout << “\nc1=” << c1.get_count(); //display cout << “\nc2=” << c2.get_count(); ++c1; //c1=1 c2 = ++c1; //c1=2, c2=2 cout << “\nc1=” << c1.get_count(); //display again cout << “\nc2=” << c2.get_count() << endl; return 0; } Here the operator++() function creates a new object of type Counter, called temp,to use as a return value. It increments the count data in its own object as before, then creates the new temp object and assigns count in the new object the same value as in its own object. Finally, it returns the temp object. This has the desired effect. Expressions like ++c1 Chapter 8324 09 3087 CH08 11/29/01 2:18 PM Page 324now return a value, so they can be used in other expressions, such as c2 = ++c1; as shown in main(), where the value returned from c1++ is assigned to c2. The output from this program is c1=0 c2=0 c1=2 c2=2 Nameless Temporary Objects In COUNTPP2 we created a temporary object of type Counter, named temp, whose sole purpose was to provide a return value for the ++ operator. This required three statements. Counter temp; // make a temporary Counter object temp.count = count; // give it same value as this object return temp; // return it There are more convenient ways to return temporary objects from functions and overloaded operators. Let’s examine another approach, as shown in the program COUNTPP3: // countpp3.cpp // increment counter variable with ++ operator // uses unnamed temporary object #include using namespace std; //////////////////////////////////////////////////////////////// class Counter { private: unsigned int count; //count public: Counter() : count(0) //constructor no args { } Counter(int c) : count(c) //constructor, one arg { } unsigned int get_count() //return count { return count; } Counter operator ++ () //increment count { ++count; // increment count, then return return Counter(count); // an unnamed temporary object } // initialized to this count }; //////////////////////////////////////////////////////////////// Operator Overloading 8 O PERATOR O VERLOADING 325 09 3087 CH08 11/29/01 2:18 PM Page 325int main() { Counter c1, c2; //c1=0, c2=0 cout << “\nc1=” << c1.get_count(); //display cout << “\nc2=” << c2.get_count(); ++c1; //c1=1 c2 = ++c1; //c1=2, c2=2 cout << “\nc1=” << c1.get_count(); //display again cout << “\nc2=” << c2.get_count() << endl; return 0; } In this program a single statement return Counter(count); does what all three statements did in COUNTPP2. This statement creates an object of type Counter. This object has no name; it won’t be around long enough to need one. This unnamed object is initialized to the value provided by the argument count. But wait: Doesn’t this require a constructor that takes one argument? It does, and to make this statement work we sneakily inserted just such a constructor into the member function list in COUNTPP3. Counter(int c) : count(c) //constructor, one arg { } Once the unnamed object is initialized to the value of count, it can then be returned. The output of this program is the same as that of COUNTPP2. The approaches in both COUNTPP2 and COUNTPP3involve making a copy of the original object (the object of which the function is a member), and returning the copy. (Another approach, as we’ll see in Chapter 11, “Virtual Functions,” is to return the value of the original object using the this pointer.) Postfix Notation So far we’ve shown the increment operator used only in its prefix form. ++c1 What about postfix, where the variable is incremented after its value is used in the expression? c1++ Chapter 8326 09 3087 CH08 11/29/01 2:18 PM Page 326To make both versions of the increment operator work, we define two overloaded ++ operators, as shown in the POSTFIX program: // postfix.cpp // overloaded ++ operator in both prefix and postfix #include using namespace std; //////////////////////////////////////////////////////////////// class Counter { private: unsigned int count; //count public: Counter() : count(0) //constructor no args { } Counter(int c) : count(c) //constructor, one arg { } unsigned int get_count() const //return count { return count; } Counter operator ++ () //increment count (prefix) { //increment count, then return return Counter(++count); //an unnamed temporary object } //initialized to this count Counter operator ++ (int) //increment count (postfix) { //return an unnamed temporary return Counter(count++); //object initialized to this } //count, then increment count }; //////////////////////////////////////////////////////////////// int main() { Counter c1, c2; //c1=0, c2=0 cout << “\nc1=” << c1.get_count(); //display cout << “\nc2=” << c2.get_count(); ++c1; //c1=1 c2 = ++c1; //c1=2, c2=2 (prefix) cout << “\nc1=” << c1.get_count(); //display cout << “\nc2=” << c2.get_count(); c2 = c1++; //c1=3, c2=2 (postfix) Operator Overloading 8 O PERATOR O VERLOADING 327 09 3087 CH08 11/29/01 2:18 PM Page 327cout << “\nc1=” << c1.get_count(); //display again cout << “\nc2=” << c2.get_count() << endl; return 0; } Now there are two different declarators for overloading the ++ operator. The one we’ve seen before, for prefix notation, is Counter operator ++ () The new one, for postfix notation, is Counter operator ++ (int) The only difference is the int in the parentheses. This int isn’t really an argument, and it doesn’t mean integer. It’s simply a signal to the compiler to create the postfix version of the operator. The designers of C++ are fond of recycling existing operators and keywords to play multiple roles, and int is the one they chose to indicate postfix. (Well, can you think of a better syntax?) Here’s the output from the program: c1=0 c2=0 c1=2 c2=2 c1=3 c2=2 We saw the first four of these output lines in COUNTPP2 and COUNTPP3. But in the last two lines we see the results of the statement c2=c1++; Here, c1 is incremented to 3, but c2 is assigned the value of c1 before it is incremented, so c2 retains the value 2. Of course, you can use this same approach with the decrement operator (--). Overloading Binary Operators Binary operators can be overloaded just as easily as unary operators. We’ll look at examples that overload arithmetic operators, comparison operators, and arithmetic assignment operators. Arithmetic Operators In the ENGLCON program in Chapter 6 we showed how two English Distance objects could be added using a member function add_dist(): Chapter 8328 09 3087 CH08 11/29/01 2:18 PM Page 328dist3.add_dist(dist1, dist2); By overloading the + operator we can reduce this dense-looking expression to dist3 = dist1 + dist2; Here’s the listing for ENGLPLUS, which does just that: // englplus.cpp // overloaded ‘+’ operator adds two Distances #include using namespace std; //////////////////////////////////////////////////////////////// class Distance //English Distance class { private: int feet; float inches; public: //constructor (no args) Distance() : feet(0), inches(0.0) { } //constructor (two args) Distance(int ft, float in) : feet(ft), inches(in) { } void getdist() //get length from user { cout << “\nEnter feet: “; cin >> feet; cout << “Enter inches: “; cin >> inches; } void showdist() const //display distance { cout << feet << “\’-” << inches << ‘\”’; } Distance operator + ( Distance ) const; //add 2 distances }; //-------------------------------------------------------------- //add this distance to d2 Distance Distance::operator + (Distance d2) const //return sum { int f = feet + d2.feet; //add the feet float i = inches + d2.inches; //add the inches if(i >= 12.0) //if total exceeds 12.0, { //then decrease inches i -= 12.0; //by 12.0 and f++; //increase feet by 1 } //return a temporary Distance return Distance(f,i); //initialized to sum } //////////////////////////////////////////////////////////////// Operator Overloading 8 O PERATOR O VERLOADING 329 09 3087 CH08 11/29/01 2:18 PM Page 329int main() { Distance dist1, dist3, dist4; //define distances dist1.getdist(); //get dist1 from user Distance dist2(11, 6.25); //define, initialize dist2 dist3 = dist1 + dist2; //single ‘+’ operator dist4 = dist1 + dist2 + dist3; //multiple ‘+’ operators //display all lengths cout << “dist1 = “; dist1.showdist(); cout << endl; cout << “dist2 = “; dist2.showdist(); cout << endl; cout << “dist3 = “; dist3.showdist(); cout << endl; cout << “dist4 = “; dist4.showdist(); cout << endl; return 0; } To show that the result of an addition can be used in another addition as well as in an assignment, another addition is performed in main(). We add dist1, dist2, and dist3 to obtain dist4 (which should be double the value of dist3), in the statement dist4 = dist1 + dist2 + dist3; Here’s the output from the program: Enter feet: 10 Enter inches: 6.5 dist1 = 10’-6.5” ← from user dist2 = 11’-6.25” ← initialized in program dist3 = 22’-0.75” ← dist1+dist2 dist4 = 44’-1.5” ← dist1+dist2+dist3 In class Distance the declaration for the operator+() function looks like this: Distance operator + ( Distance ); This function has a return type of Distance, and takes one argument of type Distance. In expressions like dist3 = dist1 + dist2; it’s important to understand how the return value and arguments of the operator relate to the objects. When the compiler sees this expression it looks at the argument types, and finding only type Distance, it realizes it must use the Distance member function operator+(). But what does this function use as its argument—dist1 or dist2? And doesn’t it need two arguments, since there are two numbers to be added? Chapter 8330 09 3087 CH08 11/29/01 2:18 PM Page 330Here’s the key: The argument on the left side of the operator (dist1 in this case) is the object of which the operator is a member. The object on the right side of the operator (dist2) must be furnished as an argument to the operator. The operator returns a value, which can be assigned or used in other ways; in this case it is assigned to dist3. Figure 8.2 shows how this looks. Operator Overloading 8 O PERATOR O VERLOADING 331 FIGURE 8.2 Overloaded binary operator: one argument. In the operator+() function, the left operand is accessed directly—since this is the object of which the operator is a member—using feet and inches. The right operand is accessed as the function’s argument, as d2.feet and d2.inches. We can generalize and say that an overloaded operator always requires one less argument than its number of operands, since one operand is the object of which the operator is a member. That’s why unary operators require no arguments. (This rule does not apply to friend functions and operators, C++ features we’ll discuss in Chapter 11.) To calculate the return value of operator+() in ENGLPLUS, we first add the feet and inches from the two operands (adjusting for a carry if necessary). The resulting values, f and i,are then used to initialize a nameless Distance object, which is returned in the statement 09 3087 CH08 11/29/01 2:18 PM Page 331return Distance(f, i); This is similar to the arrangement used in COUNTPP3, except that the constructor takes two arguments instead of one. The statement dist3 = dist1 + dist2; in main() then assigns the value of the nameless Distance object to dist3. Compare this intuitively obvious statement with the use of a function call to perform the same task, as in the ENGLCON example in Chapter 6. Similar functions could be created to overload other operators in the Distance class, so you could subtract, multiply, and divide objects of this class in natural-looking ways. Concatenating Strings The + operator cannot be used to concatenate C-strings. That is, you can’t say str3 = str1 + str2; where str1, str2, and str3 are C-string variables (arrays of type char), as in “cat” plus “bird” equals “catbird.” However, if we use our own String class, as shown in the STROBJ program in Chapter 6, we can overload the + operator to perform such concatenation. This is what the Standard C++ string class does, but it’s easier to see how it works in our less ambitious String class. Overloading the + operator to do something that isn’t strictly addition is another example of redefining the C++ language. Here’s the listing for STRPLUS: // strplus.cpp // overloaded ‘+’ operator concatenates strings #include using namespace std; #include //for strcpy(), strcat() #include //for exit() //////////////////////////////////////////////////////////////// class String //user-defined string type { private: enum { SZ=80 }; //size of String objects char str[SZ]; //holds a string public: String() //constructor, no args { strcpy(str, “”); } String( char s[] ) //constructor, one arg { strcpy(str, s); } void display() const //display the String { cout << str; } String operator + (String ss) const //add Strings Chapter 8332 09 3087 CH08 11/29/01 2:18 PM Page 332{ String temp; //make a temporary String if( strlen(str) + strlen(ss.str) < SZ ) { strcpy(temp.str, str); //copy this string to temp strcat(temp.str, ss.str); //add the argument string } else { cout << “\nString overflow”; exit(1); } return temp; //return temp String } }; //////////////////////////////////////////////////////////////// int main() { String s1 = “\nMerry Christmas! “; //uses constructor 2 String s2 = “Happy new year!”; //uses constructor 2 String s3; //uses constructor 1 s1.display(); //display strings s2.display(); s3.display(); s3 = s1 + s2; //add s2 to s1, //assign to s3 s3.display(); //display s3 cout << endl; return 0; } The program first displays three strings separately. (The third is empty at this point, so nothing is printed when it displays itself.) Then the first two strings are concatenated and placed in the third, and the third string is displayed again. Here’s the output: Merry Christmas! Happy new year! ← s1, s2, and s3 (empty) Merry Christmas! Happy new year! ← s3 after concatenation By now the basics of overloading the + operator should be somewhat familiar. The declarator String operator + (String ss) shows that the + operator takes one argument of type String and returns an object of the same type. The concatenation process in operator+() involves creating a temporary object of type String, copying the string from our own String object into it, concatenating the argument string using the library function strcat(), and returning the resulting temporary string. Note that we can’t use the Operator Overloading 8 O PERATOR O VERLOADING 333 09 3087 CH08 11/29/01 2:18 PM Page 333return String(string); approach, where a nameless temporary String is created, because we need access to the temporary String not only to initialize it, but to concatenate the argument string to it. We must be careful that we don’t overflow the fixed-length strings used in the String class. To prevent such accidents in the operator+() function, we check that the combined length of the two strings to be concatenated will not exceed the maximum string length. If they do, we print an error message instead of carrying out the concatenation operation. (We could handle errors in other ways, like returning 0 if an error occurred, or better yet, throwing an exception, as dis- cussed in Chapter 14, “Templates and Exceptions.”) Remember that using an enum to set the constant value SZ is a temporary fix. When all compilers comply with Standard C++ you can change it to static const int SZ = 80; Multiple Overloading We’ve seen different uses of the + operator: to add English distances and to concatenate strings. You could put both these classes together in the same program, and C++ would still know how to interpret the + operator: It selects the correct function to carry out the “addition” based on the type of operand. Comparison Operators Let’s see how to overload a different kind of C++ operator: comparison operators. Comparing Distances In our first example we’ll overload the less than operator (<) in the Distance class so that we can compare two distances. Here’s the listing for ENGLESS: // engless.cpp // overloaded ‘<’ operator compares two Distances #include using namespace std; //////////////////////////////////////////////////////////////// class Distance //English Distance class { private: int feet; float inches; public: //constructor (no args) Distance() : feet(0), inches(0.0) { } //constructor (two args) Chapter 8334 09 3087 CH08 11/29/01 2:18 PM Page 334Distance(int ft, float in) : feet(ft), inches(in) { } void getdist() //get length from user { cout << “\nEnter feet: “; cin >> feet; cout << “Enter inches: “; cin >> inches; } void showdist() const //display distance { cout << feet << “\’-” << inches << ‘\”’; } bool operator < (Distance) const; //compare distances }; //-------------------------------------------------------------- //compare this distance with d2 bool Distance::operator < (Distance d2) const //return the sum { float bf1 = feet + inches/12; float bf2 = d2.feet + d2.inches/12; return (bf1 < bf2) ? true : false; } //////////////////////////////////////////////////////////////// int main() { Distance dist1; //define Distance dist1 dist1.getdist(); //get dist1 from user Distance dist2(6, 2.5); //define and initialize dist2 //display distances cout << “\ndist1 = “; dist1.showdist(); cout << “\ndist2 = “; dist2.showdist(); if( dist1 < dist2 ) //overloaded ‘<’ operator cout << “\ndist1 is less than dist2”; else cout << “\ndist1 is greater than (or equal to) dist2”; cout << endl; return 0; } This program compares a distance entered by the user with a distance, 6'–2.5'', initialized by the program. Depending on the result, it then prints one of two possible sentences. Here’s some typical output: Enter feet: 5 Enter inches: 11.5 dist1 = 5’-11.5” dist2 = 6’-2.5” dist1 is less than dist2 Operator Overloading 8 O PERATOR O VERLOADING 335 09 3087 CH08 11/29/01 2:18 PM Page 335The approach used in the operator<() function in ENGLESS is similar to overloading the + operator in the ENGLPLUS program, except that here the operator<() function has a return type of bool. The return value is false or true, depending on the comparison of the two distances. The comparison is made by converting both distances to floating-point feet, and comparing them using the normal < operator. Remember that the use of the conditional operator return (bf1 < bf2) ? true : false; is the same as if(bf1 < bf2) return true; else return false; Comparing Strings Here’s another example of overloading an operator, this time the equal to (==) operator. We’ll use it to compare two of our homemade String objects, returning true if they’re the same and false if they’re different. Here’s the listing for STREQUAL: //strequal.cpp //overloaded ‘==’ operator compares strings #include using namespace std; #include //for strcmp() //////////////////////////////////////////////////////////////// class String //user-defined string type { private: enum { SZ = 80 }; //size of String objects char str[SZ]; //holds a string public: String() //constructor, no args { strcpy(str, “”); } String( char s[] ) //constructor, one arg { strcpy(str, s); } void display() const //display a String { cout << str; } void getstr() //read a string { cin.get(str, SZ); } bool operator == (String ss) const //check for equality { return ( strcmp(str, ss.str)==0 ) ? true : false; } }; //////////////////////////////////////////////////////////////// Chapter 8336 09 3087 CH08 11/29/01 2:18 PM Page 336int main() { String s1 = “yes”; String s2 = “no”; String s3; cout << “\nEnter ‘yes’ or ‘no’: “; s3.getstr(); //get String from user if(s3==s1) //compare with “yes” cout << “You typed yes\n”; else if(s3==s2) //compare with “no” cout << “You typed no\n”; else cout << “You didn’t follow instructions\n”; return 0; } The main() part of this program uses the == operator twice, once to see if a string input by the user is “yes” and once to see if it’s “no.” Here’s the output when the user types “yes”: Enter ‘yes’ or ‘no’: yes You typed yes The operator==() function uses the library function strcmp() to compare the two C-strings. This function returns 0 if the strings are equal, a negative number if the first is less than the second, and a positive number if the first is greater than the second. Here less than and greater than are used in their lexicographical sense to indicate whether the first string appears before or after the second in an alphabetized listing. Other comparison operators, such as < and >, could also be used to compare the lexicographical value of strings. Or, alternatively, these comparison operators could be redefined to compare string lengths. Since you’re the one defining how the operators are used, you can use any definition that seems appropriate to your situation. Arithmetic Assignment Operators Let’s finish up our exploration of overloaded binary operators with an arithmetic assignment operator: the += operator. Recall that this operator combines assignment and addition into one step. We’ll use this operator to add one English distance to a second, leaving the result in the first. This is similar to the ENGLPLUS example shown earlier, but there is a subtle difference. Here’s the listing for ENGLPLEQ: // englpleq.cpp // overloaded ‘+=’ assignment operator Operator Overloading 8 O PERATOR O VERLOADING 337 09 3087 CH08 11/29/01 2:18 PM Page 337#include using namespace std; //////////////////////////////////////////////////////////////// class Distance //English Distance class { private: int feet; float inches; public: //constructor (no args) Distance() : feet(0), inches(0.0) { } //constructor (two args) Distance(int ft, float in) : feet(ft), inches(in) { } void getdist() //get length from user { cout << “\nEnter feet: “; cin >> feet; cout << “Enter inches: “; cin >> inches; } void showdist() const //display distance { cout << feet << “\’-” << inches << ‘\”’; } void operator += ( Distance ); }; //-------------------------------------------------------------- //add distance to this one void Distance::operator += (Distance d2) { feet += d2.feet; //add the feet inches += d2.inches; //add the inches if(inches >= 12.0) //if total exceeds 12.0, { //then decrease inches inches -= 12.0; //by 12.0 and feet++; //increase feet } //by 1 } //////////////////////////////////////////////////////////////// int main() { Distance dist1; //define dist1 dist1.getdist(); //get dist1 from user cout << “\ndist1 = “; dist1.showdist(); Distance dist2(11, 6.25); //define, initialize dist2 cout << “\ndist2 = “; dist2.showdist(); dist1 += dist2; //dist1 = dist1 + dist2 cout << “\nAfter addition,”; Chapter 8338 09 3087 CH08 11/29/01 2:18 PM Page 338cout << “\ndist1 = “; dist1.showdist(); cout << endl; return 0; } In this program we obtain a distance from the user and add to it a second distance, initialized to 11'–6.25'' by the program. Here’s a sample of interaction with the program: Enter feet: 3 Enter inches: 5.75 dist1 = 3’-5.75” dist2 = 11’-6.25” After addition, dist1 = 15’-0” In this program the addition is carried out in main() with the statement dist1 += dist2; This causes the sum of dist1 and dist2 to be placed in dist1. Notice the difference between the function used here, operator+=(), and that used in ENGLPLUS, operator+(). In the earlier operator+() function, a new object of type Distance had to be created and returned by the function so it could be assigned to a third Distance object, as in dist3 = dist1 + dist2; In the operator+=() function in ENGLPLEQ,the object that takes on the value of the sum is the object of which the function is a member. Thus it is feet and inches that are given values, not temporary variables used only to return an object. The operator+=() function has no return value; it returns type void. A return value is not necessary with arithmetic assignment operators such as +=, because the result of the assignment operator is not assigned to anything. The operator is used alone, in expressions like the one in the program. dist1 += dist2; If you wanted to use this operator in more complex expressions, like dist3 = dist1 += dist2; then you would need to provide a return value. You can do this by ending the operator+=() function with a statement like return Distance(feet, inches); in which a nameless object is initialized to the same values as this object and returned. Operator Overloading 8 O PERATOR O VERLOADING 339 09 3087 CH08 11/29/01 2:18 PM Page 339The Subscript Operator ([]) The subscript operator, [], which is normally used to access array elements, can be overloaded. This is useful if you want to modify the way arrays work in C++. For example, you might want to make a “safe” array: One that automatically checks the index numbers you use to access the array, to ensure that they are not out of bounds. (You can also use the vector class, described in Chapter 15, “The Standard Template Library.”) To demonstrate the overloaded subscript operator, we must return to another topic, first mentioned in Chapter 5: returning values from functions by reference. To be useful, the overloaded subscript operator must return by reference. To see why this is true, we’ll show three example programs that implement a safe array, each one using a different approach to inserting and reading the array elements: • Separate put() and get() functions •A single access() function using return by reference • The overloaded [] operator using return by reference All three programs create a class called safearay, whose only member data is an array of 100 int values, and all three check to ensure that all array accesses are within bounds. The main() program in each program tests the class by filling the safe array with values (each one equal to 10 times its array index) and then displaying them all to assure the user that everything is working as it should. Separate get() and put() Functions The first program provides two functions to access the array elements: putel() to insert a value into the array, and getel() to find the value of an array element. Both functions check the value of the index number supplied to ensure it’s not out of bounds; that is, less than 0 or larger than the array size (minus 1). Here’s the listing for ARROVER1: // arrover1.cpp // creates safe array (index values are checked before access) // uses separate put and get functions #include using namespace std; #include // for exit() const int LIMIT = 100; //////////////////////////////////////////////////////////////// class safearay { private: int arr[LIMIT]; public: Chapter 8340 09 3087 CH08 11/29/01 2:18 PM Page 340void putel(int n, int elvalue) //set value of element { if( n< 0 || n>=LIMIT ) { cout << “\nIndex out of bounds”; exit(1); } arr[n] = elvalue; } int getel(int n) const //get value of element { if( n< 0 || n>=LIMIT ) { cout << “\nIndex out of bounds”; exit(1); } return arr[n]; } }; //////////////////////////////////////////////////////////////// int main() { safearay sa1; for(int j=0; j using namespace std; Operator Overloading 8 O PERATOR O VERLOADING 341 09 3087 CH08 11/29/01 2:18 PM Page 341#include //for exit() const int LIMIT = 100; //array size //////////////////////////////////////////////////////////////// class safearay { private: int arr[LIMIT]; public: int& access(int n) //note: return by reference { if( n< 0 || n>=LIMIT ) { cout << “\nIndex out of bounds”; exit(1); } return arr[n]; } }; //////////////////////////////////////////////////////////////// int main() { safearay sa1; for(int j=0; j using namespace std; #include //for exit() const int LIMIT = 100; //array size //////////////////////////////////////////////////////////////// class safearay { private: int arr[LIMIT]; public: int& operator [](int n) //note: return by reference { if( n< 0 || n>=LIMIT ) { cout << “\nIndex out of bounds”; exit(1); } return arr[n]; } }; //////////////////////////////////////////////////////////////// int main() { safearay sa1; for(int j=0; j(floatvar); Casting provides explicit conversion: It’s obvious in the listing that static_cast() is intended to convert from float to int. However, such explicit conversions use the same built-in routines as implicit conversions. Conversions Between Objects and Basic Types When we want to convert between user-defined data types and basic types, we can’t rely on built-in conversion routines, since the compiler doesn’t know anything about user-defined types besides what we tell it. Instead, we must write these routines ourselves. Our next example shows how to convert between a basic type and a user-defined type. In this example the user-defined type is (surprise!) the English Distance class from previous examples, and the basic type is float, which we use to represent meters, a unit of length in the metric measurement system. The example shows conversion both from Distance to float, and from float to Distance. Here’s the listing for ENGLCONV: // englconv.cpp // conversions: Distance to meters, meters to Distance #include using namespace std; //////////////////////////////////////////////////////////////// class Distance //English Distance class { private: const float MTF; //meters to feet int feet; float inches; public: //constructor (no args) Distance() : feet(0), inches(0.0), MTF(3.280833F) { } //constructor (one arg) Distance(float meters) : MTF(3.280833F) { //convert meters to Distance float fltfeet = MTF * meters; //convert to float feet feet = int(fltfeet); //feet is integer part inches = 12*(fltfeet-feet); //inches is what’s left } //constructor (two args) Operator Overloading 8 O PERATOR O VERLOADING 345 09 3087 CH08 11/29/01 2:18 PM Page 345Distance(int ft, float in) : feet(ft), inches(in), MTF(3.280833F) { } void getdist() //get length from user { cout << “\nEnter feet: “; cin >> feet; cout << “Enter inches: “; cin >> inches; } void showdist() const //display distance { cout << feet << “\’-” << inches << ‘\”’; } operator float() const //conversion operator { //converts Distance to meters float fracfeet = inches/12; //convert the inches fracfeet += static_cast(feet); //add the feet return fracfeet/MTF; //convert to meters } }; //////////////////////////////////////////////////////////////// int main() { float mtrs; Distance dist1 = 2.35F; //uses 1-arg constructor to //convert meters to Distance cout << “\ndist1 = “; dist1.showdist(); mtrs = static_cast(dist1); //uses conversion operator //for Distance to meters cout << “\ndist1 = “ << mtrs << “ meters\n”; Distance dist2(5, 10.25); //uses 2-arg constructor mtrs = dist2; //also uses conversion op cout << “\ndist2 = “ << mtrs << “ meters\n”; // dist2 = mtrs; //error, = won’t convert return 0; } In main() the program first converts a fixed float quantity—2.35, representing meters—to feet and inches, using the one-argument constructor: Distance dist1 = 2.35F; Going in the other direction, it converts a Distance to meters in the statements mtrs = static_cast(dist2); Chapter 8346 09 3087 CH08 11/29/01 2:18 PM Page 346and mtrs = dist2; Here’s the output: dist1 = 7’-8.51949” ← this is 2.35 meters dist1 = 2.35 meters ← this is 7’–8.51949” dist2 = 1.78435 meters ← this is 5’–10.25” We’ve seen how conversions are performed using simple assignment statements in main(). Now let’s see what goes on behind the scenes, in the Distance member functions. Converting a user-defined type to a basic type requires a different approach than converting a basic type to a user-defined type. We’ll see how both types of conversions are carried out in ENGLCONV. From Basic to User-Defined To go from a basic type—float in this case—to a user-defined type such as Distance, we use a constructor with one argument. These are sometimes called conversion constructors. Here’s how this constructor looks in ENGLCONV: Distance(float meters) { float fltfeet = MTF * meters; feet = int(fltfeet); inches = 12 * (fltfeet-feet); } This function is called when an object of type Distance is created with a single argument. The function assumes that this argument represents meters. It converts the argument to feet and inches, and assigns the resulting values to the object. Thus the conversion from meters to Distance is carried out along with the creation of an object in the statement Distance dist1 = 2.35; From User-Defined to Basic What about going the other way, from a user-defined type to a basic type? The trick here is to create something called a conversion operator. Here’s where we do that in ENGLCONV: operator float() { float fracfeet = inches/12; fracfeet += float(feet); return fracfeet/MTF; } This operator takes the value of the Distance object of which it is a member, converts it to a float value representing meters, and returns this value. Operator Overloading 8 O PERATOR O VERLOADING 347 09 3087 CH08 11/29/01 2:18 PM Page 347This operator can be called with an explicit cast mtrs = static_cast(dist1); or with a simple assignment mtrs = dist2; Both forms convert the Distance object to its equivalent float value in meters. Conversion Between C-Strings and String Objects Here’s another example that uses a one-argument constructor and a conversion operator. It operates on the String class that we saw in the STRPLUS example earlier in this chapter. // strconv.cpp // convert between ordinary strings and class String #include using namespace std; #include //for strcpy(), etc. //////////////////////////////////////////////////////////////// class String //user-defined string type { private: enum { SZ = 80 }; //size of all String objects char str[SZ]; //holds a C-string public: String() //no-arg constructor { str[0] = ‘\0’; } String( char s[] ) //1-arg constructor { strcpy(str, s); } // convert C-string to String void display() const //display the String { cout << str; } operator char*() //conversion operator { return str; } //convert String to C-string }; //////////////////////////////////////////////////////////////// int main() { String s1; //use no-arg constructor //create and initialize C-string char xstr[] = “Joyeux Noel! “; s1 = xstr; //use 1-arg constructor // to convert C-string to String s1.display(); //display String Chapter 8348 09 3087 CH08 11/29/01 2:18 PM Page 348String s2 = “Bonne Annee!”; //uses 1-arg constructor //to initialize String cout << static_cast(s2); //use conversion operator cout << endl; //to convert String to C-string return 0; //before sending to << op } The one-argument constructor converts a normal string (an array of char) to an object of class String: String(char s[]) { strcpy(str, s); } The C-string s is passed as an argument, and copied into the str data member in a newly created String object, using the strcpy() library function. This conversion will be applied when a String is created, as in String s2 = “Bonne Annee!”; or it will be applied in assignment statements, as in s1 = xstr; where s1 is type String and xstr is a C-string. A conversion operator is used to convert from a String type to a C-string: operator char*() { return str; } The asterisk in this expression means pointer to. We won’t explore pointers until Chapter 10, but its use here is not hard to figure out. It means pointer to char, which is very similar to array of type char. Thus char* is similar to char[]. It’s another way of specifying a C-string data type. The conversion operator is used by the compiler in the statement cout << static_cast(s2); Here the s2 variable is an argument supplied to the overloaded operator <<. Since the << opera- tor doesn’t know anything about our user-defined String type, the compiler looks for a way to convert s2 to a type that << does know about. We specify the type we want to convert it to with the char* cast, so it looks for a conversion from String to C-string, finds our operator char*() function, and uses it to generate a C-string, which is then sent on to << to be displayed. (The effect is similar to calling the String::display() function, but given the ease and intuitive clarity of displaying with <<, the display() function is redundant and could be removed.) Operator Overloading 8 O PERATOR O VERLOADING 349 09 3087 CH08 11/29/01 2:18 PM Page 349Here’s the output from STRCONV: Joyeux Noel! Bonne Annee! The STRCONV example demonstrates that conversions take place automatically not only in assignment statements but in other appropriate places, such as in arguments sent to operators (such as <<) or functions. If you supply an operator or a function with arguments of the wrong type, they will be converted to arguments of an acceptable type, provided you have defined such a conversion. Note that you can’t use an explicit assignment statement to convert a String to a C-string: xstr = s2; The C-string xstr is an array, and you can’t normally assign to arrays (although as we’ll see in Chapter 11, when you overload the assignment operator, all sorts of things are possible). Conversions Between Objects of Different Classes What about converting between objects of different user-defined classes? The same two methods just shown for conversions between basic types and user-defined types also apply to conversions between two user-defined types. That is, you can use a one-argument constructor or you can use a conversion operator. The choice depends on whether you want to put the conversion routine in the class declaration of the source object or of the destination object. For example, suppose you say objecta = objectb; where objecta is a member of class A and objectb is a member of class B. Is the conversion routine located in class A (the destination class, since objecta receives the value) or class B (the source class)? We’ll look at both cases. Two Kinds of Time Our example programs will convert between two ways of measuring time: 12-hour time and 24-hour time. These methods of telling time are sometimes called civilian time and military time. Our time12 class will represent civilian time, as used in digital clocks and airport flight departure displays. We’ll assume that in this context there is no need for seconds, so time12 uses only hours (from 1 to 12), minutes, and an “a.m.” or “p.m.” designation. Our time24 class, which is for more exacting applications such as air navigation, uses hours (from 00 to 23), minutes, and seconds. Table 8.1 shows the differences. Chapter 8350 09 3087 CH08 11/29/01 2:18 PM Page 350TABLE 8.1 12-Hour and 24-Hour Time 12-Hour Time 24-Hour Time 12:00 a.m. (midnight) 00:00 12:01 a.m. 00:01 1:00 a.m. 01:00 6:00 a.m. 06:00 11:59 a.m 11:59 12:00 p.m. (noon) 12:00 12:01 p.m. 12:01 6:00 p.m. 18:00 11:59 p.m. 23:59 Note that 12 a.m. (midnight) in civilian time is 00 hours in military time. There is no 0 hour in civilian time. Routine in Source Object The first example program shows a conversion routine located in the source class. When the conversion routine is in the source class, it is commonly implemented as a conversion operator. Here’s the listing for TIMES1: //times1.cpp //converts from time24 to time12 using operator in time24 #include #include using namespace std; //////////////////////////////////////////////////////////////// class time12 { private: bool pm; //true = pm, false = am int hrs; //1 to 12 int mins; //0 to 59 public: //no-arg constructor time12() : pm(true), hrs(0), mins(0) { } //3-arg constructor time12(bool ap, int h, int m) : pm(ap), hrs(h), mins(m) { } void display() const //format: 11:59 p.m. { cout << hrs << ‘:’; Operator Overloading 8 O PERATOR O VERLOADING 351 09 3087 CH08 11/29/01 2:18 PM Page 351if(mins < 10) cout << ‘0’; //extra zero for “01” cout << mins << ‘ ‘; string am_pm = pm ? “p.m.” : “a.m.”; cout << am_pm; } }; //////////////////////////////////////////////////////////////// class time24 { private: int hours; //0 to 23 int minutes; //0 to 59 int seconds; //0 to 59 public: //no-arg constructor time24() : hours(0), minutes(0), seconds(0) { } time24(int h, int m, int s) : //3-arg constructor hours(h), minutes(m), seconds(s) { } void display() const //format: 23:15:01 { if(hours < 10) cout << ‘0’; cout << hours << ‘:’; if(minutes < 10) cout << ‘0’; cout << minutes << ‘:’; if(seconds < 10) cout << ‘0’; cout << seconds; } operator time12() const; //conversion operator }; //-------------------------------------------------------------- time24::operator time12() const //conversion operator { int hrs24 = hours; bool pm = hours < 12 ? false : true; //find am/pm //round secs int roundMins = seconds < 30 ? minutes : minutes+1; if(roundMins == 60) //carry mins? { roundMins=0; ++hrs24; if(hrs24 == 12 || hrs24 == 24) //carry hrs? pm = (pm==true) ? false : true; //toggle am/pm } int hrs12 = (hrs24 < 13) ? hrs24 : hrs24-12; Chapter 8352 09 3087 CH08 11/29/01 2:18 PM Page 352if(hrs12==0) //00 is 12 a.m. { hrs12=12; pm=false; } return time12(pm, hrs12, roundMins); } //////////////////////////////////////////////////////////////// int main() { int h, m, s; while(true) { //get 24-hr time from user cout << “Enter 24-hour time: \n”; cout << “ Hours (0 to 23): “; cin >> h; if(h > 23) //quit if hours > 23 return(1); cout << “ Minutes: “; cin >> m; cout << “ Seconds: “; cin >> s; time24 t24(h, m, s); //make a time24 cout << “You entered: “; //display the time24 t24.display(); time12 t12 = t24; //convert time24 to time12 cout << “\n12-hour time: “; //display equivalent time12 t12.display(); cout << “\n\n”; } return 0; } In the main() part of TIMES1 we define an object of type time24,called t24, and give it values for hours, minutes, and seconds obtained from the user. We also define an object of type time12, called t12, and initialize it to t24 in the statement time12 t12 = t24; Since these objects are from different classes, the assignment involves a conversion, and—as we specified—in this program the conversion operator is a member of the time24 class. Here’s its declarator: time24::operator time12() const //conversion operator } This function transforms the object of which it is a member to a time12 object, and returns this object, which main() then assigns to t12. Here’s some interaction with TIMES1: Operator Overloading 8 O PERATOR O VERLOADING 353 09 3087 CH08 11/29/01 2:18 PM Page 353Enter 24-hour time: Hours (0 to 23): 17 Minutes: 59 Seconds: 45 You entered: 17:59:45 12-hour time: 6:00 p.m. The seconds value is rounded up, pushing the 12-hour time from 5:59 p.m. to 6:00 p.m. Entering an hours value greater than 23 causes the program to exit. Routine in Destination Object Let’s see how the same conversion is carried out when the conversion routine is in the destination class. In this situation it’s common to use a one-argument constructor. However, things are complicated by the fact that the constructor in the destination class must be able to access the data in the source class to perform the conversion. The data in time24—hours, minutes and seconds—is private, so we must provide special member functions in time24 to allow direct access to it. These are called getHrs(), getMins(), and getSecs(). Here’s the listing for TIMES2: //times2.cpp //converts from time24 to time12 using constructor in time12 #include #include using namespace std; //////////////////////////////////////////////////////////////// class time24 { private: int hours; //0 to 23 int minutes; //0 to 59 int seconds; //0 to 59 public: //no-arg constructor time24() : hours(0), minutes(0), seconds(0) { } time24(int h, int m, int s) : //3-arg constructor hours(h), minutes(m), seconds(s) { } void display() const //format 23:15:01 { if(hours < 10) cout << ‘0’; cout << hours << ‘:’; if(minutes < 10) cout << ‘0’; cout << minutes << ‘:’; if(seconds < 10) cout << ‘0’; cout << seconds; } Chapter 8354 09 3087 CH08 11/29/01 2:18 PM Page 354int getHrs() const { return hours; } int getMins() const { return minutes; } int getSecs() const { return seconds; } }; //////////////////////////////////////////////////////////////// class time12 { private: bool pm; //true = pm, false = am int hrs; //1 to 12 int mins; //0 to 59 public: //no-arg constructor time12() : pm(true), hrs(0), mins(0) { } time12(time24); //1-arg constructor //3-arg constructor time12(bool ap, int h, int m) : pm(ap), hrs(h), mins(m) { } void display() const { cout << hrs << ‘:’; if(mins < 10) cout << ‘0’; //extra zero for “01” cout << mins << ‘ ‘; string am_pm = pm ? “p.m.” : “a.m.”; cout << am_pm; } }; //-------------------------------------------------------------- time12::time12( time24 t24 ) //1-arg constructor { //converts time24 to time12 int hrs24 = t24.getHrs(); //get hours //find am/pm pm = t24.getHrs() < 12 ? false : true; mins = (t24.getSecs() < 30) ? //round secs t24.getMins() : t24.getMins()+1; if(mins == 60) //carry mins? { mins=0; ++hrs24; if(hrs24 == 12 || hrs24 == 24) //carry hrs? pm = (pm==true) ? false : true; //toggle am/pm } hrs = (hrs24 < 13) ? hrs24 : hrs24-12; //convert hrs if(hrs==0) //00 is 12 a.m. Operator Overloading 8 O PERATOR O VERLOADING 355 09 3087 CH08 11/29/01 2:18 PM Page 355{ hrs=12; pm=false; } } //////////////////////////////////////////////////////////////// int main() { int h, m, s; while(true) { //get 24-hour time from user cout << “Enter 24-hour time: \n”; cout << “ Hours (0 to 23): “; cin >> h; if(h > 23) //quit if hours > 23 return(1); cout << “ Minutes: “; cin >> m; cout << “ Seconds: “; cin >> s; time24 t24(h, m, s); //make a time24 cout << “You entered: “; //display the time24 t24.display(); time12 t12 = t24; //convert time24 to time12 cout << “\n12-hour time: “; //display equivalent time12 t12.display(); cout << “\n\n”; } return 0; } The conversion routine is the one-argument constructor from the time12 class. This function sets the object of which it is a member to values that correspond to the time24 values of the object received as an argument. It works in much the same way as the conversion operator in TIMES1, except that it must work a little harder to access the data in the time24 object, using getHrs() and similar functions. The main() part of TIMES2 is the same as that in TIMES1. The one-argument constructor again allows the time24-to-time12 conversion to take place in the statement time12 t12 = t24; The output is similar as well. The difference is behind the scenes, where the conversion is handled by a constructor in the destination object rather than a conversion operator in the source object. Chapter 8356 09 3087 CH08 11/29/01 2:18 PM Page 356Conversions: When to Use What When should you use the one-argument constructor in the destination class, as opposed to the conversion operator in the source class? Mostly you can take your pick. However, sometimes the choice is made for you. If you have purchased a library of classes, you may not have access to their source code. If you use an object of such a class as the source in a conversion, you’ll have access only to the destination class, and you’ll need to use a one-argument constructor. If the library class object is the destination, you must use a conversion operator in the source. UML Class Diagrams We introduced the UML in Chapter 1, “The Big Picture.” Now that you know something about classes, let’s take a look at our first UML feature: the class diagram. This diagram offers a new way of looking at object-oriented programs, and may throw some additional light on the workings of the TIMES1 and TIMES2programs. Looking at the listing for TIMES1 we can see that there are two classes: time12 and time24. In a UML class diagram, classes are represented by rectangles, as shown in Figure 8.3. Operator Overloading 8 O PERATOR O VERLOADING 357 time12 time24 FIGURE 8.3 UML class diagram of the TIMES1program. Each class rectangle is divided into sections by horizontal lines. The class name goes in the top section. We don’t show them here, but you can include sections for member data (called attrib- utes in the UML) and member functions (called operations). Associations Classes may have various kinds of relationships with each other. The classes in TIMES1are related by association. We indicate this with a line connecting their rectangles. (We’ll see what another kind of class relationship, generalization, looks like in Chapter 9, “Inheritance.”) What constitutes an association? Conceptually, the real-world entities that are represented by classes in the program have some kind of obvious relationship. Drivers are related to cars, books are related to libraries, race horses are related to race tracks. If such entities were classes in a program, they would be related by association. 09 3087 CH08 11/29/01 2:18 PM Page 357In the TIMES2.CPP program, we can see that class time12 is associatd with class time24 because we are converting objects of one class into objects of the other. A class association actually implies that objects of the classes, rather than the classes themselves, have some kind of relationship. Typically, two classes are associated if an object of one class calls a member function (an operation) of an object of the other class. An association might also exist if an attribute of one class is an object of the other class. In the TIMES1program, an object of the time12 class, called t12,calls the conversion routine operator time12() in the object t24 of the time24 class. This happens in main() in the statement time12 t12 = t24; //convert time24 to time12 Such a call is represented by an association line between the two classes. Navigability We can add an open arrowhead to indicate the direction or navigability of the association. (As we’ll see later, closed arrowheads have a different meaning.) Because time12 calls time24,the arrow points from time12 to time24. It’s called a unidirectional association because it only goes one way. If each of two classes called an operation in the other, there would be arrowheads on both ends of the line and it would be called a bidirectional association. As are many things in the UML, navigability arrows are optional. Pitfalls of Operator Overloading and Conversion Operator overloading and type conversions give you the opportunity to create what amounts to an entirely new language. When a, b, and c are objects from user-defined classes, and + is overloaded, the statement a = b + c; can mean something quite different than it does when a, b, and c are variables of basic data types. The ability to redefine the building blocks of the language can be a blessing in that it can make your listing more intuitive and readable. It can also have the opposite effect, making your listing more obscure and hard to understand. Here are some guidelines. Use Similar Meanings Use overloaded operators to perform operations that are as similar as possible to those performed on basic data types. You could overload the + sign to perform subtraction, for example, but that would hardly make your listings more comprehensible. Chapter 8358 09 3087 CH08 11/29/01 2:18 PM Page 358Overloading an operator assumes that it makes sense to perform a particular operation on objects of a certain class. If we’re going to overload the + operator in class X,the result of adding two objects of class X should have a meaning at least somewhat similar to addition. For example, in this chapter we showed how to overload the + operator for the English Distance class. Adding two distances is clearly meaningful. We also overloaded + for the String class. Here we interpret the addition of two strings to mean placing one string after another to form a third. This also has an intuitively satisfying interpretation. But for many classes it may not be reasonable to talk about “adding” their objects. You probably wouldn’t want to add two objects of a class called employee that held personal data, for example. Use Similar Syntax Use overloaded operators in the same way you use basic types. For example, if alpha and beta are basic types, the assignment operator in the statement alpha += beta; sets alpha to the sum of alpha and beta. Any overloaded version of this operator should do something analogous. It should probably do the same thing as alpha = alpha + beta; where the + is overloaded. If you overload one arithmetic operator, you may for consistency want to overload all of them. This will prevent confusion. Some syntactical characteristics of operators can’t be changed. As you may have discovered, you can’t overload a binary operator to be a unary operator, or vice versa. Show Restraint Remember that if you have overloaded the + operator, anyone unfamiliar with your listing will need to do considerable research to find out what a statement like a = b + c; really means. If the number of overloaded operators grows too large, and if the operators are used in nonintuitive ways, the whole point of using them is lost, and reading the listing becomes harder instead of easier. Use overloaded operators sparingly, and only when the usage is obvious. When in doubt, use a function instead of an overloaded operator, since a function name can state its own purpose. If you write a function to find the left side of a string, for example, you’re better off calling it getleft() than trying to overload some operator such as && to do the same thing. Operator Overloading 8 O PERATOR O VERLOADING 359 09 3087 CH08 11/29/01 2:18 PM Page 359Avoid Ambiguity Suppose you use both a one-argument constructor and a conversion operator to perform the same conversion (time24 to time12,for example). How will the compiler know which conversion to use? It won’t. The compiler does not like to be placed in a situation where it doesn’t know what to do, and it will signal an error. So avoid doing the same conversion in more than one way. Not All Operators Can Be Overloaded The following operators cannot be overloaded: the member access or dot operator (.), the scope resolution operator (::), and the conditional operator (?:). Also, the pointer-to-member operator (->), which we have not yet encountered, cannot be overloaded. In case you wondered, no, you can’t create new operators (like *&) and try to overload them; only existing operators can be overloaded. Keywords explicit and mutable Let’s look at two unusual keywords: explicit and mutable. They have quite different effects, but are grouped together here because they both modify class members. The explicit keyword relates to data conversion, but mutable has a more subtle purpose. Preventing Conversions with explicit There may be some specific conversions you have decided are a good thing, and you’ve taken steps to make them possible by installing appropriate conversion operators and one-argument constructors, as shown in the TIME1 and TIME2examples. However, there may be other conversions that you don’t want to happen. You should actively discourage any conversion that you don’t want. This prevents unpleasant surprises. It’s easy to prevent a conversion performed by a conversion operator: just don’t define the operator. However, things aren’t so easy with constructors. You may want to construct objects using a single value of another type, but you may not want the implicit conversions a one- argument constructor makes possible in other situations. What to do? Standard C++ includes a keyword, explicit,to solve this problem. It’s placed just before the declaration of a one-argument constructor. The EXPLICIT example program (based on the ENGLCON program) shows how this looks. //explicit.cpp #include using namespace std; //////////////////////////////////////////////////////////////// class Distance //English Distance class Chapter 8360 09 3087 CH08 11/29/01 2:18 PM Page 360{ private: const float MTF; //meters to feet int feet; float inches; public: //no-args constructor Distance() : feet(0), inches(0.0), MTF(3.280833F) { } //EXPLICIT one-arg constructor explicit Distance(float meters) : MTF(3.280833F) { float fltfeet = MTF * meters; feet = int(fltfeet); inches = 12*(fltfeet-feet); } void showdist() //display distance { cout << feet << “\’-” << inches << ‘\”’; } }; //////////////////////////////////////////////////////////////// int main() { void fancyDist(Distance); //declaration Distance dist1(2.35F); //uses 1-arg constructor to //convert meters to Distance // Distance dist1 = 2.35F; //ERROR if ctor is explicit cout << “\ndist1 = “; dist1.showdist(); float mtrs = 3.0F; cout << “\ndist1 “; // fancyDist(mtrs); //ERROR if ctor is explicit return 0; } //-------------------------------------------------------------- void fancyDist(Distance d) { cout << “(in feet and inches) = “; d.showdist(); cout << endl; } This program includes a function (fancyDist()) that embellishes the output of a Distance object by printing the phrase “(in feet and inches)” before the feet and inches figures. The argument to this function is a Distance variable, and you can call fancyDist() with such a variable with no problem. Operator Overloading 8 O PERATOR O VERLOADING 361 09 3087 CH08 11/29/01 2:18 PM Page 361The tricky part is that, unless you take some action to prevent it, you can also call fancyDist() with a variable of type float as the argument: fancyDist(mtrs); The compiler will realize it’s the wrong type and look for a conversion operator. Finding a Distance constructor that takes type float as an argument, it will arrange for this constructor to convert float to Distance and pass the Distance value to the function. This is an implicit conversion, one which you may not have intended to make possible. However, if we make the constructor explicit, we prevent implicit conversions. You can check this by removing the comment symbol from the call to fancyDist() in the program: the compiler will tell you it can’t perform the conversion. Without the explicit keyword, this call is perfectly legal. As a side effect of the explicit constructor, note that you can’t use the form of object initialization that uses an equal sign Distance dist1 = 2.35F; whereas the form with parentheses Distance dist1(2.35F); works as it always has. Changing const Object Data Using mutable Ordinarily, when you create a const object (as described in Chapter 6), you want a guarantee that none of its member data can be changed. However, a situation occasionally arises where you want to create const objects that have some specific member data item that needs to be modified despite the object’s constness. As an example, let’s imagine a window (the kind that Windows programs commonly draw on the screen). It may be that some of the features of the window, such as its scrollbars and menus, are owned by the window. Ownership is common in various programming situations, and indicates a greater degree of independence than when one object is an attribute of another. In such a situation an object may remain unchanged, except that its owner may change. A scrollbar retains the same size, color, and orientation, but its ownership may be transferred from one window to another. It’s like what happens when your bank sells your mortgage to another bank; all the terms of the mortgage are the same, but the owner is different. Chapter 8362 09 3087 CH08 11/29/01 2:18 PM Page 362Let’s say we want to be able to create const scrollbars in which attributes remain unchanged, except for their ownership. That’s where the mutable keyword comes in. The MUTABLE program shows how this looks. //mutable.cpp #include #include using namespace std; //////////////////////////////////////////////////////////////// class scrollbar { private: int size; //related to constness mutable string owner; //not relevant to constness public: scrollbar(int sz, string own) : size(sz), owner(own) { } void setSize(int sz) //changes size { size = sz; } void setOwner(string own) const //changes owner { owner = own; } int getSize() const //returns size { return size; } string getOwner() const //returns owner { return owner; } }; //////////////////////////////////////////////////////////////// int main() { const scrollbar sbar(60, “Window1”); // sbar.setSize(100); //can’t do this to const obj sbar.setOwner(“Window2”); //this is OK //these are OK too cout << sbar.getSize() << “, “ << sbar.getOwner() << endl; return 0; } The size attribute represents the scrollbar data that cannot be modified in const objects. The owner attribute, however, can change, even if the object is const. To permit this, it’s made mutable. In main() we create a const object sbar. Its size cannot be modified, but its owner can, using the setOwner() function. (In a non-const object, of course, both attributes could be modified.) In this situation, sbar is said to have logical constness. That means that in theory it can’t be modified, but in practice it can, in a limited way. Operator Overloading 8 O PERATOR O VERLOADING 363 09 3087 CH08 11/29/01 2:18 PM Page 363Summary In this chapter we’ve seen how the normal C++ operators can be given new meanings when applied to user-defined data types. The keyword operator is used to overload an operator, and the resulting operator will adopt the meaning supplied by the programmer. Closely related to operator overloading is the issue of type conversion. Some conversions take place between user-defined types and basic types. Two approaches are used in such conversions: A one-argument constructor changes a basic type to a user-defined type, and a conversion operator converts a user-defined type to a basic type. When one user-defined type is converted to another, either approach can be used. Table 8.2 summarizes these conversions. TABLE 8.2 Type Conversions Routine in Destination Routine in Source Basic to basic (Built-In Conversion Operators) Basic to class Constructor N/A Class to basic N/A Conversion operator Class to class Constructor Conversion operator A constructor given the keyword explicit cannot be used in implicit data conversion situations. A data member given the keyword mutable can be changed, even if its object is const. UML class diagrams show classes and relationships between classes. An association represents a conceptual relationship between the real-world objects that the program’s classes represent. Associations can have a direction from one class to another; this is called navigability. Questions Answers to these questions can be found in Appendix G. 1. Operator overloading is a. making C++ operators work with objects. b. giving C++ operators more than they can handle. c. giving new meanings to existing C++ operators. d. making new C++ operators. 2. Assuming that class X does not use any overloaded operators, write a statement that subtracts an object of class X, x1,from another such object, x2, and places the result in x3. Chapter 8364 09 3087 CH08 11/29/01 2:18 PM Page 3643. Assuming that class X includes a routine to overload the - operator, write a statement that would perform the same task as that specified in Question 2. 4. True or false: The >= operator can be overloaded. 5. Write a complete definition for an overloaded operator for the Counter class of the COUNTPP1example that, instead of incrementing the count, decrements it. 6. How many arguments are required in the definition of an overloaded unary operator? 7. Assume a class C with objects obj1, obj2, and obj3. For the statement obj3 = obj1 - obj2 to work correctly, the overloaded - operator must a. take two arguments. b. return a value. c. create a named temporary object. d. use the object of which it is a member as an operand. 8. Write a complete definition for an overloaded ++ operator for the Distance class from the ENGLPLUS example. It should add 1 to the feet member data, and make possible statements like dist1++; 9. Repeat Question 8, but allow statements like the following: dist2 = dist1++; 10. When used in prefix form, what does the overloaded ++ operator do differently from what it does in postfix form? 11. Here are two declarators that describe ways to add two string objects: void add(String s1, String s2) String operator + (String s) Match the following from the first declarator with the appropriate selection from the second: function name (add) matches _________. return value (type void) matches _________. first argument (s1) matches _________. second argument (s2) matches _________. object of which function is a member matches _________. a. argument (s) b. object of which operator is a member c. operator (+) d. return value (type String) e. no match for this item Operator Overloading 8 O PERATOR O VERLOADING 365 09 3087 CH08 11/29/01 2:18 PM Page 36512. True or false: An overloaded operator always requires one less argument than its number of operands. 13. When you overload an arithmetic assignment operator, the result a. goes in the object to the right of the operator. b. goes in the object to the left of the operator. c. goes in the object of which the operator is a member. d. must be returned. 14. Write the complete definition of an overloaded ++ operator that works with the String class from the STRPLUS example and has the effect of changing its operand to uppercase. You can use the library function toupper() (header file CCTYPE), which takes as its only argument the character to be changed and returns the changed character (or the same character if no change is necessary). 15. To convert from a user-defined class to a basic type, you would most likely use a. a built-in conversion operator. b. a one-argument constructor. c. an overloaded = operator. d. a conversion operator that’s a member of the class. 16. True or false: The statement objA=objB; will cause a compiler error if the objects are of different classes. 17. To convert from a basic type to a user-defined class, you would most likely use a. a built-in conversion operator. b. a one-argument constructor. c. an overloaded = operator. d. a conversion operator that’s a member of the class. 18. True or false: If you’ve defined a constructor to handle definitions like aclass obj = intvar; you can also make statements like obj = intvar;. 19. If objA is in class A, and objB is in class B,and you want to say objA = objB;, and you want the conversion routine to go in class A, what type of conversion routine might you use? 20. True or false: The compiler won’t object if you overload the * operator to perform divi- sion. 21. In a UML class diagram, an association arises whenever a. two classes are in the same program. b. one class is descended from another. c. two classes use the same global variable. d. one class calls a member function in the other class. Chapter 8366 09 3087 CH08 11/29/01 2:19 PM Page 36622. In the UML, member data items are called _________ and member functions are called ___________. 23. True or false: rectangles that symbolize classes have rounded corners. 24. Navigability from class A to class B means that a. an object of class A can call an operation in an object of class B. b. there is a relationship between class A and class B. c. objects can go from class A to class B. d. messages from class B are received by class A. Exercises Answers to starred exercises can be found in Appendix G. *1. To the Distance class in the ENGLPLUS program in this chapter, add an overloaded - operator that subtracts two distances. It should allow statements like dist3= dist1-dist2;. Assume that the operator will never be used to subtract a larger number from a smaller one (that is, negative distances are not allowed). *2. Write a program that substitutes an overloaded += operator for the overloaded + operator in the STRPLUS program in this chapter. This operator should allow statements like s1 += s2; where s2 is added (concatenated) to s1 and the result is left in s1. The operator should also permit the results of the operation to be used in other calculations, as in s3 = s1 += s2; *3. Modify the time class from Exercise 3 in Chapter 6 so that instead of a function add_time() it uses the overloaded + operator to add two times. Write a program to test this class. *4. Create a class Int based on Exercise 1 in Chapter 6. Overload four integer arithmetic operators (+, -, *, and /) so that they operate on objects of type Int. If the result of any such arithmetic operation exceeds the normal range of ints (in a 32-bit environment)— from 2,147,483,648 to –2,147,483,647—have the operator print a warning and terminate the program. Such a data type might be useful where mistakes caused by arithmetic over- flow are unacceptable. Hint: To facilitate checking for overflow, perform the calculations using type long double. Write a program to test this class. 5. Augment the time class referred to in Exercise 3 to include overloaded increment (++) and decrement (--) operators that operate in both prefix and postfix notation and return values. Add statements to main() to test these operators. Operator Overloading 8 O PERATOR O VERLOADING 367 09 3087 CH08 11/29/01 2:19 PM Page 3676. Add to the time class of Exercise 5 the ability to subtract two time values using the overloaded (-) operator, and to multiply a time value by a number of type float,using the overloaded (*) operator. 7. Modify the fraction class in the four-function fraction calculator from Exercise 11 in Chapter 6 so that it uses overloaded operators for addition, subtraction, multiplication, and division. (Remember the rules for fraction arithmetic in Exercise 12 in Chapter 3, “Loops and Decisions.”) Also overload the == and != comparison operators, and use them to exit from the loop if the user enters 0/1, 0/1 for the values of the two input fractions. You may want to modify the lowterms() function so that it returns the value of its argument reduced to lowest terms. This makes it more useful in the arithmetic functions, where it can be applied just before the answer is returned. 8. Modify the bMoney class from Exercise 12 in Chapter 7, “Arrays and Strings,” to include the following arithmetic operations, performed with overloaded operators: bMoney = bMoney + bMoney bMoney = bMoney - bMoney bMoney = bMoney * long double (price per widget times number of widgets) long double = bMoney / bMoney (total price divided by price per widget) bMoney = bMoney / long double (total price divided by number of widgets) Notice that the / operator is overloaded twice. The compiler can distinguish between the two usages because the arguments are different. Remember that it’s easy to perform arithmetic operations on bMoney objects by performing the same operation on their long double data. Make sure the main() program asks the user to enter two money strings and a floating- point number. It should then carry out all five operations and display the results. This should happen in a loop, so the user can enter more numbers if desired. Some money operations don’t make sense: bMoney * bMoney doesn’t represent anything real, since there is no such thing as square money; and you can’t add bMoney to long double (what’s dollars plus widgets?). To make it impossible to compile such illegal operations, don’t include conversion operators for bMoney to long double or long double to bMoney. If you do, and you write an expression like bmon2 = bmon1 + widgets; // doesn’t make sense then the compiler will automatically convert widgets to bMoney and carry out the addition. Without them, the compiler will flag such conversions as errors, making it easier to catch conceptual mistakes. Also, make any conversion constructors explicit. There are some other plausible money operations that we don’t yet know how to perform with overloaded operators, since they require an object on the right side of the operator but not the left: long double * bMoney // can’t do this yet: bMoney only on right long double / bMoney // can’t do this yet: bMoney only on right We’ll learn how to handle this situation when we discuss friend functions in Chapter 11. Chapter 8368 09 3087 CH08 11/29/01 2:19 PM Page 3689. Augment the safearay class in the ARROVER3program in this chapter so that the user can specify both the upper and lower bound of the array (indexes running from 100 to 200, for example). Have the overloaded subscript operator check the index each time the array is accessed to ensure that it is not out of bounds. You’ll need to add a two- argument constructor that specifies the upper and lower bounds. Since we have not yet learned how to allocate memory dynamically, the member data will still be an array that starts at 0 and runs up to 99, but perhaps you can map the indexes for the safearay into different indexes in the real int array. For example, if the client selects a range from 100 to 175, you could map this into the range from arr[0] to arr[75]. 10. For math buffs only: Create a class Polar that represents the points on the plain as polar coordinates (radius and angle). Create an overloaded +operator for addition of two Polar quantities. “Adding” two points on the plain can be accomplished by adding their X coordinates and then adding their Y coordinates. This gives the X and Y coordinates of the “answer.” Thus you’ll need to convert two sets of polar coordinates to rectangular coordinates, add them, then convert the resulting rectangular representation back to polar. 11. Remember the sterling structure? We saw it in Exercise 10 in Chapter 2, “C++ Programming Basics,” and in Exercise 11 in Chapter 5, among other places. Turn it into a class, with pounds (type long), shillings (type int), and pence (type int) data items. Create the following member functions: • no-argument constructor • one-argument constructor, taking type double (for converting from decimal pounds) •three-argument constructor, taking pounds, shillings, and pence • getSterling() to get an amount in pounds, shillings, and pence from the user, format £9.19.11 • putSterling() to display an amount in pounds, shillings, and pence, format £9.19.11 • addition (sterling + sterling) using overloaded + operator •subtraction (sterling - sterling) using overloaded - operator • multiplication (sterling * double) using overloaded * operator •division (sterling / sterling) using overloaded / operator •division (sterling / double) using overloaded / operator • operator double (to convert to double) To perform arithmetic, you could (for example) add each object’s data separately: Add the pence, carry, add the shillings, carry, and so on. However, it’s easier to use the conversion operator to convert both sterling objects to type double, perform the arithmetic on the doubles, and convert back to sterling. Thus the overloaded + operator looks like this: Operator Overloading 8 O PERATOR O VERLOADING 369 09 3087 CH08 11/29/01 2:19 PM Page 369sterling sterling::operator + (sterling s2) { return sterling( double(sterling(pounds, shillings, pence)) + double(s2) ); } This creates two temporary double variables, one derived from the object of which the function is a member, and one derived from the argument s2. These double variables are then added, and the result is converted back to sterling and returned. Notice that we use a different philosophy with the sterling class than with the bMoney class. With sterling we use conversion operators, thus giving up the ability to catch illegal math operations but gaining simplicity in writing the overloaded math operators. 12. Write a program that incorporates both the bMoney class from Exercise 8 and the sterling class from Exercise 11. Write conversion operators to convert between bMoney and sterling,assuming that one pound (£1.0.0) equals fifty dollars ($50.00). This was the approximate exchange rate in the 19th century when the British Empire was at its height and the pounds-shillings-pence format was in use. Write a main() program that allows the user to enter an amount in either currency and then converts it to the other currency and displays the result. Minimize any modifications to the existing bMoney and sterling classes. Chapter 8370 09 3087 CH08 11/29/01 2:19 PM Page 370CHAPTER 9 Inheritance IN THIS CHAPTER • Derived Class and Base Class 373 • Derived Class Constructors 380 • Overriding Member Functions 382 • Which Function Is Used? 383 •Inheritance in the English Distance Class 384 •Class Hierarchies 388 •Inheritance and Graphics Shapes 393 • Public and Private Inheritance 396 • Levels of Inheritance 399 • Multiple Inheritance 403 • private Derivation in EMPMULT 409 • Ambiguity in Multiple Inheritance 413 • Aggregation: Classes Within Classes 414 •Inheritance and Program Development 420 10 3087 CH09 11/29/01 2:22 PM Page 371Chapter 9372 Inheritance is probably the most powerful feature of object-oriented programming, after classes themselves. Inheritance is the process of creating new classes, called derived classes,from existing or base classes. The derived class inherits all the capabilities of the base class but can add embellishments and refinements of its own. The base class is unchanged by this process. The inheritance relationship is shown in Figure 9.1. FIGURE 9.1 Inheritance. The arrow in Figure 9.1 goes in the opposite direction of what you might expect. If it pointed down we would label it inheritance. However, the more common approach is to point the arrow up, from the derived class to the base class, and to think of it as a “derived from” arrow. 10 3087 CH09 11/29/01 2:22 PM Page 372Inheritance is an essential part of OOP. Its big payoff is that it permits code reusability. Once a base class is written and debugged, it need not be touched again, but, using inheritance, can nevertheless be adapted to work in different situations. Reusing existing code saves time and money and increases a program’s reliability. Inheritance can also help in the original conceptu- alization of a programming problem, and in the overall design of the program. An important result of reusability is the ease of distributing class libraries. A programmer can use a class created by another person or company, and, without modifying it, derive other classes from it that are suited to particular situations. We’ll examine these features of inheritance in more detail after we’ve seen some specific instances of inheritance at work. Derived Class and Base Class Remember the COUNTPP3example from Chapter 8, “Operator Overloading”? This program used a class Counter as a general-purpose counter variable. A count could be initialized to 0 or to a specified number with constructors, incremented with the ++ operator, and read with the get_count() operator. Let’s suppose that we have worked long and hard to make the Counter class operate just the way we want, and we’re pleased with the results, except for one thing. We really need a way to decrement the count. Perhaps we’re counting people entering a bank, and we want to increment the count when they come in and decrement it when they go out, so that the count represents the number of people in the bank at any moment. We could insert a decrement routine directly into the source code of the Counter class. However, there are several reasons that we might not want to do this. First, the Counter class works very well and has undergone many hours of testing and debugging. (Of course that’s an exaggeration in this case, but it would be true in a larger and more complex class.) If we start fooling around with the source code for Counter,the testing process will need to be carried out again, and of course we may foul something up and spend hours debugging code that worked fine before we modified it. In some situations there might be another reason for not modifying the Counter class: We might not have access to its source code, especially if it was distributed as part of a class library. (We’ll discuss this issue further in Chapter 13, “Multifile Programs.”) To avoid these problems we can use inheritance to create a new class based on Counter, without modifying Counter itself. Here’s the listing for COUNTEN, which includes a new class, CountDn, that adds a decrement operator to the Counter class: Inheritance 9 I NHERITANCE 373 10 3087 CH09 11/29/01 2:22 PM Page 373// counten.cpp // inheritance with Counter class #include using namespace std; //////////////////////////////////////////////////////////////// class Counter //base class { protected: //NOTE: not private unsigned int count; //count public: Counter() : count(0) //no-arg constructor { } Counter(int c) : count(c) //1-arg constructor { } unsigned int get_count() const //return count { return count; } Counter operator ++ () //incr count (prefix) { return Counter(++count); } }; //////////////////////////////////////////////////////////////// class CountDn : public Counter //derived class { public: Counter operator -- () //decr count (prefix) { return Counter(--count); } }; //////////////////////////////////////////////////////////////// int main() { CountDn c1; //c1 of class CountDn cout << “\nc1=” << c1.get_count(); //display c1 ++c1; ++c1; ++c1; //increment c1, 3 times cout << “\nc1=” << c1.get_count(); //display it --c1; --c1; //decrement c1, twice cout << “\nc1=” << c1.get_count(); //display it cout << endl; return 0; } The listing starts off with the Counter class, which (with one small exception, which we’ll look at later) has not changed since its appearance in COUNTPP3. Notice that, for simplicity, we haven’t modeled this program on the POSTFIX program, which incorporated the second overloaded ++ operator to provide postfix notation. Chapter 9374 10 3087 CH09 11/29/01 2:22 PM Page 374Specifying the Derived Class Following the Counter class in the listing is the specification for a new class, CountDn. This class incorporates a new function, operator--(), which decrements the count. However—and here’s the key point—the new CountDn class inherits all the features of the Counter class. CountDn doesn’t need a constructor or the get_count() or operator++() functions, because these already exist in Counter. The first line of CountDn specifies that it is derived from Counter: class CountDn : public Counter Here we use a single colon (not the double colon used for the scope resolution operator), followed by the keyword public and the name of the base class Counter. This sets up the relationship between the classes. This line says that CountDn is derived from the base class Counter. (We’ll explore the effect of the keyword public later.) Generalization in UML Class Diagrams In the UML, inheritance is called generalization, because the parent class is a more general form of the child class. Or to put it another way, the child is more specific version of the parent. (We introduced the UML in Chapter 1, “The Big Picture,” and encountered class diagrams in Chapter 8, “Operator Overloading.”) The generalization in the COUNTEN program is shown in Figure 9.2. Inheritance 9 I NHERITANCE 375 Counter count counter() counter(int) get_count() operator++() CountDn operator--() FIGURE 9.2 UML class diagram for COUNTEN. 10 3087 CH09 11/29/01 2:22 PM Page 375In UML class diagrams, generalization is indicated by a triangular arrowhead on the line connecting the parent and child classes. Remember that the arrow means inherited from or derived from or is a more specific version of. The direction of the arrow emphasizes that the derived class refers to functions and data in the base class, while the base class has no access to the derived class. Notice that we’ve added attributes (member data) and operations (member functions) to the classes in the diagram. The top area holds the class title, the middle area holds attributes, and the bottom area is for operations. Accessing Base Class Members An important topic in inheritance is knowing when a member function in the base class can be used by objects of the derived class. This is called accessibility. Let’s see how the compiler handles the accessibility issue in the COUNTEN example. Substituting Base Class Constructors In the main() part of COUNTEN we create an object of class CountDn: CountDn c1; This causes c1 to be created as an object of class CountDn and initialized to 0. But wait—how is this possible? There is no constructor in the CountDn class specifier, so what entity carries out the initialization? It turns out that—at least under certain circumstances—if you don’t spec- ify a constructor, the derived class will use an appropriate constructor from the base class. In COUNTEN there’s no constructor in CountDn,so the compiler uses the no-argument constructor from Count. This flexibility on the part of the compiler—using one function because another isn’t available— appears regularly in inheritance situations. Generally, the substitution is what you want, but sometimes it can be unnerving. Substituting Base Class Member Functions The object c1 of the CountDn class also uses the operator++() and get_count() functions from the Counter class. The first is used to increment c1: ++c1; The second is used to display the count in c1: cout << “\nc1=” << c1.get_count(); Again the compiler, not finding these functions in the class of which c1 is a member, uses member functions from the base class. Chapter 9376 10 3087 CH09 11/29/01 2:22 PM Page 376Output of COUNTEN In main() we increment c1 three times, print out the resulting value, decrement c1 twice, and finally print out its value again. Here’s the output: c1=0 ← after initialization c1=3 ← after ++c1, ++c1, ++c1 c1=1 ← after --c1, --c1 The ++ operator, the constructors, the get_count() function in the Counter class, and the -- operator in the CountDn class all work with objects of type CountDn. The protected Access Specifier We have increased the functionality of a class without modifying it. Well, almost without modifying it. Let’s look at the single change we made to the Counter class. The data in the classes we’ve looked at so far, including count in the Counter class in the earlier COUNTPP3program, have used the private access specifier. In the Counter class in COUNTEN, count is given a new specifier: protected. What does this do? Let’s first review what we know about the access specifiers private and public. A member function of a class can always access class members, whether they are public or private. But an object declared externally can only invoke (using the dot operator, for example) public members of the class. It’s not allowed to use private members. For instance, suppose an object objA is an instance of class A, and function funcA() is a member function of A. Then in main() (or any other function that is not a member of A) the statement objA.funcA(); will not be legal unless funcA() is public. The object objA cannot invoke private members of class A. Private members are, well, private. This is shown in Figure 9.3. This is all we need to know if we don’t use inheritance. With inheritance, however, there is a whole raft of additional possibilities. The question that concerns us at the moment is, can member functions of the derived class access members of the base class? In other words, can operator--() in CountDn access count in Counter? The answer is that member functions can access members of the base class if the members are public,or if they are protected. They can’t access private members. We don’t want to make count public, since that would allow it to be accessed by any function anywhere in the program and eliminate the advantages of data hiding. A protected member, on the other hand, can be accessed by member functions in its own class or—and here’s the key—in any class derived from its own class. It can’t be accessed from functions outside these classes, such as main(). This is just what we want. The situation is shown in Figure 9.4. Inheritance 9 I NHERITANCE 377 10 3087 CH09 11/29/01 2:22 PM Page 377FIGURE 9.3 Access specifiers without inheritance. Chapter 9378 class A ObjA Object of class A can access only public members of A. Member function of class A can access both private and public members. private Not allowed public FIGURE 9.4 Access specifiers with inheritance. 10 3087 CH09 11/29/01 2:22 PM Page 378Table 9.1 summarizes the situation in a different way. TABLE 9.1 Inheritance and Accessibility Access Accessible from Accessible from Accessible from Specifier Own Class Derived Class Objects Outside Class public yes yes yes protected yes yes no private yes no no The moral is that if you are writing a class that you suspect might be used, at any point in the future, as a base class for other classes, then any member data that the derived classes might need to access should be made protected rather than private. This ensures that the class is “inheritance ready.” Dangers of protected You should know that there’s a disadvantage to making class members protected. Say you’ve written a class library, which you’re distributing to the public. Any programmer who buys this library can access protected members of your classes simply by deriving other classes from them. This makes protected members considerably less secure than private members. To avoid corrupted data, it’s often safer to force derived classes to access data in the base class using only public functions in the base class, just as ordinary main() programs must do. Using the protected specifier leads to simpler programming, so we rely on it—perhaps a bit too much— in the examples in this book. You’ll need to weigh the advantages of protected against its disadvantages in your own programs. Base Class Unchanged Remember that, even if other classes have been derived from it, the base class remains unchanged. In the main() part of COUNTEN, we could define objects of type Counter: Counter c2; ← object of base class Such objects would behave just as they would if CountDn didn’t exist. Note also that inheritance doesn’t work in reverse. The base class and its objects don’t know anything about any classes derived from the base class. In this example that means that objects of class Counter,such as c2, can’t use the operator--() function in CountDn. If you want a counter that you can decrement, it must be of class CountDn, not Counter. Inheritance 9 I NHERITANCE 379 10 3087 CH09 11/29/01 2:22 PM Page 379Other Terms In some languages the base class is called the superclass and the derived class is called the subclass. Some writers also refer to the base class as the parent and the derived class as the child. Derived Class Constructors There’s a potential glitch in the COUNTEN program. What happens if we want to initialize a CountDn object to a value? Can the one-argument constructor in Counter be used? The answer is no. As we saw in COUNTEN, the compiler will substitute a no-argument constructor from the base class, but it draws the line at more complex constructors. To make such a definition work we must write a new set of constructors for the derived class. This is shown in the COUNTEN2program. // counten2.cpp // constructors in derived class #include using namespace std; //////////////////////////////////////////////////////////////// class Counter { protected: //NOTE: not private unsigned int count; //count public: Counter() : count() //constructor, no args { } Counter(int c) : count(c) //constructor, one arg { } unsigned int get_count() const //return count { return count; } Counter operator ++ () //incr count (prefix) { return Counter(++count); } }; //////////////////////////////////////////////////////////////// class CountDn : public Counter { public: CountDn() : Counter() //constructor, no args { } CountDn(int c) : Counter(c) //constructor, 1 arg { } CountDn operator -- () //decr count (prefix) { return CountDn(--count); } }; //////////////////////////////////////////////////////////////// int main() { Chapter 9380 10 3087 CH09 11/29/01 2:22 PM Page 380CountDn c1; //class CountDn CountDn c2(100); cout << “\nc1=” << c1.get_count(); //display cout << “\nc2=” << c2.get_count(); //display ++c1; ++c1; ++c1; //increment c1 cout << “\nc1=” << c1.get_count(); //display it --c2; --c2; //decrement c2 cout << “\nc2=” << c2.get_count(); //display it CountDn c3 = --c2; //create c3 from c2 cout << “\nc3=” << c3.get_count(); //display c3 cout << endl; return 0; } This program uses two new constructors in the CountDn class. Here is the no-argument constructor: CountDn() : Counter() { } This constructor has an unfamiliar feature: the function name following the colon. This construction causes the CountDn() constructor to call the Counter() constructor in the base class. In main(), when we say CountDn c1; the compiler will create an object of type CountDn and then call the CountDn constructor to initialize it. This constructor will in turn call the Counter constructor, which carries out the work. The CountDn() constructor could add additional statements of its own, but in this case it doesn’t need to, so the function body between the braces is empty. Calling a constructor from the initialization list may seem odd, but it makes sense. You want to initialize any variables, whether they’re in the derived class or the base class, before any statements in either the derived or base-class constructors are executed. By calling the base- class constructor before the derived-class constructor starts to execute, we accomplish this. The statement CountDn c2(100); in main() uses the one-argument constructor in CountDn. This constructor also calls the corresponding one-argument constructor in the base class: CountDn(int c) : Counter(c) ← argument c is passed to Counter { } Inheritance 9 I NHERITANCE 381 10 3087 CH09 11/29/01 2:22 PM Page 381This construction causes the argument c to be passed from CountDn() to Counter(), where it is used to initialize the object. In main(), after initializing the c1 and c2 objects, we increment one and decrement the other and then print the results. The one-argument constructor is also used in an assignment statement. CountDn c3 = --c2; Overriding Member Functions You can use member functions in a derived class that override—that is, have the same name as—those in the base class. You might want to do this so that calls in your program work the same way for objects of both base and derived classes. Here’s an example based on the STAKARAY program from Chapter 7, “Arrays and Strings.” That program modeled a stack, a simple data storage device. It allowed you to push integers onto the stack and pop them off. However, STAKARAY had a potential flaw. If you tried to push too many items onto the stack, the program might bomb, since data would be placed in memory beyond the end of the st[] array. Or if you tried to pop too many items, the results would be meaningless, since you would be reading data from memory locations outside the array. To cure these defects we’ve created a new class, Stack2,derived from Stack. Objects of Stack2 behave in exactly the same way as those of Stack,except that you will be warned if you attempt to push too many items on the stack or if you try to pop an item from an empty stack. Here’s the listing for STAKEN: // staken.cpp // overloading functions in base and derived classes #include using namespace std; #include //for exit() //////////////////////////////////////////////////////////////// class Stack { protected: //NOTE: can’t be private enum { MAX = 3 }; //size of stack array int st[MAX]; //stack: array of integers int top; //index to top of stack public: Stack() //constructor { top = -1; } void push(int var) //put number on stack { st[++top] = var; } int pop() //take number off stack { return st[top--]; } Chapter 9382 10 3087 CH09 11/29/01 2:22 PM Page 382}; //////////////////////////////////////////////////////////////// class Stack2 : public Stack { public: void push(int var) //put number on stack { if(top >= MAX-1) //error if stack full { cout << “\nError: stack is full”; exit(1); } Stack::push(var); //call push() in Stack class } int pop() //take number off stack { if(top < 0) //error if stack empty { cout << “\nError: stack is empty\n”; exit(1); } return Stack::pop(); //call pop() in Stack class } }; //////////////////////////////////////////////////////////////// int main() { Stack2 s1; s1.push(11); //push some values onto stack s1.push(22); s1.push(33); cout << endl << s1.pop(); //pop some values from stack cout << endl << s1.pop(); cout << endl << s1.pop(); cout << endl << s1.pop(); //oops, popped one too many... cout << endl; return 0; } In this program the Stack class is just the same as it was in the STAKARAY program, except that the data members have been made protected. Which Function Is Used? The Stack2 class contains two functions, push() and pop(). These functions have the same names, and the same argument and return types, as the functions in Stack. When we call these functions from main(),in statements like s1.push(11); Inheritance 9 I NHERITANCE 383 10 3087 CH09 11/29/01 2:22 PM Page 383how does the compiler know which of the two push() functions to use? Here’s the rule: When the same function exists in both the base class and the derived class, the function in the derived class will be executed. (This is true of objects of the derived class. Objects of the base class don’t know anything about the derived class and will always use the base class functions.) We say that the derived class function overrides the base class function. So in the preceding state- ment, since s1 is an object of class Stack2,the push() function in Stack2 will be executed, not the one in Stack. The push() function in Stack2 checks to see whether the stack is full. If it is, it displays an error message and causes the program to exit. If it isn’t, it calls the push() function in Stack. Similarly, the pop() function in Stack2 checks to see whether the stack is empty. If it is, it prints an error message and exits; otherwise, it calls the pop() function in Stack. In main() we push three items onto the stack, but we pop four. The last pop elicits an error message 33 22 11 Error: stack is empty and terminates the program. Scope Resolution with Overridden Functions How do push() and pop() in Stack2 access push() and pop() in Stack? They use the scope resolution operator, ::,in the statements Stack::push(var); and return Stack::pop(); These statements specify that the push() and pop() functions in Stack are to be called. Without the scope resolution operator, the compiler would think the push() and pop() functions in Stack2 were calling themselves, which—in this case—would lead to program failure. Using the scope resolution operator allows you to specify exactly what class the function is a member of. Inheritance in the English Distance Class Here’s a somewhat more complex example of inheritance. So far in this book the various programs that used the English Distance class assumed that the distances to be represented would always be positive. This is usually the case in architectural drawings. However, if we were Chapter 9384 10 3087 CH09 11/29/01 2:22 PM Page 384measuring, say, the water level of the Pacific Ocean as the tides varied, we might want to be able to represent negative feet-and-inches quantities. (Tide levels below mean-lower-low-water are called minus tides; they prompt clam diggers to take advantage of the larger area of exposed beach.) Let’s derive a new class from Distance. This class will add a single data item to our feet-and- inches measurements: a sign, which can be positive or negative. When we add the sign, we’ll also need to modify the member functions so they can work with signed distances. Here’s the listing for ENGLEN: // englen.cpp // inheritance using English Distances #include using namespace std; enum posneg { pos, neg }; //for sign in DistSign //////////////////////////////////////////////////////////////// class Distance //English Distance class { protected: //NOTE: can’t be private int feet; float inches; public: //no-arg constructor Distance() : feet(0), inches(0.0) { } //2-arg constructor) Distance(int ft, float in) : feet(ft), inches(in) { } void getdist() //get length from user { cout << “\nEnter feet: “; cin >> feet; cout << “Enter inches: “; cin >> inches; } void showdist() const //display distance { cout << feet << “\’-” << inches << ‘\”’; } }; //////////////////////////////////////////////////////////////// class DistSign : public Distance //adds sign to Distance { private: posneg sign; //sign is pos or neg public: //no-arg constructor DistSign() : Distance() //call base constructor { sign = pos; } //set the sign to + Inheritance 9 I NHERITANCE 385 10 3087 CH09 11/29/01 2:22 PM Page 385//2- or 3-arg constructor DistSign(int ft, float in, posneg sg=pos) : Distance(ft, in) //call base constructor { sign = sg; } //set the sign void getdist() //get length from user { Distance::getdist(); //call base getdist() char ch; //get sign from user cout << “Enter sign (+ or -): “; cin >> ch; sign = (ch==’+’) ? pos : neg; } void showdist() const //display distance { cout << ( (sign==pos) ? “(+)” : “(-)” ); //show sign Distance::showdist(); //ft and in } }; //////////////////////////////////////////////////////////////// int main() { DistSign alpha; //no-arg constructor alpha.getdist(); //get alpha from user DistSign beta(11, 6.25); //2-arg constructor DistSign gamma(100, 5.5, neg); //3-arg constructor //display all distances cout << “\nalpha = “; alpha.showdist(); cout << “\nbeta = “; beta.showdist(); cout << “\ngamma = “; gamma.showdist(); cout << endl; return 0; } Here the DistSign class adds the functionality to deal with signed numbers. The Distance class in this program is just the same as in previous programs, except that the data is protected. Actually in this case it could be private, because none of the derived-class functions accesses it. However, it’s safer to make it protected so that a derived-class function could access it if necessary. Chapter 9386 10 3087 CH09 11/29/01 2:22 PM Page 386Operation of ENGLEN The main() program declares three different signed distances. It gets a value for alpha from the user and initializes beta to (+)11'–6.25'' and gamma to (–)100'–5.5''. In the output we use parentheses around the sign to avoid confusion with the hyphen separating feet and inches. Here’s some sample output: Enter feet: 6 Enter inches: 2.5 Enter sign (+ or -): - alpha = (-)6’-2.5” beta = (+)11’-6.25” gamma = (-)100’-5.5” The DistSign class is derived from Distance. It adds a single variable, sign, which is of type posneg. The sign variable will hold the sign of the distance. The posneg type is defined in an enum statement to have two possible values: pos and neg. Constructors in DistSign DistSign has two constructors, mirroring those in Distance. The first takes no arguments, the second takes either two or three arguments. The third, optional, argument in the second constructor is a sign, either pos or neg. Its default value is pos. These constructors allow us to define variables (objects) of type DistSign in several ways. Both constructors in DistSign call the corresponding constructors in Distance to set the feet- and-inches values. They then set the sign variable. The no-argument constructor always sets it to pos. The second constructor sets it to pos if no third-argument value has been provided, or to a value (pos or neg) if the argument is specified. The arguments ft and in, passed from main() to the second constructor in DistSign,are simply forwarded to the constructor in Distance. Member Functions in DistSign Adding a sign to Distance has consequences for both of its member functions. The getdist() function in the DistSign class must ask the user for the sign as well as for feet-and-inches values, and the showdist() function must display the sign along with the feet and inches. These functions call the corresponding functions in Distance,in the lines Distance::getdist(); and Distance::showdist(); Inheritance 9 I NHERITANCE 387 10 3087 CH09 11/29/01 2:22 PM Page 387These calls get and display the feet and inches values. The body of getdist() and showdist() in DistSign then go on to deal with the sign. Abetting Inheritance C++ is designed to make it efficient to create a derived class. Where we want to use parts of the base class, it’s easy to do so, whether these parts are data, constructors, or member functions. Then we add the functionality we need to create the new improved class. Notice that in ENGLEN we didn’t need to duplicate any code; instead we made use of the appropriate functions in the base class. Class Hierarchies In the examples so far in this chapter, inheritance has been used to add functionality to an existing class. Now let’s look at an example where inheritance is used for a different purpose: as part of the original design of a program. Our example models a database of employees of a widget company. We’ve simplified the situation so that only three kinds of employees are represented. Managers manage, scientists perform research to develop better widgets, and laborers operate the dangerous widget-stamping presses. The database stores a name and an employee identification number for all employees, no matter what their category. However, for managers, it also stores their titles and golf club dues. For scientists, it stores the number of scholarly articles they have published. Laborers need no additional data beyond their names and numbers. Our example program starts with a base class employee. This class handles the employee’s last name and employee number. From this class three other classes are derived: manager, scientist, and laborer. The manager and scientist classes contain additional information about these categories of employee, and member functions to handle this information, as shown in Figure 9.5. Chapter 9388 10 3087 CH09 11/29/01 2:22 PM Page 388FIGURE 9.5 UML class diagram for EMPLOY. Here’s the listing for EMPLOY: // employ.cpp // models employee database using inheritance #include using namespace std; const int LEN = 80; //maximum length of names //////////////////////////////////////////////////////////////// class employee //employee class { private: char name[LEN]; //employee name unsigned long number; //employee number public: void getdata() { cout << “\n Enter last name: “; cin >> name; cout << “ Enter number: “; cin >> number; } Inheritance 9 I NHERITANCE 389 employee name number scientist publications manager title club dues laborer 10 3087 CH09 11/29/01 2:22 PM Page 389void putdata() const { cout << “\n Name: “ << name; cout << “\n Number: “ << number; } }; //////////////////////////////////////////////////////////////// class manager : public employee //management class { private: char title[LEN]; //”vice-president” etc. double dues; //golf club dues public: void getdata() { employee::getdata(); cout << “ Enter title: “; cin >> title; cout << “ Enter golf club dues: “; cin >> dues; } void putdata() const { employee::putdata(); cout << “\n Title: “ << title; cout << “\n Golf club dues: “ << dues; } }; //////////////////////////////////////////////////////////////// class scientist : public employee //scientist class { private: int pubs; //number of publications public: void getdata() { employee::getdata(); cout << “ Enter number of pubs: “; cin >> pubs; } void putdata() const { employee::putdata(); cout << “\n Number of publications: “ << pubs; } }; //////////////////////////////////////////////////////////////// class laborer : public employee //laborer class { Chapter 9390 10 3087 CH09 11/29/01 2:22 PM Page 390}; //////////////////////////////////////////////////////////////// int main() { manager m1, m2; scientist s1; laborer l1; cout << endl; //get data for several employees cout << “\nEnter data for manager 1”; m1.getdata(); cout << “\nEnter data for manager 2”; m2.getdata(); cout << “\nEnter data for scientist 1”; s1.getdata(); cout << “\nEnter data for laborer 1”; l1.getdata(); //display data for several employees cout << “\nData on manager 1”; m1.putdata(); cout << “\nData on manager 2”; m2.putdata(); cout << “\nData on scientist 1”; s1.putdata(); cout << “\nData on laborer 1”; l1.putdata(); cout << endl; return 0; } The main() part of the program declares four objects of different classes: two managers, a scientist, and a laborer. (Of course many more employees of each type could be defined, but the output would become rather large.) It then calls the getdata() member functions to obtain information about each employee, and the putdata() function to display this information. Here’s a sample interaction with EMPLOY. First the user supplies the data. Enter data for manager 1 Enter last name: Wainsworth Enter number: 10 Enter title: President Inheritance 9 I NHERITANCE 391 10 3087 CH09 11/29/01 2:22 PM Page 391Enter golf club dues: 1000000 Enter data on manager 2 Enter last name: Bradley Enter number: 124 Enter title: Vice-President Enter golf club dues: 500000 Enter data for scientist 1 Enter last name: Hauptman-Frenglish Enter number: 234234 Enter number of pubs: 999 Enter data for laborer 1 Enter last name: Jones Enter number: 6546544 The program then plays it back. Data on manager 1 Name: Wainsworth Number: 10 Title: President Golf club dues: 1000000 Data on manager 2 Name: Bradley Number: 124 Title: Vice-President Golf club dues: 500000 Data on scientist 1 Name: Hauptman-Frenglish Number: 234234 Number of publications: 999 Data on laborer 1 Name: Jones Number: 6546544 A more sophisticated program would use an array or some other container to arrange the data so that a large number of employee objects could be accommodated. “Abstract” Base Class Notice that we don’t define any objects of the base class employee. We use this as a general class whose sole purpose is to act as a base from which other classes are derived. The laborer class operates identically to the employee class, since it contains no additional data or functions. It may seem that the laborer class is unnecessary, but by making it a separate class we emphasize that all classes are descended from the same source, employee. Also, if in the future we decided to modify the laborer class, we would not need to change the declaration for employee. Chapter 9392 10 3087 CH09 11/29/01 2:22 PM Page 392Classes used only for deriving other classes, as employee is in EMPLOY,are sometimes loosely called abstract classes, meaning that no actual instances (objects) of this class are created. However, the term abstract has a more precise definition that we’ll look at in Chapter 11, “Virtual Functions.” Constructors and Member Functions There are no constructors in either the base or derived classes, so the compiler creates objects of the various classes automatically when it encounters definitions like manager m1, m2; using the default constructor for manager calling the default constructor for employee. The getdata() and putdata() functions in employee accept a name and number from the user and display a name and number. Functions also called getdata() and putdata() in the manager and scientist classes use the functions in employee, and also do their own work. In manager,the getdata() function asks the user for a title and the amount of golf club dues, and putdata() displays these values. In scientist,these functions handle the number of publications. Inheritance and Graphics Shapes In the CIRCLES program in Chapter 6, “Objects and Classes,” we saw a program in which a class represented graphics circles that could be displayed on the screen. Of course, there are other kinds of shapes besides circles, such as squares and triangles. The very phrase “kinds of shapes” implies an inheritance relationship between something called a “shape” and specific kinds of shapes like circles and squares. We can use this relationship to make a program that is more robust and easier to understand than a program that treats different shapes as being unre- lated. In particular we’ll make a shape class that’s a base class (parent) of three derived classes: a circle class, a rect (for rectangle) class, and a tria (for triangle) class. As with other pro- grams that use the Console Graphics Lite functions, you may need to read Appendix E, “Console Graphics Lite,” and either Appendix C, “Microsoft Visual C++,” or Appendix D, “Borland C++Builder” for your specific compiler to learn how to build the graphics files into your program. Here’s the listing for MULTSHAP: // multshap.cpp // balls, rects, and polygons #include “msoftcon.h” //for graphics functions //////////////////////////////////////////////////////////////// class shape //base class Inheritance 9 I NHERITANCE 393 10 3087 CH09 11/29/01 2:22 PM Page 393{ protected: int xCo, yCo; //coordinates of shape color fillcolor; //color fstyle fillstyle; //fill pattern public: //no-arg constructor shape() : xCo(0), yCo(0), fillcolor(cWHITE), fillstyle(SOLID_FILL) { } //4-arg constructor shape(int x, int y, color fc, fstyle fs) : xCo(x), yCo(y), fillcolor(fc), fillstyle(fs) { } void draw() const //set color and fill style { set_color(fillcolor); set_fill_style(fillstyle); } }; //////////////////////////////////////////////////////////////// class circle : public shape { private: int radius; //(xCo, yCo) is center public: circle() : shape() //no-arg constr { } //5-arg constructor circle(int x, int y, int r, color fc, fstyle fs) : shape(x, y, fc, fs), radius(r) { } void draw() const //draw the circle { shape::draw(); draw_circle(xCo, yCo, radius); } }; //////////////////////////////////////////////////////////////// class rect : public shape { private: int width, height; //(xCo, yCo) is upper-left corner public: rect() : shape(), height(0), width(0) //no-arg ctor { } //6-arg ctor rect(int x, int y, int h, int w, color fc, fstyle fs) : shape(x, y, fc, fs), height(h), width(w) Chapter 9394 10 3087 CH09 11/29/01 2:22 PM Page 394{ } void draw() const //draw the rectangle { shape::draw(); draw_rectangle(xCo, yCo, xCo+width, yCo+height); set_color(cWHITE); //draw diagonal draw_line(xCo, yCo, xCo+width, yCo+height); } }; //////////////////////////////////////////////////////////////// class tria : public shape { private: int height; //(xCo, yCo) is tip of pyramid public: tria() : shape(), height(0) //no-arg constructor { } //5-arg constructor tria(int x, int y, int h, color fc, fstyle fs) : shape(x, y, fc, fs), height(h) { } void draw() const //draw the triangle { shape::draw(); draw_pyramid(xCo, yCo, height); } }; //////////////////////////////////////////////////////////////// int main() { init_graphics(); //initialize graphics system circle cir(40, 12, 5, cBLUE, X_FILL); //create circle rect rec(12, 7, 10, 15, cRED, SOLID_FILL); //create rectangle tria tri(60, 7, 11, cGREEN, MEDIUM_FILL); //create triangle cir.draw(); //draw all shapes rec.draw(); tri.draw(); set_cursor_pos(1, 25); //lower-left corner return 0; } When executed, this program produces three different shapes: a blue circle, a red rectangle, and a green triangle. Figure 9.6 shows the output of MULTSHAP. Inheritance 9 I NHERITANCE 395 10 3087 CH09 11/29/01 2:22 PM Page 395FIGURE 9.6 Output of the MULTSHAP program. The characteristics that are common to all shapes, such as their location, color, and fill pattern, are placed in the shape class. Individual shapes have more specific attributes. A circle has a radius, for example, while a rectangle has a height and width. A draw() routine in shape handles the tasks specific to all shapes: setting their color and fill pattern. Overloaded draw() functions in the circle, rect, and tria classes take care of drawing their specific shapes once the color and pattern are determined. As in the last example, the base class shape is an example of an abstract class, in that there is no meaning to instantiating an object of this class. What shape does a shape object display? The question doesn’t make sense. Only a specific shape can display itself. The shape class exists only as a repository of attributes and actions that are common to all shapes. Public and Private Inheritance C++ provides a wealth of ways to fine-tune access to class members. One such access-control mechanism is the way derived classes are declared. Our examples so far have used publicly derived classes, with declarations like class manager : public employee which appeared in the EMPLOY example. What is the effect of the public keyword in this statement, and what are the alternatives? Listen up: The keyword public specifies that objects of the derived class are able to access public member functions of the base class. The alternative is the keyword private. When this keyword is used, objects of the derived class cannot access public member functions of the base class. Since objects can never access private or protected members of a class, the result is that no member of the base class is accessible to objects of the derived class. Chapter 9396 10 3087 CH09 11/29/01 2:22 PM Page 396Access Combinations There are so many possibilities for access that it’s instructive to look at an example program that shows what works and what doesn’t. Here’s the listing for PUBPRIV: // pubpriv.cpp // tests publicly- and privately-derived classes #include using namespace std; //////////////////////////////////////////////////////////////// class A //base class { private: int privdataA; //(functions have the same access protected: //rules as the data shown here) int protdataA; public: int pubdataA; }; //////////////////////////////////////////////////////////////// class B : public A //publicly-derived class { public: void funct() { int a; a = privdataA; //error: not accessible a = protdataA; //OK a = pubdataA; //OK } }; //////////////////////////////////////////////////////////////// class C : private A //privately-derived class { public: void funct() { int a; a = privdataA; //error: not accessible a = protdataA; //OK a = pubdataA; //OK } }; //////////////////////////////////////////////////////////////// int main() { int a; Inheritance 9 I NHERITANCE 397 10 3087 CH09 11/29/01 2:22 PM Page 397B objB; a = objB.privdataA; //error: not accessible a = objB.protdataA; //error: not accessible a = objB.pubdataA; //OK (A public to B) C objC; a = objC.privdataA; //error: not accessible a = objC.protdataA; //error: not accessible a = objC.pubdataA; //error: not accessible (A private to C) return 0; } The program specifies a base class, A,with private, protected, and public data items. Two classes, B and C,are derived from A. B is publicly derived and C is privately derived. As we’ve seen before, functions in the derived classes can access protected and public data in the base class. Objects of the derived classes cannot access private or protected members of the base class. What’s new is the difference between publicly derived and privately derived classes. Objects of the publicly derived class B can access public members of the base class A, while objects of the privately derived class C cannot; they can only access the public members of their own derived class. This is shown in Figure 9.7. Chapter 9398 FIGURE 9.7 Public and private derivation. If you don’t supply any access specifier when creating a class, private is assumed. 10 3087 CH09 11/29/01 2:22 PM Page 398Access Specifiers: When to Use What How do you decide when to use private as opposed to public inheritance? In most cases a derived class exists to offer an improved—or a more specialized—version of the base class. We’ve seen examples of such derived classes (for instance, the CountDn class that adds the decrement operator to the Counter class and the manager class that is a more specialized ver- sion of the employee class). In such cases it makes sense for objects of the derived class to access the public functions of the base class if they want to perform a basic operation, and to access functions in the derived class to perform the more specialized operations that the derived class provides. In such cases public derivation is appropriate. In some situations, however, the derived class is created as a way of completely modifying the operation of the base class, hiding or disguising its original interface. For example, imagine that you have already created a really nice Array class that acts like an array but provides protection against out-of-bounds array indexes. Then suppose you want to use this Array class as the basis for a Stack class, instead of using a basic array. You might derive Stack from Array,but you wouldn’t want the users of Stack objects to treat them as if they were arrays, using the [] operator to access data items, for example. Objects of Stack should always be treated as if they were stacks, using push() and pop(). That is, you want to disguise the Array class as a Stack class. In this situation, private derivation would allow you to conceal all the Array class functions from objects of the derived Stack class. Levels of Inheritance Classes can be derived from classes that are themselves derived. Here’s a miniprogram that shows the idea: class A { }; class B : public A { }; class C : public B { }; Here B is derived from A, and C is derived from B. The process can be extended to an arbitrary number of levels—D could be derived from C, and so on. As a more concrete example, suppose that we decided to add a special kind of laborer called a foreman to the EMPLOY program. We’ll create a new program, EMPLOY2, that incorporates objects of class foreman. Since a foreman is a kind of laborer, the foreman class is derived from the laborer class, as shown in Figure 9.8. Inheritance 9 I NHERITANCE 399 10 3087 CH09 11/29/01 2:22 PM Page 399FIGURE 9.8 UML class diagram for EMPLOY2. Foremen oversee the widget-stamping operation, supervising groups of laborers. They are responsible for the widget production quota for their group. A foreman’s ability is measured by the percentage of production quotas successfully met. The quotas data item in the foreman class represents this percentage. Here’s the listing for EMPLOY2: // employ2.cpp // multiple levels of inheritance #include using namespace std; const int LEN = 80; //maximum length of names //////////////////////////////////////////////////////////////// class employee { private: char name[LEN]; //employee name unsigned long number; //employee number public: void getdata() Chapter 9400 foreman employee scientistmanager laborer 10 3087 CH09 11/29/01 2:22 PM Page 400{ cout << “\n Enter last name: “; cin >> name; cout << “ Enter number: “; cin >> number; } void putdata() const { cout << “\n Name: “ << name; cout << “\n Number: “ << number; } }; //////////////////////////////////////////////////////////////// class manager : public employee //manager class { private: char title[LEN]; //”vice-president” etc. double dues; //golf club dues public: void getdata() { employee::getdata(); cout << “ Enter title: “; cin >> title; cout << “ Enter golf club dues: “; cin >> dues; } void putdata() const { employee::putdata(); cout << “\n Title: “ << title; cout << “\n Golf club dues: “ << dues; } }; //////////////////////////////////////////////////////////////// class scientist : public employee //scientist class { private: int pubs; //number of publications public: void getdata() { employee::getdata(); cout << “ Enter number of pubs: “; cin >> pubs; } void putdata() const { employee::putdata(); cout << “\n Number of publications: “ << pubs; } Inheritance 9 I NHERITANCE 401 10 3087 CH09 11/29/01 2:22 PM Page 401}; //////////////////////////////////////////////////////////////// class laborer : public employee //laborer class { }; //////////////////////////////////////////////////////////////// class foreman : public laborer //foreman class { private: float quotas; //percent of quotas met successfully public: void getdata() { laborer::getdata(); cout << “ Enter quotas: “; cin >> quotas; } void putdata() const { laborer::putdata(); cout << “\n Quotas: “ << quotas; } }; //////////////////////////////////////////////////////////////// int main() { laborer l1; foreman f1; cout << endl; cout << “\nEnter data for laborer 1”; l1.getdata(); cout << “\nEnter data for foreman 1”; f1.getdata(); cout << endl; cout << “\nData on laborer 1”; l1.putdata(); cout << “\nData on foreman 1”; f1.putdata(); cout << endl; return 0; } Chapter 9402 10 3087 CH09 11/29/01 2:22 PM Page 402Notice that a class hierarchy is not the same as an organization chart. An organization chart shows lines of command. A class hierarchy results from generalizing common characteristics. The more general the class, the higher it is on the chart. Thus a laborer is more general than a foreman, who is a specialized kind of laborer, so laborer is shown above foreman in the class hierarchy, although a foreman is probably paid more than a laborer. Multiple Inheritance A class can be derived from more than one base class. This is called multiple inheritance. Figure 9.9 shows how this looks when a class C is derived from base classes A and B. Inheritance 9 I NHERITANCE 403 A B C FIGURE 9.9 UML class diagram for multiple inheritance. The syntax for multiple inheritance is similar to that for single inheritance. In the situation shown in Figure 9.9, the relationship is expressed like this: class A // base class A { }; class B // base class B { }; class C : public A, public B // C is derived from A and B { }; The base classes from which C is derived are listed following the colon in C’s specification; they are separated by commas. 10 3087 CH09 11/29/01 2:22 PM Page 403Member Functions in Multiple Inheritance As an example of multiple inheritance, suppose that we need to record the educational experience of some of the employees in the EMPLOY program. Let’s also suppose that, perhaps in a different project, we’ve already developed a class called student that models students with different educational backgrounds. We decide that instead of modifying the employee class to incorporate educational data, we will add this data by multiple inheritance from the student class. The student class stores the name of the school or university last attended and the highest degree received. Both these data items are stored as strings. Two member functions, getedu() and putedu(),ask the user for this information and display it. Educational information is not relevant to every class of employee. Let’s suppose, somewhat undemocratically, that we don’t need to record the educational experience of laborers; it’s only relevant for managers and scientists. We therefore modify manager and scientist so that they inherit from both the employee and student classes, as shown in Figure 9.10. Chapter 9404 employee student manager scientist laborer FIGURE 9.10 UML class diagram for EMPMULT. 10 3087 CH09 11/29/01 2:22 PM Page 404Here’s a miniprogram that shows these relationships (but leaves out everything else): class student { }; class employee { }; class manager : private employee, private student { }; class scientist : private employee, private student { }; class laborer : public employee { }; And here, featuring considerably more detail, is the listing for EMPMULT: //empmult.cpp //multiple inheritance with employees and degrees #include using namespace std; const int LEN = 80; //maximum length of names //////////////////////////////////////////////////////////////// class student //educational background { private: char school[LEN]; //name of school or university char degree[LEN]; //highest degree earned public: void getedu() { cout << “ Enter name of school or university: “; cin >> school; cout << “ Enter highest degree earned \n”; cout << “ (Highschool, Bachelor’s, Master’s, PhD): “; cin >> degree; } void putedu() const { cout << “\n School or university: “ << school; cout << “\n Highest degree earned: “ << degree; } }; //////////////////////////////////////////////////////////////// class employee { private: char name[LEN]; //employee name unsigned long number; //employee number Inheritance 9 I NHERITANCE 405 10 3087 CH09 11/29/01 2:22 PM Page 405public: void getdata() { cout << “\n Enter last name: “; cin >> name; cout << “ Enter number: “; cin >> number; } void putdata() const { cout << “\n Name: “ << name; cout << “\n Number: “ << number; } }; //////////////////////////////////////////////////////////////// class manager : private employee, private student //management { private: char title[LEN]; //”vice-president” etc. double dues; //golf club dues public: void getdata() { employee::getdata(); cout << “ Enter title: “; cin >> title; cout << “ Enter golf club dues: “; cin >> dues; student::getedu(); } void putdata() const { employee::putdata(); cout << “\n Title: “ << title; cout << “\n Golf club dues: “ << dues; student::putedu(); } }; //////////////////////////////////////////////////////////////// class scientist : private employee, private student //scientist { private: int pubs; //number of publications public: void getdata() { employee::getdata(); cout << “ Enter number of pubs: “; cin >> pubs; student::getedu(); } Chapter 9406 10 3087 CH09 11/29/01 2:22 PM Page 406void putdata() const { employee::putdata(); cout << “\n Number of publications: “ << pubs; student::putedu(); } }; //////////////////////////////////////////////////////////////// class laborer : public employee //laborer { }; //////////////////////////////////////////////////////////////// int main() { manager m1; scientist s1, s2; laborer l1; cout << endl; cout << “\nEnter data for manager 1”; //get data for m1.getdata(); //several employees cout << “\nEnter data for scientist 1”; s1.getdata(); cout << “\nEnter data for scientist 2”; s2.getdata(); cout << “\nEnter data for laborer 1”; l1.getdata(); cout << “\nData on manager 1”; //display data for m1.putdata(); //several employees cout << “\nData on scientist 1”; s1.putdata(); cout << “\nData on scientist 2”; s2.putdata(); cout << “\nData on laborer 1”; l1.putdata(); cout << endl; return 0; } Inheritance 9 I NHERITANCE 407 10 3087 CH09 11/29/01 2:22 PM Page 407The getdata() and putdata() functions in the manager and scientist classes incorporate calls to functions in the student class, such as student::getedu(); and student::putedu(); These routines are accessible in manager and scientist because these classes are descended from student. Here’s some sample interaction with EMPMULT: Enter data for manager 1 Enter last name: Bradley Enter number: 12 Enter title: Vice-President Enter golf club dues: 100000 Enter name of school or university: Yale Enter highest degree earned (Highschool, Bachelor’s, Master’s, PhD): Bachelor’s Enter data for scientist 1 Enter last name: Twilling Enter number: 764 Enter number of pubs: 99 Enter name of school or university: MIT Enter highest degree earned (Highschool, Bachelor’s, Master’s, PhD): PhD Enter data for scientist 2 Enter last name: Yang Enter number: 845 Enter number of pubs: 101 Enter name of school or university: Stanford Enter highest degree earned (Highschool, Bachelor’s, Master’s, PhD): Master’s Enter data for laborer 1 Enter last name: Jones Enter number: 48323 As we saw in the EMPLOY and EMPLOY2 examples, the program then displays this information in roughly the same form. Chapter 9408 10 3087 CH09 11/29/01 2:22 PM Page 408private Derivation in EMPMULT The manager and scientist classes in EMPMULT are privately derived from the employee and student classes. There is no need to use public derivation because objects of manager and scientist never call routines in the employee and student base classes. However, the laborer class must be publicly derived from employer, since it has no member functions of its own and relies on those in employee. Constructors in Multiple Inheritance EMPMULT has no constructors. Let’s look at an example that does use constructors, and see how they’re handled in multiple inheritance. Imagine that we’re writing a program for building contractors, and that this program models lumber-supply items. It uses a class that represents a quantity of lumber of a certain type: 100 8-foot-long construction grade 2×4s, for example. The class should store various kinds of data about each such lumber item. We need to know the length (3'–6'', for example) and we need to store the number of such pieces of lumber and their unit cost. We also need to store a description of the lumber we’re talking about. This has two parts. The first is the nominal dimensions of the cross-section of the lumber. This is given in inches. For instance, lumber 2 inches by 4 inches (for you metric folks, about 5 cm by 10 cm) is called a two-by-four. This is usually written 2×4. We also need to know the grade of lumber—rough-cut, construction grade, surfaced-four-sides, and so on. We find it convenient to create a Type class to hold this data. This class incorporates member data for the nominal dimensions and the grade of the lumber, both expressed as strings, such as 2×6 and construction. Member functions get this information from the user and display it. We’ll use the Distance class from previous examples to store the length. Finally we create a Lumber class that inherits both the Type and Distance classes. Here’s the listing for ENGLMULT: // englmult.cpp // multiple inheritance with English Distances #include #include using namespace std; //////////////////////////////////////////////////////////////// class Type //type of lumber { private: string dimensions; string grade; Inheritance 9 I NHERITANCE 409 10 3087 CH09 11/29/01 2:22 PM Page 409public: //no-arg constructor Type() : dimensions(“N/A”), grade(“N/A”) { } //2-arg constructor Type(string di, string gr) : dimensions(di), grade(gr) { } void gettype() //get type from user { cout << “ Enter nominal dimensions (2x4 etc.): “; cin >> dimensions; cout << “ Enter grade (rough, const, etc.): “; cin >> grade; } void showtype() const //display type { cout << “\n Dimensions: “ << dimensions; cout << “\n Grade: “ << grade; } }; //////////////////////////////////////////////////////////////// class Distance //English Distance class { private: int feet; float inches; public: //no-arg constructor Distance() : feet(0), inches(0.0) { } //constructor (two args) Distance(int ft, float in) : feet(ft), inches(in) { } void getdist() //get length from user { cout << “ Enter feet: “; cin >> feet; cout << “ Enter inches: “; cin >> inches; } void showdist() const //display distance { cout << feet << “\’-” << inches << ‘\”’; } }; //////////////////////////////////////////////////////////////// class Lumber : public Type, public Distance { private: int quantity; //number of pieces double price; //price of each piece public: //constructor (no args) Lumber() : Type(), Distance(), quantity(0), price(0.0) Chapter 9410 10 3087 CH09 11/29/01 2:22 PM Page 410{ } //constructor (6 args) Lumber( string di, string gr, //args for Type int ft, float in, //args for Distance int qu, float prc ) : //args for our data Type(di, gr), //call Type ctor Distance(ft, in), //call Distance ctor quantity(qu), price(prc) //initialize our data { } void getlumber() { Type::gettype(); Distance::getdist(); cout << “ Enter quantity: “; cin >> quantity; cout << “ Enter price per piece: “; cin >> price; } void showlumber() const { Type::showtype(); cout << “\n Length: “; Distance::showdist(); cout << “\n Price for “ << quantity << “ pieces: $” << price * quantity; } }; //////////////////////////////////////////////////////////////// int main() { Lumber siding; //constructor (no args) cout << “\nSiding data:\n”; siding.getlumber(); //get siding from user //constructor (6 args) Lumber studs( “2x4”, “const”, 8, 0.0, 200, 4.45F ); //display lumber data cout << “\nSiding”; siding.showlumber(); cout << “\nStuds”; studs.showlumber(); cout << endl; return 0; } The major new feature in this program is the use of constructors in the derived class Lumber. These constructors call the appropriate constructors in Type and Distance. Inheritance 9 I NHERITANCE 411 10 3087 CH09 11/29/01 2:22 PM Page 411No-Argument Constructor The no-argument constructor in Type looks like this: Type() { strcpy(dimensions, “N/A”); strcpy(grade, “N/A”); } This constructor fills in “N/A” (not available) for the dimensions and grade variables so the user will be made aware if an attempt is made to display data for an uninitialized lumber object. You’re already familiar with the no-argument constructor in the Distance class: Distance() : feet(0), inches(0.0) { } The no-argument constructor in Lumber calls both of these constructors. Lumber() : Type(), Distance(), quantity(0), price(0.0) { } The names of the base-class constructors follow the colon and are separated by commas. When the Lumber() constructor is invoked, these base-class constructors—Type() and Distance()— will be executed. The quantity and price attributes are also initialized. Multi-Argument Constructors Here is the two-argument constructor for Type: Type(string di, string gr) : dimensions(di), grade(gr) { } This constructor copies string arguments to the dimensions and grade member data items. Here’s the constructor for Distance, which is again familiar from previous programs: Distance(int ft, float in) : feet(ft), inches(in) { } The constructor for Lumber calls both of these constructors, so it must supply values for their arguments. In addition it has two arguments of its own: the quantity of lumber and the unit price. Thus this constructor has six arguments. It makes two calls to the two constructors, each of which takes two arguments, and then initializes its own two data items. Here’s what it looks like: Lumber( string di, string gr, //args for Type int ft, float in, //args for Distance int qu, float prc ) : //args for our data Type(di, gr), //call Type ctor Distance(ft, in), //call Distance ctor quantity(qu), price(prc) //initialize our data { } Chapter 9412 10 3087 CH09 11/29/01 2:22 PM Page 412Ambiguity in Multiple Inheritance Odd sorts of problems may surface in certain situations involving multiple inheritance. Here’s a common one. Two base classes have functions with the same name, while a class derived from both base classes has no function with this name. How do objects of the derived class access the correct base class function? The name of the function alone is insufficient, since the compiler can’t figure out which of the two functions is meant. Here’s an example, AMBIGU,that demonstrates the situation: // ambigu.cpp // demonstrates ambiguity in multiple inheritance #include using namespace std; //////////////////////////////////////////////////////////////// class A { public: void show() { cout << “Class A\n”; } }; class B { public: void show() { cout << “Class B\n”; } }; class C : public A, public B { }; //////////////////////////////////////////////////////////////// int main() { C objC; //object of class C // objC.show(); //ambiguous--will not compile objC.A::show(); //OK objC.B::show(); //OK return 0; } The problem is resolved using the scope-resolution operator to specify the class in which the function lies. Thus objC.A::show(); refers to the version of show() that’s in the A class, while objC.B::show(); Inheritance 9 I NHERITANCE 413 10 3087 CH09 11/29/01 2:22 PM Page 413refers to the function in the B class. Stroustrup (see Appendix H, “Bibliography”) calls this dis- ambiguation. Another kind of ambiguity arises if you derive a class from two classes that are each derived from the same class. This creates a diamond-shaped inheritance tree. The DIAMOND program shows how this looks. //diamond.cpp //investigates diamond-shaped multiple inheritance #include using namespace std; //////////////////////////////////////////////////////////////// class A { public: void func(); }; class B : public A { }; class C : public A { }; class D : public B, public C { }; //////////////////////////////////////////////////////////////// int main() { D objD; objD.func(); //ambiguous: won’t compile return 0; } Classes B and C are both derived from class A, and class D is derived by multiple inheritance from both B and C. Trouble starts if you try to access a member function in class A from an object of class D. In this example objD tries to access func(). However, both B and C contain a copy of func(), inherited from A. The compiler can’t decide which copy to use, and signals an error. There are various advanced ways of coping with this problem, but the fact that such ambiguities can arise causes many experts to recommend avoiding multiple inheritance altogether. You should certainly not use it in serious programs unless you have considerable experience. Aggregation: Classes Within Classes We’ll discuss aggregation here because, while it is not directly related to inheritance, both aggregation and inheritance are class relationships that are more specialized than associations. It is instructive to compare and contrast them. Chapter 9414 10 3087 CH09 11/29/01 2:22 PM Page 414If a class B is derived by inheritance from a class A, we can say that “B is a kind of A.” This is because B has all the characteristics of A, and in addition some of its own. It’s like saying that a starling is a kind of bird: A starling has the characteristics shared by all birds (wings, feathers, and so on) but has some distinctive characteristics of its own (such as dark iridescent plumage). For this reason inheritance is often called a “kind of” relationship. Aggregation is called a “has a” relationship. We say a library has a book or an invoice has an item line. Aggregation is also called a “part-whole” relationship: the book is part of the library. In object-oriented programming, aggregation may occur when one object is an attribute of another. Here’s a case where an object of class A is an attribute of class B: class A { }; class B { A objA; // define objA as an object of class A }; In the UML, aggregation is considered a special kind of association. Sometimes it’s hard to tell when an association is also an aggregation. It’s always safe to call a relationship an association, but if class A contains objects of class B, and is organizationally superior to class B, it’s a good candidate for aggregation. A company might have an aggregation of employees, or a stamp collection might have an aggregation of stamps. Aggregation is shown in the same way as association in UML class diagrams, except that the “whole” end of the association line has an open diamond-shaped arrowhead. Figure 9.11 shows how this looks. Inheritance 9 I NHERITANCE 415 Library Publications Staff FIGURE 9.11 UML class diagram showing aggregation. 10 3087 CH09 11/29/01 2:22 PM Page 415Aggregation in the EMPCONT Program Let’s rearrange the EMPMULT program to use aggregation instead of inheritance. In EMPMULT the manager and scientist classes are derived from the employee and student classes using the inheritance relationship. In our new program, EMPCONT,the manager and scientist classes contain instances of the employee and student classes as attributes. This aggregation relationship is shown in Figure 9.12. Chapter 9416 manager scientistemployee student laborer FIGURE 9.12 UML class diagram for EMPCONT. The following miniprogram shows these relationships in a different way: class student {}; class employee {}; class manager { student stu; // stu is an object of class student employee emp; // emp is an object of class employee }; class scientist { student stu; // stu is an object of class student employee emp; // emp is an object of class employee }; class laborer { employee emp; // emp is an object of class employee }; 10 3087 CH09 11/29/01 2:22 PM Page 416Here’s the full-scale listing for EMPCONT: // empcont.cpp // containership with employees and degrees #include #include using namespace std; //////////////////////////////////////////////////////////////// class student //educational background { private: string school; //name of school or university string degree; //highest degree earned public: void getedu() { cout << “ Enter name of school or university: “; cin >> school; cout << “ Enter highest degree earned \n”; cout << “ (Highschool, Bachelor’s, Master’s, PhD): “; cin >> degree; } void putedu() const { cout << “\n School or university: “ << school; cout << “\n Highest degree earned: “ << degree; } }; //////////////////////////////////////////////////////////////// class employee { private: string name; //employee name unsigned long number; //employee number public: void getdata() { cout << “\n Enter last name: “; cin >> name; cout << “ Enter number: “; cin >> number; } void putdata() const { cout << “\n Name: “ << name; cout << “\n Number: “ << number; } }; //////////////////////////////////////////////////////////////// class manager //management Inheritance 9 I NHERITANCE 417 10 3087 CH09 11/29/01 2:22 PM Page 417{ private: string title; //”vice-president” etc. double dues; //golf club dues employee emp; //object of class employee student stu; //object of class student public: void getdata() { emp.getdata(); cout << “ Enter title: “; cin >> title; cout << “ Enter golf club dues: “; cin >> dues; stu.getedu(); } void putdata() const { emp.putdata(); cout << “\n Title: “ << title; cout << “\n Golf club dues: “ << dues; stu.putedu(); } }; //////////////////////////////////////////////////////////////// class scientist //scientist { private: int pubs; //number of publications employee emp; //object of class employee student stu; //object of class student public: void getdata() { emp.getdata(); cout << “ Enter number of pubs: “; cin >> pubs; stu.getedu(); } void putdata() const { emp.putdata(); cout << “\n Number of publications: “ << pubs; stu.putedu(); } }; //////////////////////////////////////////////////////////////// class laborer //laborer { Chapter 9418 10 3087 CH09 11/29/01 2:22 PM Page 418private: employee emp; //object of class employee public: void getdata() { emp.getdata(); } void putdata() const { emp.putdata(); } }; //////////////////////////////////////////////////////////////// int main() { manager m1; scientist s1, s2; laborer l1; cout << endl; cout << “\nEnter data for manager 1”; //get data for m1.getdata(); //several employees cout << “\nEnter data for scientist 1”; s1.getdata(); cout << “\nEnter data for scientist 2”; s2.getdata(); cout << “\nEnter data for laborer 1”; l1.getdata(); cout << “\nData on manager 1”; //display data for m1.putdata(); //several employees cout << “\nData on scientist 1”; s1.putdata(); cout << “\nData on scientist 2”; s2.putdata(); cout << “\nData on laborer 1”; l1.putdata(); cout << endl; return 0; } The student and employee classes are the same in EMPCONT as they were in EMPMULT,but they are related in a different way to the manager and scientist classes. Inheritance 9 I NHERITANCE 419 10 3087 CH09 11/29/01 2:22 PM Page 419Composition: A Stronger Aggregation Composition is a stronger form of aggregation. It has all the characteristics of aggregation, plus two more: • The part may belong to only one whole. • The lifetime of the part is the same as the lifetime of the whole. A car is composed of doors (among other things). The doors can’t belong to some other car, and they are born and die along with the car. A room is composed of a floor, ceiling, and walls. While aggregation is a “has a” relationship, composition is a “consists of” relationship. In UML diagrams, composition is shown in the same way as aggregation, except that the diamond-shaped arrowhead is solid instead of open. This is shown in Figure 9.13. Chapter 9420 Car Doors Engine FIGURE 9.13 UML class diagram showing composition. Even a single object can be related to a class by composition. In a car there is only one engine. Inheritance and Program Development The program-development process, as practiced for decades by programmers everywhere, is being fundamentally altered by object-oriented programming. This is due not only to the use of classes in OOP but to inheritance as well. Let’s see how this comes about. Programmer A creates a class. Perhaps it’s something like the Distance class, with a complete set of member functions for arithmetic operations on a user-defined data type. 10 3087 CH09 11/29/01 2:22 PM Page 420Programmer B likes the Distance class but thinks it could be improved by using signed distances. The solution is to create a new class, like DistSign in the ENGLEN example, that is derived from Distance but incorporates the extensions necessary to implement signed distances. Programmers C and D then write applications that use the DistSign class. Programmer B may not have access to the source code for the Distance member functions, and programmers C and D may not have access to the source code for DistSign. Yet, because of the software reusability feature of C++, B can modify and extend the work of A, and C and D can make use of the work of B (and A). Notice that the distinction between software tool developers and application writers is becoming blurred. Programmer A creates a general-purpose programming tool, the Distance class. Programmer B creates a specialized version of this class, the DistSign class. Programmers C and D create applications. A is a tool developer, and C and D are applications developers. B is somewhere in between. In any case OOP is making the programming scene more flexible and at the same time more complex. In Chapter 13 we’ll see how a class can be divided into a client-accessible part and a part that is distributed only in object form, so it can be used by other programmers without the distribution of source code. Summary A class, called the derived class, can inherit the features of another class, called the base class. The derived class can add other features of its own, so it becomes a specialized version of the base class. Inheritance provides a powerful way to extend the capabilities of existing classes, and to design programs using hierarchical relationships. Accessibility of base class members from derived classes and from objects of derived classes is an important issue. Data or functions in the base class that are prefaced by the keyword pro- tected can be accessed from derived classes but not by any other objects, including objects of derived classes. Classes may be publicly or privately derived from base classes. Objects of a publicly derived class can access public members of the base class, while objects of a privately derived class cannot. A class can be derived from more than one base class. This is called multiple inheritance. A class can also be contained within another class. In the UML, inheritance is called generalization. This relationship is represented in class diagrams by an open triangle pointing to the base (parent) class. Inheritance 9 I NHERITANCE 421 10 3087 CH09 11/29/01 2:22 PM Page 421Aggregation is a “has a” or “part-whole” relationship: one class contains objects of another class. Aggregation is represented in UML class diagrams by an open diamond pointing to the “whole” part of the part-whole pair. Composition is a strong form of aggregation. Its arrowhead is solid rather than open. Inheritance permits the reusability of software: Derived classes can extend the capabilities of base classes with no need to modify—or even access the source code of—the base class. This leads to new flexibility in the software development process, and to a wider range of roles for software developers. Questions Answers to these questions can be found in Appendix G. 1. Inheritance is a way to a. make general classes into more specific classes. b. pass arguments to objects of classes. c. add features to existing classes without rewriting them. d. improve data hiding and encapsulation. 2. A “child” class is said to be _________ from a base class. 3. Advantages of inheritance include a. providing class growth through natural selection. b. facilitating class libraries. c. avoiding the rewriting of code. d. providing a useful conceptual framework. 4. Write the first line of the specifier for a class Bosworth that is publicly derived from a class Alphonso. 5. True or false: Adding a derived class to a base class requires fundamental changes to the base class. 6. To be accessed from a member function of the derived class, data or functions in the base class must be public or _________. 7. If a base class contains a member function basefunc(), and a derived class does not contain a function with this name, can an object of the derived class access basefunc()? 8. Assume that the classes mentioned in Question 4 and the class Alphonso contain a member function called alfunc(). Write a statement that allows object BosworthObj of class Bosworth to access alfunc(). Chapter 9422 10 3087 CH09 11/29/01 2:22 PM Page 4229. True or false: If no constructors are specified for a derived class, objects of the derived class will use the constructors in the base class. 10. If a base class and a derived class each include a member function with the same name, which member function will be called by an object of the derived class, assuming the scope-resolution operator is not used? 11. Write a declarator for a no-argument constructor of the derived class Bosworth of Question 4 that calls a no-argument constructor in the base class Alphonso. 12. The scope-resolution operator usually a. limits the visibility of variables to a certain function. b. tells what base class a class is derived from. c. specifies a particular class. d. resolves ambiguities. 13. True or false: It is sometimes useful to specify a class from which no objects will ever be created. 14. Assume that there is a class Derv that is derived from a base class Base. Write the declarator for a derived-class constructor that takes one argument and passes this argu- ment along to the constructor in the base class. 15. Assume a class Derv that is privately derived from class Base. An object of class Derv located in main() can access a. public members of Derv. b. protected members of Derv. c. private members of Derv. d. public members of Base. e. protected members of Base. f. private members of Base. 16. True or false: A class D can be derived from a class C, which is derived from a class B, which is derived from a class A. 17. A class hierarchy a. shows the same relationships as an organization chart. b. describes “has a” relationships. c. describes “is a kind of” relationships. d. shows the same relationships as a family tree. 18. Write the first line of a specifier for a class Tire that is derived from class Wheel and from class Rubber. Inheritance 9 I NHERITANCE 423 10 3087 CH09 11/29/01 2:22 PM Page 42319. Assume a class Derv derived from a base class Base. Both classes contain a member function func() that takes no arguments. Write a statement to go in a member function of Derv that calls func() in the base class. 20. True or false: It is illegal to make objects of one class members of another class. 21. In the UML, inheritance is called _____________. 22. Aggregation is a. a stronger form of instantiation. b. a stronger form of generalization. c. a stronger form of composition. d. a “has a” relationship. 23. True or false: the arrow representing generalization points to the more specific class. 24. Composition is a ___________ form of ____________. Exercises Answers to starred exercises can be found in Appendix G. *1. Imagine a publishing company that markets both book and audiocassette versions of its works. Create a class publication that stores the title (a string) and price (type float) of a publication. From this class derive two classes: book, which adds a page count (type int), and tape, which adds a playing time in minutes (type float). Each of these three classes should have a getdata() function to get its data from the user at the keyboard, and a putdata() function to display its data. Write a main() program to test the book and tape classes by creating instances of them, asking the user to fill in data with getdata(), and then displaying the data with putdata(). *2. Recall the STRCONV example from Chapter 8. The String class in this example has a flaw: It does not protect itself if its objects are initialized to have too many characters. (The SZ constant has the value 80.) For example, the definition String s = “This string will surely exceed the width of the “ “screen, which is what the SZ constant represents.”; will cause the str array in s to overflow, with unpredictable consequences, such as crashing the system. With String as a base class, derive a class Pstring (for “protected string”) that prevents buffer overflow when too long a string constant is used in a definition. A new constructor in the derived class should copy only SZ–1 characters into str if the string constant is longer, but copy the entire constant if it’s shorter. Write a main() program to test different lengths of strings. Chapter 9424 10 3087 CH09 11/29/01 2:22 PM Page 424*3. Start with the publication, book, and tape classes of Exercise 1. Add a base class sales that holds an array of three floats so that it can record the dollar sales of a particular publication for the last three months. Include a getdata() function to get three sales amounts from the user, and a putdata() function to display the sales figures. Alter the book and tape classes so they are derived from both publication and sales. An object of class book or tape should input and output sales data along with its other data. Write a main() function to create a book object and a tape object and exercise their input/output capabilities. 4. Assume that the publisher in Exercises 1 and 3 decides to add a third way to distribute books: on computer disk, for those who like to do their reading on their laptop. Add a disk class that, like book and tape, is derived from publication. The disk class should incorporate the same member functions as the other classes. The data item unique to this class is the disk type: either CD or DVD. You can use an enum type to store this item. The user could select the appropriate type by typing c or d. 5. Derive a class called employee2 from the employee class in the EMPLOY program in this chapter. This new class should add a type double data item called compensation, and also an enum type called period to indicate whether the employee is paid hourly, weekly, or monthly. For simplicity you can change the manager, scientist, and laborer classes so they are derived from employee2 instead of employee. However, note that in many circumstances it might be more in the spirit of OOP to create a separate base class called compensation and three new classes manager2, scientist2, and laborer2, and use multiple inheritance to derive these three classes from the original manager, scientist, and laborer classes and from compensation. This way none of the original classes needs to be modified. 6. Start with the ARROVER3 program in Chapter 8. Keep the safearay class the same as in that program, and, using inheritance, derive the capability for the user to specify both the upper and lower bounds of the array in a constructor. This is similar to Exercise 9 in Chapter 8, except that inheritance is used to derive a new class (you can call it safehilo) instead of modifying the original class. 7. Start with the COUNTEN2 program in this chapter. It can increment or decrement a counter, but only using prefix notation. Using inheritance, add the ability to use postfix notation for both incrementing and decrementing. (See Chapter 8 for a description of postfix notation.) 8. Operators in some computer languages, such as Visual Basic, allow you to select parts of an existing string and assign them to other strings. (The Standard C++ string class offers a different approach.) Using inheritance, add this capability to the Pstring class of Exercise 2. In the derived class, Pstring2,incorporate three new functions: left(), mid(), and right(). Inheritance 9 I NHERITANCE 425 10 3087 CH09 11/29/01 2:22 PM Page 425s2.left(s1, n) // s2 is assigned the leftmost n characters // from s1 s2.mid(s1, s, n) // s2 is assigned the middle n characters // from s1, starting at character number s // (leftmost character is 0) s2.right(s1, n) // s2 is assigned the rightmost n characters // from s1 You can use for loops to copy the appropriate parts of s1, character by character, to a temporary Pstring2 object, which is then returned. For extra credit, have these functions return by reference, so they can be used on the left side of the equal sign to change parts of an existing string. 9. Start with the publication, book, and tape classes of Exercise 1. Suppose you want to add the date of publication for both books and tapes. From the publication class, derive a new class called publication2 that includes this member data. Then change book and tape so they are derived from publication2 instead of publication. Make all the necessary changes in member functions so the user can input and output dates along with the other data. For the dates, you can use the date class from Exercise 5 in Chapter 6, which stores a date as three ints, for month, day, and year. 10. There is only one kind of manager in the EMPMULT program in this chapter. Any serious company has executives as well as managers. From the manager class derive a class called executive. (We’ll assume an executive is a high-end kind of manager.) The addi- tional data in the executive class will be the size of the employee’s yearly bonus and the number of shares of company stock held in his or her stock-option plan. Add the appropriate member functions so these data items can be input and displayed along with the other manager data. 11. Various situations require that pairs of numbers be treated as a unit. For example, each screen coordinate has an x (horizontal) component and a y (vertical) component. Represent such a pair of numbers as a structure called pair that comprises two int member variables. Now, assume you want to be able to store pair variables on a stack. That is, you want to be able to place a pair (which contains two integers) onto a stack using a single call to a push() function with a structure of type pair as an argument, and retrieve a pair using a single call to a pop() function, which will return a structure of type pair. Start with the Stack2 class in the STAKEN program in this chapter, and from it derive a new class called pairStack. This new class need contain only two members: the overloaded push() and pop() functions. The pairStack::push() function will need to make two calls to Stack2::push() to store the two integers in its pair, and the pairStack::pop() function will need to make two calls to Stack2::pop() (although not necessarily in the same order). Chapter 9426 10 3087 CH09 11/29/01 2:22 PM Page 42612. Amazing as it may seem, the old British pounds-shillings-pence money notation (£9.19.11—see Exercise 10 in Chapter 4, “Structures”) isn’t the whole story. A penny was further divided into halfpennies and farthings, with a farthing being worth 1/4 of a penny. There was a halfpenny coin, a farthing coin, and a halffarthing coin. Fortunately all this can be expressed numerically in eighths of a penny: 1/8 penny is a halffarthing 1/4 penny is a farthing 3/8 penny is a farthing and a half 1/2 penny is a halfpenny (pronounced ha’penny) 5/8 penny is a halfpenny plus a halffarthing 3/4 penny is a halfpenny plus a farthing 7/8 penny is a halfpenny plus a farthing and a half Let’s assume we want to add to the sterling class the ability to handle such fractional pennies. The I/O format can be something like £1.1.1-1/4 or £9.19.11-7/8, where the hyphen separates the fraction from the pennies. Derive a new class called sterfrac from sterling. It should be able to perform the four arithmetic operations on sterling quantities that include eighths of a penny. Its only mem- ber data is an int indicating the number of eighths; you can call it eighths. You’ll need to overload many of the functions in sterling to handle the eighths. The user should be able to type any fraction in lowest terms, and the display should also show fractions in lowest terms. It’s not necessary to use the full-scale fraction class (see Exercise 11 in Chapter 6), but you could try that for extra credit. Inheritance 9 I NHERITANCE 427 10 3087 CH09 11/29/01 2:22 PM Page 42710 3087 CH09 11/29/01 2:22 PM Page 428CHAPTER 10 Pointers IN THIS CHAPTER • Addresses and Pointers 430 • The Address-of Operator & 431 • Pointers and Arrays 440 • Pointers and Functions 443 • Pointers and C-Type Strings 452 • Memory Management: new and delete 458 • Pointers to Objects 464 •A Linked List Example 469 • Pointers to Pointers 474 •A Parsing Example 479 •Simulation: A Horse Race 484 • UML State Diagrams 490 • Debugging Pointers 492 11 3087 CH10 11/29/01 2:21 PM Page 429Chapter 10430 Pointers are the hobgoblin of C++ (and C) programming; seldom has such a simple idea inspired so much perplexity for so many. But fear not. In this chapter we will try to demystify pointers and show practical uses for them in C++ programming. What are pointers for? Here are some common uses: • Accessing array elements •Passing arguments to a function when the function needs to modify the original argument •Passing arrays and strings to functions • Obtaining memory from the system •Creating data structures such as linked lists Pointers are an important feature of C++ (and C), while many other languages, such as Visual Basic and Java, have no pointers at all. (Java has references, which are sort of watered-down pointers.) Is this emphasis on pointers really necessary? You can do a lot without them, as their absence from the preceding chapters demonstrates. Some operations that use pointers in C++ can be carried out in other ways. For example, array elements can be accessed with array nota- tion rather than pointer notation (we’ll see the difference soon), and a function can modify arguments passed by reference, as well as those passed by pointers. However, in some situations pointers provide an essential tool for increasing the power of C++. A notable example is the creation of data structures such as linked lists and binary trees. In fact, several key features of C++, such as virtual functions, the new operator, and the this pointer (discussed in Chapter 11, “Virtual Functions”), require the use of pointers. So, although you can do a lot of programming in C++ without using pointers, you will find them essential to obtaining the most from the language. In this chapter we’ll introduce pointers gradually, starting with fundamental concepts and working up to complex pointer applications. If you already know C, you can probably skim over the first half of the chapter. However, you should read the sections in the second half on the new and delete operators, accessing member functions using pointers, arrays of pointers to objects, and linked-list objects. Addresses and Pointers The ideas behind pointers are not complicated. Here’s the first key concept: Every byte in the computer’s memory has an address. Addresses are numbers, just as they are for houses on a street. The numbers start at 0 and go up from there—1, 2, 3, and so on. If you have 1MB of memory, the highest address is 1,048,575. (Of course you have much more.) Your program, when it is loaded into memory, occupies a certain range of these addresses. That means that every variable and every function in your program starts at a particular address. Figure 10.1 shows how this looks. 11 3087 CH10 11/29/01 2:21 PM Page 430FIGURE 10.1 Memory addresses. The Address-of Operator & You can find the address occupied by a variable by using the address-of operator &. Here’s a short program, VARADDR,that demonstrates how to do this: // varaddr.cpp // addresses of variables #include using namespace std; int main() { int var1 = 11; //define and initialize int var2 = 22; //three variables int var3 = 33; Pointers 10 P OINTERS 431 11 3087 CH10 11/29/01 2:21 PM Page 431cout << &var1 << endl //print the addresses << &var2 << endl //of these variables << &var3 << endl; return 0; } This simple program defines three integer variables and initializes them to the values 11, 22, and 33. It then prints out the addresses of these variables. The actual addresses occupied by the variables in a program depend on many factors, such as the computer the program is running on, the size of the operating system, and whether any other programs are currently in memory. For these reasons you probably won’t get the same addresses we did when you run this program. (You may not even get the same results twice in a row.) Here’s the output on our machine: 0x8f4ffff4 ← address of var1 0x8f4ffff2 ← address of var2 0x8f4ffff0 ← address of var3 Remember that the address of a variable is not at all the same as its contents. The contents of the three variables are 11, 22, and 33. Figure 10.2 shows the three variables in memory. Chapter 10432 FIGURE 10.2 Addresses and contents of variables. 11 3087 CH10 11/29/01 2:21 PM Page 432The << insertion operator interprets the addresses in hexadecimal arithmetic, as indicated by the prefix 0x before each number. This is the usual way to show memory addresses. If you aren’t familiar with the hexadecimal number system, don’t worry. All you really need to know is that each variable starts at a unique address. However, you might note in the output that each address differs from the next by exactly 2 bytes. That’s because integers occupy 2 bytes of memory (on a 16-bit system). If we had used variables of type char,they would have adjacent addresses, since a char occupies 1 byte; and if we had used type double,the addresses would have differed by 8 bytes. The addresses appear in descending order because local variables are stored on the stack, which grows downward in memory. If we had used global variables, they would have ascend- ing addresses, since global variables are stored on the heap, which grows upward. Again, you don’t need to worry too much about these considerations, since the compiler keeps track of the details for you. Don’t confuse the address-of operator &, which precedes a variable name in a variable declara- tion, with the reference operator &, which follows the type name in a function prototype or def- inition. (References were discussed in Chapter 5, “Functions.”) Pointer Variables Addresses by themselves are rather limited. It’s nice to know that we can find out where things are in memory, as we did in VARADDR,but printing out address values is not all that useful. The potential for increasing our programming power requires an additional idea: variables that hold address values. We’ve seen variable types that store characters, integers, floating-point numbers, and so on. Addresses are stored similarly. A variable that holds an address value is called a pointer variable,or simply a pointer. What is the data type of pointer variables? It’s not the same as the variable whose address is being stored; a pointer to int is not type int. You might think a pointer data type would be called something like pointer or ptr. However, things are slightly more complicated. The next program, PTRVAR,shows the syntax for pointer variables. // ptrvar.cpp // pointers (address variables) #include using namespace std; int main() { int var1 = 11; //two integer variables int var2 = 22; Pointers 10 P OINTERS 433 11 3087 CH10 11/29/01 2:21 PM Page 433cout << &var1 << endl //print addresses of variables << &var2 << endl << endl; int* ptr; //pointer to integers ptr = &var1; //pointer points to var1 cout << ptr << endl; //print pointer value ptr = &var2; //pointer points to var2 cout << ptr << endl; //print pointer value return 0; } This program defines two integer variables, var1 and var2, and initializes them to the values 11 and 22. It then prints out their addresses. The program next defines a pointer variable in the line int* ptr; To the uninitiated this is a rather bizarre syntax. The asterisk means pointer to. Thus the state- ment defines the variable ptr as a pointer to int. This is another way of saying that this vari- able can hold the addresses of integer variables. What’s wrong with the idea of a general-purpose pointer type that holds pointers to any data type? If we called it type pointer we could write declarations like pointer ptr; The problem is that the compiler needs to know what kind of variable the pointer points to. (We’ll see why when we talk about pointers and arrays.) The syntax used in C++ allows point- ers to any type to be declared: char* cptr; // pointer to char int* iptr; // pointer to int float* fptr; // pointer to float Distance* distptr; // pointer to user-defined Distance class and so on. Syntax Quibbles We should note that it is common to write pointer definitions with the asterisk closer to the variable name than to the type. char *charptr; It doesn’t matter to the compiler, but placing the asterisk next to the type helps emphasize that the asterisk is part of the variable type (pointer to char), not part of the name itself. Chapter 10434 11 3087 CH10 11/29/01 2:21 PM Page 434If you define more than one pointer of the same type on one line, you need only insert the type-pointed-to once, but you need to place an asterisk before each variable name. char* ptr1, * ptr2, * ptr3; // three variables of type char* Or you can use the asterisk-next-to-the-name approach. char *ptr1, *ptr2, *ptr3; // three variables of type char* Pointers Must Have a Value An address like 0x8f4ffff4 can be thought of as a pointer constant. A pointer like ptr can be thought of as a pointer variable. Just as the integer variable var1 can be assigned the constant value 11, so can the pointer variable ptr be assigned the constant value 0x8f4ffff4. When we first define a variable, it holds no value (unless we initialize it at the same time). It may hold a garbage value, but this has no meaning. In the case of pointers, a garbage value is the address of something in memory, but probably not of something that we want. So before a pointer is used, a specific address must be placed in it. In the PTRVAR program, ptr is first assigned the address of var1 in the line ptr = &var1; ← put address of var1 in ptr Following this, the program prints out the value contained in ptr, which should be the same address printed for &var1. The same pointer variable ptr is then assigned the address of var2, and this value is printed out. Figure 10.3 shows the operation of the PTRVAR program. Here’s the output of PTRVAR: 0x8f51fff4 ← address of var1 0x8f51fff2 ← address of var2 0x8f51fff4 ← ptr set to address of var1 0x8f51fff2 ← ptr set to address of var2 To summarize: A pointer can hold the address of any variable of the correct type; it’s a recepta- cle awaiting an address. However, it must be given some value, or it will point to an address we don’t want it to point to, such as into our program code or the operating system. Rogue pointer values can result in system crashes and are difficult to debug, since the compiler gives no warning. The moral: Make sure you give every pointer variable a valid address value before using it. Pointers 10 P OINTERS 435 11 3087 CH10 11/29/01 2:21 PM Page 435Chapter 10436 FIGURE 10.3 Changing values in ptr. Accessing the Variable Pointed To Suppose that we don’t know the name of a variable but we do know its address. Can we access the contents of the variable? (It may seem like mismanagement to lose track of variable names, but we’ll soon see that there are many variables whose names we don’t know.) There is a special syntax to access the value of a variable using its address instead of its name. Here’s an example program, PTRACC,that shows how it’s done: // ptracc.cpp // accessing the variable pointed to #include using namespace std; int main() { int var1 = 11; //two integer variables int var2 = 22; 11 3087 CH10 11/29/01 2:21 PM Page 436int* ptr; //pointer to integers ptr = &var1; //pointer points to var1 cout << *ptr << endl; //print contents of pointer (11) ptr = &var2; //pointer points to var2 cout << *ptr << endl; //print contents of pointer (22) return 0; } This program is very similar to PTRVAR,except that instead of printing the address values in ptr, we print the integer value stored at the address that’s stored in ptr. Here’s the output: 11 22 The expression that accesses the variables var1 and var2 is *ptr, which occurs in each of the two cout statements. When an asterisk is used in front of a variable name, as it is in the *ptr expression, it is called the dereference operator (or sometimes the indirection operator). It means the value of the variable pointed to by. Thus the expression *ptr represents the value of the variable pointed to by ptr. When ptr is set to the address of var1,the expression *ptr has the value 11, since var1 is 11. When ptr is changed to the address of var2,the expression *ptr acquires the value 22, since var2 is 22. Another name for the dereference operator is the contents of operator, which is another way to say the same thing. Figure 10.4 shows how this looks. You can use a pointer not only to display a variable’s value, but also to perform any operation you would perform on the variable directly. Here’s a program, PTRTO,that uses a pointer to assign a value to a variable, and then to assign that value to another variable: // ptrto.cpp // other access using pointers #include using namespace std; int main() { int var1, var2; //two integer variables int* ptr; //pointer to integers ptr = &var1; //set pointer to address of var1 *ptr = 37; //same as var1=37 var2 = *ptr; //same as var2=var1 cout << var2 << endl; //verify var2 is 37 return 0; } Pointers 10 P OINTERS 437 11 3087 CH10 11/29/01 2:21 PM Page 437FIGURE 10.4 Access via pointer. Remember that the asterisk used as the dereference operator has a different meaning than the asterisk used to declare pointer variables. The dereference operator precedes the variable and means value of the variable pointed to by. The asterisk used in a declaration means pointer to. int* ptr; //declaration: pointer to int *ptr = 37; //indirection: value of variable pointed to by ptr Using the dereference operator to access the value stored in an address is called indirect addressing,or sometimes dereferencing,the pointer. Here’s a capsule summary of what we’ve learned so far: int v; //defines variable v of type int int* p; //defines p as a pointer to int p = &v; //assigns address of variable v to pointer p v = 3; //assigns 3 to v *p = 3; //also assigns 3 to v Chapter 10438 11 3087 CH10 11/29/01 2:21 PM Page 438The last two statements show the difference between normal or direct addressing, where we refer to a variable by name, and pointer or indirect addressing, where we refer to the same variable using its address. In the example programs we’ve shown so far in this chapter, there’s really no advantage to using the pointer expression to access variables, since we can access them directly. The value of pointers becomes evident when you can’t access a variable directly, as we’ll see later. Pointer to void Before we go on to see pointers at work, we should note one peculiarity of pointer data types. Ordinarily, the address that you put in a pointer must be the same type as the pointer. You can’t assign the address of a float variable to a pointer to int,for example: float flovar = 98.6; int* ptrint = &flovar; //ERROR: can’t assign float* to int* However, there is an exception to this. There is a sort of general-purpose pointer that can point to any data type. This is called a pointer to void, and is defined like this: void* ptr; //ptr can point to any data type Such pointers have certain specialized uses, such as passing pointers to functions that operate independently of the data type pointed to. The next example uses a pointer to void and also shows that, if you don’t use void, you must be careful to assign pointers an address of the same type as the pointer. Here’s the listing for PTRVOID: // ptrvoid.cpp // pointers to type void #include using namespace std; int main() { int intvar; //integer variable float flovar; //float variable int* ptrint; //define pointer to int float* ptrflo; //define pointer to float void* ptrvoid; //define pointer to void ptrint = &intvar; //ok, int* to int* // ptrint = &flovar; //error, float* to int* // ptrflo = &intvar; //error, int* to float* ptrflo = &flovar; //ok, float* to float* Pointers 10 P OINTERS 439 11 3087 CH10 11/29/01 2:21 PM Page 439ptrvoid = &intvar; //ok, int* to void* ptrvoid = &flovar; //ok, float* to void* return 0; } You can assign the address of intvar to ptrint because they are both type int*,but you can’t assign the address of flovar to ptrint because the first is type float* and the second is type int*. However, ptrvoid can be given any pointer value, such as int*, because it is a pointer to void. If for some unusual reason you really need to assign one kind of pointer type to another, you can use the reinterpret_cast. For the lines commented out in PTRVOID,that would look like this: ptrint = reinterpret_cast(flovar); ptrflo = reinterpret_cast(intvar); The use of reinterpret_cast in this way is not recommended, but occasionally it’s the only way out of a difficult situation. Static casts won’t work with pointers. Old-style C casts can be used, but are always a bad idea in C++. We’ll see examples of reinterpret_cast in Chapter 12, “Streams and Files,” where it’s used to alter the way a data buffer is interpreted. Pointers and Arrays There is a close association between pointers and arrays. We saw in Chapter 7, “Arrays and Strings,” how array elements are accessed. The following program, ARRNOTE,provides a review. // arrnote.cpp // array accessed with array notation #include using namespace std; int main() { //array int intarray[5] = { 31, 54, 77, 52, 93 }; for(int j=0; j<5; j++) //for each element, cout << intarray[j] << endl; //print value return 0; } The cout statement prints each array element in turn. For instance, when j is 3, the expression intarray[j] takes on the value intarray[3] and accesses the fourth array element, the inte- ger 52. Here’s the output of ARRNOTE: Chapter 10440 11 3087 CH10 11/29/01 2:21 PM Page 44031 54 77 52 93 Surprisingly, array elements can be accessed using pointer notation as well as array notation. The next example, PTRNOTE, is similar to ARRNOTE except that it uses pointer notation. // ptrnote.cpp // array accessed with pointer notation #include using namespace std; int main() { //array int intarray[5] = { 31, 54, 77, 52, 93 }; for(int j=0; j<5; j++) //for each element, cout << *(intarray+j) << endl; //print value return 0; } The expression *(intarray+j) in PTRNOTE has exactly the same effect as intarray[j] in ARRNOTE, and the output of the programs is identical. But how do we interpret the expression *(intarray+j)? Suppose j is 3, so the expression is equivalent to *(intarray+3). We want this to represent the contents of the fourth element of the array (52). Remember that the name of an array is its address. The expression intarray+j is thus an address with something added to it. You might expect that intarray+3 would cause 3 bytes to be added to intarray. But that doesn’t produce the result we want: intarray is an array of integers, and 3 bytes into this array is the middle of the second element, which is not very useful. We want to obtain the fourth integer in the array, not the fourth byte, as shown in Figure 10.5. (This figure assumes 2-byte integers.) The C++ compiler is smart enough to take the size of the data into account when it performs arithmetic on data addresses. It knows that intarray is an array of type int because it was declared that way. So when it sees the expression intarray+3, it interprets it as the address of the fourth integer in intarray, not the fourth byte. But we want the value of this fourth array element, not the address. To take the value, we use the dereference operator (*). The resulting expression, when j is 3, is *(intarray+3), which is the content of the fourth array element, or 52. Pointers 10 P OINTERS 441 11 3087 CH10 11/29/01 2:21 PM Page 441FIGURE 10.5 Counting by integers. Now we see why a pointer declaration must include the type of the variable pointed to. The compiler needs to know whether a pointer is a pointer to int or a pointer to double so that it can perform the correct arithmetic to access elements of the array. It multiplies the index value by 2 in the case of type int,but by 8 in the case of double. Pointer Constants and Pointer Variables Suppose that, instead of adding j to intarray to step through the array addresses, you wanted to use the increment operator. Could you write *(intarray++)? The answer is no, and the reason is that you can’t increment a constant (or indeed change it in any way). The expression intarray is the address where the system has chosen to place your array, and it will stay at this address until the program terminates. intarray is a pointer con- stant. You can’t say intarray++ any more than you can say 7++. (In a multitasking system, variable addresses may change during program execution. An active program may be swapped out to disk and then reloaded at a different memory location. However, this process is invisible to your program.) But while you can’t increment an address, you can increment a pointer that holds an address. The next example, PTRINC,shows how: Chapter 10442 11 3087 CH10 11/29/01 2:21 PM Page 442// ptrinc.cpp // array accessed with pointer #include using namespace std; int main() { int intarray[] = { 31, 54, 77, 52, 93 }; //array int* ptrint; //pointer to int ptrint = intarray; //points to intarray for(int j=0; j<5; j++) //for each element, cout << *(ptrint++) << endl; //print value return 0; } Here we define a pointer to int—ptrint—and give it the value intarray,the address of the array. Now we can access the contents of the array elements with the expression *(ptrint++) The variable ptrint starts off with the same address value as intarray,thus allowing the first array element, intarray[0], which has the value 31, to be accessed as before. But, because ptrint is a variable and not a constant, it can be incremented. After it is incremented, it points to the second array element, intarray[1]. The expression *(ptrint++) then represents the contents of the second array element, or 54. The loop causes the expression to access each array element in turn. The output of PTRINC is the same as that for PTRNOTE. Pointers and Functions In Chapter 5 we noted that there are three ways to pass arguments to a function: by value, by reference, and by pointer. If the function is intended to modify variables in the calling pro- gram, these variables cannot be passed by value, since the function obtains only a copy of the variable. However, either a reference argument or a pointer can be used in this situation. Passing Simple Variables We’ll first review how arguments are passed by reference, and then compare this to passing pointer arguments. The PASSREF program shows passing by reference. // passref.cpp // arguments passed by reference #include using namespace std; int main() { Pointers 10 P OINTERS 443 11 3087 CH10 11/29/01 2:21 PM Page 443void centimize(double&); //prototype double var = 10.0; //var has value of 10 inches cout << “var = “ << var << “ inches” << endl; centimize(var); //change var to centimeters cout << “var = “ << var << “ centimeters” << endl; return 0; } //-------------------------------------------------------------- void centimize(double& v) { v *= 2.54; //v is the same as var } Here we want to convert a variable var in main() from inches to centimeters. We pass the vari- able by reference to the function centimize(). (Remember that the & following the data type double in the prototype for this function indicates that the argument is passed by reference.) The centimize() function multiplies the original variable by 2.54. Notice how the function refers to the variable. It simply uses the argument name v; v and var are different names for the same thing. Once it has converted var to centimeters, main() displays the result. Here’s the output of PASSREF: var = 10 inches var = 25.4 centimeters The next example, PASSPTR,shows an equivalent situation when pointers are used: // passptr.cpp // arguments passed by pointer #include using namespace std; int main() { void centimize(double*); //prototype double var = 10.0; //var has value of 10 inches cout << “var = “ << var << “ inches” << endl; centimize(&var); //change var to centimeters cout << “var = “ << var << “ centimeters” << endl; return 0; } //-------------------------------------------------------------- void centimize(double* ptrd) Chapter 10444 11 3087 CH10 11/29/01 2:21 PM Page 444{ *ptrd *= 2.54; //*ptrd is the same as var } The output of PASSPTR is the same as that of PASSREF. The function centimize() is declared as taking an argument that is a pointer to double: void centimize(double*) // argument is pointer to double When main() calls the function, it supplies the address of the variable as the argument: centimize(&var); Remember that this is not the variable itself, as it is in passing by reference, but the variable’s address. Because the centimize() function is passed an address, it must use the dereference operator, *ptrd,to access the value stored at this address: *ptrd *= 2.54; // multiply the contents of ptrd by 2.54 Of course this is the same as *ptrd = *ptrd * 2.54; // multiply the contents of ptrd by 2.54 where the standalone asterisk means multiplication. (This operator really gets around.) Since ptrd contains the address of var,anything done to *ptrd is actually done to var. Figure 10.6 shows how changing *ptrd in the function changes var in the calling program. Pointers 10 P OINTERS 445 FIGURE 10.6 Pointer passed to function. 11 3087 CH10 11/29/01 2:21 PM Page 445Passing a pointer as an argument to a function is in some ways similar to passing a reference. They both permit the variable in the calling program to be modified by the function. However, the mechanism is different. A reference is an alias for the original variable, while a pointer is the address of the variable. Passing Arrays We’ve seen numerous examples, starting in Chapter 7, of arrays passed as arguments to func- tions, and their elements being accessed by the function. Until this chapter, since we had not yet learned about pointers, this was done using array notation. However, it’s more common to use pointer notation instead of array notation when arrays are passed to functions. The PASSARR program shows how this looks: // passarr.cpp // array passed by pointer #include using namespace std; const int MAX = 5; //number of array elements int main() { void centimize(double*); //prototype double varray[MAX] = { 10.0, 43.1, 95.9, 59.7, 87.3 }; centimize(varray); //change elements of varray to cm for(int j=0; j using namespace std; int main() { void order(int*, int*); //prototype int n1=99, n2=11; //one pair ordered, one not int n3=22, n4=88; order(&n1, &n2); //order each pair of numbers order(&n3, &n4); cout << “n1=” << n1 << endl; //print out all numbers cout << “n2=” << n2 << endl; cout << “n3=” << n3 << endl; cout << “n4=” << n4 << endl; return 0; } //-------------------------------------------------------------- void order(int* numb1, int* numb2) //orders two numbers { if(*numb1 > *numb2) //if 1st larger than 2nd, { int temp = *numb1; //swap them *numb1 = *numb2; Chapter 10448 11 3087 CH10 11/29/01 2:21 PM Page 448*numb2 = temp; } } The function order() works the same as it did in REFORDER,except that it is passed the addresses of the numbers to be ordered, and it accesses the numbers using pointers. That is, *numb1 accesses the number in main() passed as the first argument, and *numb2 accesses the second. Here’s the output from PTRORDER: n1=11 ← this and n2=99 ← this are swapped, since they weren’t in order n3=22 ← this and n4=88 ← this are not swapped, since they were in order We’ll use the order() function from PTRORDER in our next example program, PTRSORT, which sorts an array of integers. // ptrsort.cpp // sorts an array using pointers #include using namespace std; int main() { void bsort(int*, int); //prototype const int N = 10; //array size //test array int arr[N] = { 37, 84, 62, 91, 11, 65, 57, 28, 19, 49 }; bsort(arr, N); //sort the array for(int j=0; j *numb2) //if 1st larger than 2nd, { int temp = *numb1; //swap them *numb1 = *numb2; *numb2 = temp; } } The array arr of integers in main() is initialized to unsorted values. The address of the array, and the number of elements, are passed to the bsort() function. This sorts the array, and the sorted values are then printed. Here’s the output of the PTRSORT: 11 19 28 37 49 57 62 65 84 91 The Bubble Sort The bsort() function sorts the array using a variation of the bubble sort. This is a simple (although notoriously slow) approach to sorting. Here’s how it works, assuming we want to arrange the numbers in the array in ascending order. First the first element of the array (arr[0]) is compared in turn with each of the other elements (starting with the second). If it’s greater than any of them, the two are swapped. When this is done we know that at least the first element is in order; it’s now the smallest element. Next the second element is compared in turn with all the other elements, starting with the third, and again swapped if it’s bigger. When we’re done we know that the second element has the second-smallest value. This process is continued for all the elements until the next-to-the-last, at which time the array is assumed to be ordered. Figure 10.8 shows the bubble sort in action (with fewer items than in PTRSORT). In PTRSORT,the number in the first position, 37, is compared with each element in turn, and swapped with 11. The number in the second position, which starts off as 84, is compared with each element. It’s swapped with 62; then 62 (which is now in the second position) is swapped with 37, 37 is swapped with 28, and 28 is swapped with 19. The number in the third position, which is 84 again, is swapped with 62, 62 is swapped with 57, 57 with 37, and 37 with 28. The process continues until the array is sorted. The bsort() function in PTRSORT consists of two nested loops, each of which controls a pointer. The outer loop uses the loop variable j, and the inner one uses k. The expressions ptr+j and ptr+k point to various elements of the array, as determined by the loop variables. The expression ptr+j moves down the array, starting at the first element (the top) and stepping down integer by integer until one short of the last element (the bottom). For each position taken by ptr+j in the outer loop, the expression ptr+k in the inner loop starts pointing one below ptr+j and moves down to the bottom of the array. Each time through the inner loop, the Chapter 10450 11 3087 CH10 11/29/01 2:21 PM Page 450elements pointed to by ptr+j and ptr+k are compared, using the order() function, and if the first is greater than the second, they’re swapped. Figure 10.9 shows this process. Pointers 10 P OINTERS 451 FIGURE 10.8 Operation of the bubble sort. The PTRSORT example begins to reveal the power of pointers. They provide a consistent and efficient way to operate on array elements and other variables whose names aren’t known to a particular function. 11 3087 CH10 11/29/01 2:21 PM Page 451FIGURE 10.9 Operation of PTRSORT. Pointers and C-Type Strings As we noted in Chapter 7, C-type strings are simply arrays of type char. Thus pointer notation can be applied to the characters in strings, just as it can to the elements of any array. Pointers to String Constants Here’s an example, TWOSTR,in which two strings are defined, one using array notation as we’ve seen in previous examples, and one using pointer notation: // twostr.cpp // strings defined using array and pointer notation #include using namespace std; int main() { char str1[] = “Defined as an array”; char* str2 = “Defined as a pointer”; cout << str1 << endl; // display both strings cout << str2 << endl; // str1++; // can’t do this; str1 is a constant str2++; // this is OK, str2 is a pointer cout << str2 << endl; // now str2 starts “efined...” return 0; } Chapter 10452 11 3087 CH10 11/29/01 2:21 PM Page 452In many ways these two types of definition are equivalent. You can print out both strings as the example shows, use them as function arguments, and so on. But there is a subtle difference: str1 is an address—that is, a pointer constant—while str2 is a pointer variable. So str2 can be changed, while str1 cannot, as shown in the program. Figure 10.10 shows how these two kinds of strings look in memory. Pointers 10 P OINTERS 453 FIGURE 10.10 Strings as arrays and pointers. We can increment str2, since it is a pointer, but once we do, it no longer points to the first character in the string. Here’s the output of TWOSTR: Defined as an array Defined as a pointer efined as a pointer ← following str2++ (‘D’ is gone) A string defined as a pointer is considerably more flexible than one defined as an array. The following examples will make use of this flexibility. Strings as Function Arguments Here’s an example that shows a string used as a function argument. The function simply prints the string, by accessing each character in turn. Here’s the listing for PTRSTR: // ptrstr.cpp // displays a string with pointer notation #include using namespace std; 11 3087 CH10 11/29/01 2:21 PM Page 453int main() { void dispstr(char*); //prototype char str[] = “Idle people have the least leisure.”; dispstr(str); //display the string return 0; } //-------------------------------------------------------------- void dispstr(char* ps) { while( *ps ) //until null character, cout << *ps++; //print characters cout << endl; } The array address str is used as the argument in the call to function dispstr(). This address is a constant, but since it is passed by value, a copy of it is created in dispstr(). This copy is a pointer, ps. A pointer can be changed, so the function increments ps to display the string. The expression *ps++ returns the successive characters of the string. The loop cycles until it finds the null character (‘\0’) at the end of the string. Since this has the value 0, which repre- sents false,the while loop terminates at this point. Copying a String Using Pointers We’ve seen examples of pointers used to obtain values from an array. Pointers can also be used to insert values into an array. The next example, COPYSTR, demonstrates a function that copies one string to another: // copystr.cpp // copies one string to another with pointers #include using namespace std; int main() { void copystr(char*, const char*); //prototype char* str1 = “Self-conquest is the greatest victory.”; char str2[80]; //empty string copystr(str2, str1); //copy str1 to str2 cout << str2 << endl; //display str2 return 0; } Chapter 10454 11 3087 CH10 11/29/01 2:21 PM Page 454//-------------------------------------------------------------- void copystr(char* dest, const char* src) { while( *src ) //until null character, *dest++ = *src++; //copy chars from src to dest *dest = ‘\0’; //terminate dest } Here the main() part of the program calls the function copystr() to copy str1 to str2. In this function the expression *dest++ = *src++; takes the value at the address pointed to by src and places it in the address pointed to by dest. Both pointers are then incremented, so the next time through the loop the next character will be transferred. The loop terminates when a null character is found in src; at this point a null is inserted in dest and the function returns. Figure 10.11 shows how the pointers move through the strings. Pointers 10 P OINTERS 455 FIGURE 10.11 Operation of COPYSTR. 11 3087 CH10 11/29/01 2:21 PM Page 455Library String Functions Many of the library functions we have already used for strings have string arguments that are specified using pointer notation. As an example you can look at the description of strcpy() in your compiler’s documentation (or in the STRING.H header file). This function copies one string to another; we can compare it with our homemade copystr() function in the COPYSTR exam- ple. Here’s the syntax for the strcpy() library function: char* strcpy(char* dest, const char* src); This function takes two arguments of type char*. (The next section, “The const Modifier and Pointers,” explains the meaning of const in this context.) The strcpy() function also returns a pointer to char; this is the address of the dest string. In other respects, this function works very much like our homemade copystr() function. The const Modifier and Pointers The use of the const modifier with pointer declarations can be confusing, because it can mean one of two things, depending on where it’s placed. The following statements show the two pos- sibilities: const int* cptrInt; //cptrInt is a pointer to constant int int* const ptrcInt; //ptrcInt is a constant pointer to int Following the first declaration, you cannot change the value of whatever cptrInt points to, although you can change cptrInt itself. Following the second declaration, you can change what ptrcInt points to, but you cannot change the value of ptrcInt itself. You can remember the difference by reading from right to left, as indicated in the comments. You can use const in both positions to make the pointer and what it points to constant. In the declaration of strcpy() just shown, the argument const char* src specifies that the characters pointed to by src cannot be changed by strcpy(). It does not imply that the src pointer itself cannot be modified. To do that the argument declaration would need to be char* const src. Arrays of Pointers to Strings Just as there are arrays of variables of type int or type float,there can also be arrays of pointers. A common use for this construction is an array of pointers to strings. In Chapter 7 the STRARAY program demonstrated an array of char* strings. As we noted, there is a disadvantage to using an array of strings, in that the subarrays that hold the strings must all be the same length, so space is wasted when strings are shorter than the length of the subarrays (see Figure 7.10 in Chapter 7). Chapter 10456 11 3087 CH10 11/29/01 2:21 PM Page 456Let’s see how to use pointers to solve this problem. We will modify STRARAY to create an array of pointers to strings, rather than an array of strings. Here’s the listing for PTRTOSTR: // ptrtostr.cpp // an array of pointers to strings #include using namespace std; const int DAYS = 7; //number of pointers in array int main() { //array of pointers to char char* arrptrs[DAYS] = { “Sunday”, “Monday”, “Tuesday”, “Wednesday”, “Thursday”, “Friday”, “Saturday” }; for(int j=0; j> size; // get size from user int arr[size]; // error; array size must be a constant The compiler requires the array size to be a constant. But in many situations we don’t know how much memory we need until runtime. We might want to store a string that was typed in by the user, for example. In this situation we can define an array sized to hold the largest string we expect, but this wastes memory. (As we’ll learn in Chapter 15, “The Standard Template Library,” you can also use a vector, which is a sort of expandable array.) The new Operator C++ provides a different approach to obtaining blocks of memory: the new operator. This ver- satile operator obtains memory from the operating system and returns a pointer to its starting point. The NEWINTRO example shows how new is used: // newintro.cpp // introduces operator new #include #include //for strlen using namespace std; int main() { char* str = “Idle hands are the devil’s workshop.”; int len = strlen(str); //get length of str char* ptr; //make a pointer to char ptr = new char[len+1]; //set aside memory: string + ‘\0’ strcpy(ptr, str); //copy str to new memory area ptr cout << “ptr=” << ptr << endl; //show that ptr is now in str delete[] ptr; //release ptr’s memory return 0; } The expression ptr = new char[len+1]; returns a pointer to a section of memory just large enough to hold the string str, whose length len we found with the strlen() library function, plus an extra byte for the null character ‘\0’ at the end of the string. Figure 10.13 shows the syntax of a statement using the new operator. Remember to use brackets around the size; the compiler won’t object if you mistakenly use parentheses, but the results will be incorrect. Pointers 10 P OINTERS 459 11 3087 CH10 11/29/01 2:21 PM Page 459FIGURE 10.14 Memory obtained by the new operator. Chapter 10460 FIGURE 10.13 Syntax of the new operator. Figure 10.14 shows the memory obtained by new and the pointer to it. 11 3087 CH10 11/29/01 2:21 PM Page 460In NEWINTRO we use strcpy() to copy string str to the newly created memory area pointed to by ptr. Since we made this area equal in size to the length of str,the string fits exactly. The output of NEWINTRO is ptr=Idle hands are the devil’s workshop. C programmers will recognize that new plays a role similar to the malloc() family of library functions. The new approach is superior in that it returns a pointer to the appropriate data type, while malloc()’s pointer must be cast to the appropriate type. There are other advantages as well. C programmers may wonder whether there is a C++ equivalent to realloc() for changing the size of memory that has already been reallocated. Sorry, there’s no renew in C++. You’ll need to fall back on the ploy of creating a larger (or smaller) space with new, and copying your data from the old area to the new one. The delete Operator If your program reserves many chunks of memory using new,eventually all the available mem- ory will be reserved and the system will crash. To ensure safe and efficient use of memory, the new operator is matched by a corresponding delete operator that returns memory to the oper- ating system. In NEWINTRO the statement delete[] ptr; returns to the system whatever memory was pointed to by ptr. Actually, there is no need for this operator in NEWINTRO, since memory is automatically returned when the program terminates. However, suppose you use new in a function. If the function uses a local variable as a pointer to this memory, the pointer will be destroyed when the function terminates, but the memory will be left as an orphan, taking up space that is inac- cessible to the rest of the program. Thus it is always good practice, and often essential, to delete memory when you’re through with it. Deleting the memory doesn’t delete the pointer that points to it (str in NEWINTRO), and doesn’t change the address value in the pointer. However, this address is no longer valid; the memory it points to may be changed to something entirely different. Be careful that you don’t use pointers to memory that has been deleted. The brackets following delete indicate that we’re deleting an array. If you create a single object with new, you don’t need the brackets when you delete it. ptr = new SomeClass; // allocate a single object . . . delete ptr; // no brackets following delete Pointers 10 P OINTERS 461 11 3087 CH10 11/29/01 2:21 PM Page 461However, don’t forget the brackets when deleting arrays of objects. Using them ensures that all the members of the array are deleted, and that the destructor is called for each one. A String Class Using new The new operator often appears in constructors. As an example, we’ll modify the String class, last seen in examples such as STRPLUS in Chapter 8, “Operator Overloading.” You may recall that a potential defect of that class was that all String objects occupied the same fixed amount of memory. A string shorter than this fixed length wasted memory, and a longer string—if one were mistakenly generated—could crash the system by extending beyond the end of the array. Our next example uses new to obtain exactly the right amount of memory. Here’s the listing for NEWSTR: // newstr.cpp // using new to get memory for strings #include #include //for strcpy(), etc using namespace std; //////////////////////////////////////////////////////////////// class String //user-defined string type { private: char* str; //pointer to string public: String(char* s) //constructor, one arg { int length = strlen(s); //length of string argument str = new char[length+1]; //get memory strcpy(str, s); //copy argument to it } ~String() //destructor { cout << “Deleting str.\n”; delete[] str; //release memory } void display() //display the String { cout << str << endl; } }; //////////////////////////////////////////////////////////////// int main() { //uses 1-arg constructor String s1 = “Who knows nothing doubts nothing.”; Chapter 10462 11 3087 CH10 11/29/01 2:21 PM Page 462cout << “s1=”; //display string s1.display(); return 0; } The output from this program is s1=Who knows nothing doubts nothing. Deleting str. The String class has only one data item: a pointer to char called str. This pointer will point to the string held by the String object. There is no array within the object to hold the string. The string is stored elsewhere; only the pointer to it is a member of String. Constructor in NEWSTR The constructor in this example takes a normal char* string as its argument. It obtains space in memory for this string with new; str points to the newly obtained memory. The constructor then uses strcpy() to copy the string into this new space. Destructor in NEWSTR We haven’t seen many destructors in our examples so far, but now that we’re allocating mem- ory with new, destructors become important. If we allocate memory when we create an object, it’s reasonable to deallocate the memory when the object is no longer needed. As you may recall from Chapter 6, a destructor is a routine that is called automatically when an object is destroyed. The destructor in NEWSTR looks like this: ~String() { cout << “Deleting str.”; delete[] str; } This destructor gives back to the system the memory obtained when the object was created. You can tell from the program’s output that the destructor executed at the end of the program. Objects (like other variables) are typically destroyed when the function in which they were defined terminates. This destructor ensures that memory obtained by the String object will be returned to the system, and not left in limbo, when the object is destroyed. We should note a potential glitch in using destructors as shown in NEWSTR. If you copy one String object to another, say with a statement like s2 = s1, you’re really only copying the pointer to the actual (char*) string. Both objects now point to the same string in memory. But if you now delete one string, the destructor will delete the char* string, leaving the other object with an invalid pointer. This can be subtle, because objects can be deleted in non- obvious ways, such as when a function in which a local object has been created returns. In Chapter 11 we’ll see how to make a smarter destructor that counts how many String objects are pointing to a string. Pointers 10 P OINTERS 463 11 3087 CH10 11/29/01 2:21 PM Page 463Pointers to Objects Pointers can point to objects as well as to simple data types and arrays. We’ve seen many examples of objects defined and given a name, in statements like Distance dist; where an object called dist is defined to be of the Distance class. Sometimes, however, we don’t know, at the time that we write the program, how many objects we want to create. When this is the case we can use new to create objects while the program is running. As we’ve seen, new returns a pointer to an unnamed object. Let’s look at a short exam- ple program, ENGLPTR,that compares the two approaches to creating objects. // englptr.cpp // accessing member functions by pointer #include using namespace std; //////////////////////////////////////////////////////////////// class Distance //English Distance class { private: int feet; float inches; public: void getdist() //get length from user { cout << “\nEnter feet: “; cin >> feet; cout << “Enter inches: “; cin >> inches; } void showdist() //display distance { cout << feet << “\’-” << inches << ‘\”’; } }; //////////////////////////////////////////////////////////////// int main() { Distance dist; //define a named Distance object dist.getdist(); //access object members dist.showdist(); // with dot operator Distance* distptr; //pointer to Distance distptr = new Distance; //points to new Distance object distptr->getdist(); //access object members distptr->showdist(); // with -> operator cout << endl; return 0; } Chapter 10464 11 3087 CH10 11/29/01 2:21 PM Page 464This program uses a variation of the English Distance class seen in previous chapters. The main() function defines dist,uses the Distance member function getdist() to get a distance from the user, and then uses showdist() to display it. Referring to Members ENGLPTR then creates another object of type Distance using the new operator, and returns a pointer to it called distptr. The question is, how do we refer to the member functions in the object pointed to by distptr? You might guess that we would use the dot (.) membership-access operator, as in distptr.getdist(); // won’t work; distptr is not a variable but this won’t work. The dot operator requires the identifier on its left to be a variable. Since distptr is a pointer to a variable, we need another syntax. One approach is to dereference (get the contents of the variable pointed to by) the pointer: (*distptr).getdist(); // ok but inelegant However, this is slightly cumbersome because of the parentheses. (The parentheses are neces- sary because the dot operator (.) has higher precedence than the dereference operator (*). An equivalent but more concise approach is furnished by the membership-access operator, which consists of a hyphen and a greater-than sign: distptr->getdist(); // better approach As you can see in ENGLPTR,the -> operator works with pointers to objects in just the same way that the . operator works with objects. Here’s the output of the program: Enter feet: 10 ← this object uses the dot operator Enter inches: 6.25 10’-6.25” Enter feet: 6 ← this object uses the -> operator Enter inches: 4.75 6’-4.75” Another Approach to new You may come across another—less common—approach to using new to obtain memory for objects. Since new can return a pointer to an area of memory that holds an object, we should be able to refer to the original object by dereferencing the pointer. The ENGLREF example shows how this is done. Pointers 10 P OINTERS 465 11 3087 CH10 11/29/01 2:21 PM Page 465// englref.cpp // dereferencing the pointer returned by new #include using namespace std; //////////////////////////////////////////////////////////////// class Distance // English Distance class { private: int feet; float inches; public: void getdist() // get length from user { cout << “\nEnter feet: “; cin >> feet; cout << “Enter inches: “; cin >> inches; } void showdist() // display distance { cout << feet << “\’-” << inches << ‘\”’; } }; //////////////////////////////////////////////////////////////// int main() { Distance& dist = *(new Distance); // create Distance object // alias is “dist” dist.getdist(); // access object members dist.showdist(); // with dot operator cout << endl; return 0; } The expression new Distance returns a pointer to a memory area large enough for a Distance object, so we can refer to the original object as *(new Distance) This is the object pointed to by the pointer. Using a reference, we define dist to be an object of type Distance, and we set it equal to *(new Distance). Now we can refer to members of dist using the dot membership operator, rather than ->. This approach is less common than using pointers to objects obtained with new,or simply declaring an object, but it works in a similar way. Chapter 10466 11 3087 CH10 11/29/01 2:21 PM Page 466An Array of Pointers to Objects A common programming construction is an array of pointers to objects. This arrangement allows easy access to a group of objects, and is more flexible than placing the objects them- selves in an array. (For instance, in the PERSORT example in this chapter we’ll see how a group of objects can be sorted by sorting an array of pointers to them, rather than sorting the objects themselves.) Our next example, PTROBJS,creates an array of pointers to the person class. Here’s the listing: // ptrobjs.cpp // array of pointers to objects #include using namespace std; //////////////////////////////////////////////////////////////// class person //class of persons { protected: char name[40]; //person’s name public: void setName() //set the name { cout << “Enter name: “; cin >> name; } void printName() //get the name { cout << “\n Name is: “ << name; } }; //////////////////////////////////////////////////////////////// int main() { person* persPtr[100]; //array of pointers to persons int n = 0; //number of persons in array char choice; do //put persons in array { persPtr[n] = new person; //make new object persPtr[n]->setName(); //set person’s name n++; //count new person cout << “Enter another (y/n)? “; //enter another cin >> choice; //person? } while( choice==’y’ ); //quit on ‘n’ Pointers 10 P OINTERS 467 11 3087 CH10 11/29/01 2:21 PM Page 467for(int j=0; jprintName(); } cout << endl; return 0; } //end main() The class person has a single data item, name, which holds a string representing a person’s name. Two member functions, setName() and printName(),allow the name to be set and dis- played. Program Operation The main() function defines an array, persPtr,of 100 pointers to type person. In a do loop it then asks the user to enter a name. With this name it creates a person object using new, and stores a pointer to this object in the array persPtr. To demonstrate how easy it is to access the objects using the pointers, it then prints out the name data for each person object. Here’s a sample interaction with the program: Enter name: Stroustrup ← user enters names Enter another (y/n)? y Enter name: Ritchie Enter another (y/n)? y Enter name: Kernighan Enter another (y/n)? n Person number 1 ← program displays all names stored Name is: Stroustrup Person number 2 Name is: Ritchie Person number 3 Name is: Kernighan Accessing Member Functions We need to access the member functions setName() and printName() in the person objects pointed to by the pointers in the array persPtr. Each of the elements of the array persPtr is specified in array notation to be persPtr[j] (or equivalently by pointer notation to be *(persPtr+j)). The elements are pointers to objects of type person. To access a member of an object using a pointer, we use the -> operator. Putting this all together, we have the following syntax for getname(): persPtr[j]->getName() This executes the getname() function in the person object pointed to by element j of the persPtr array. (It’s a good thing we don’t have to program using English syntax.) Chapter 10468 11 3087 CH10 11/29/01 2:21 PM Page 468A Linked List Example Our next example shows a simple linked list. What is a linked list? It’s another way to store data. You’ve seen numerous examples of data stored in arrays. Another data structure is an array of pointers to data members, as in the PTRTOSTRS and PTROBJS examples. Both the array and the array of pointers suffer from the necessity to declare a fixed-size array before running the program. A Chain of Pointers The linked list provides a more flexible storage system in that it doesn’t use arrays at all. Instead, space for each data item is obtained as needed with new,and each item is connected, or linked,to the next data item using a pointer. The individual items don’t need to be located contiguously in memory the way array elements are; they can be scattered anywhere. In our example the entire linked list is an object of class linklist. The individual data items, or links, are represented by structures of type link. Each such structure contains an integer— representing the object’s single data item—and a pointer to the next link. The list itself stores a pointer to the link at the head of the list. This arrangement is shown in Figure 10.15. Pointers 10 P OINTERS 469 FIGURE 10.15 A linked list. Here’s the listing for LINKLIST: // linklist.cpp // linked list #include using namespace std; 11 3087 CH10 11/29/01 2:21 PM Page 469//////////////////////////////////////////////////////////////// struct link //one element of list { int data; //data item link* next; //pointer to next link }; //////////////////////////////////////////////////////////////// class linklist //a list of links { private: link* first; //pointer to first link public: linklist() //no-argument constructor { first = NULL; } //no first link void additem(int d); //add data item (one link) void display(); //display all links }; //-------------------------------------------------------------- void linklist::additem(int d) //add data item { link* newlink = new link; //make a new link newlink->data = d; //give it data newlink->next = first; //it points to next link first = newlink; //now first points to this } //-------------------------------------------------------------- void linklist::display() //display all links { link* current = first; //set ptr to first link while( current != NULL ) //quit on last link { cout << current->data << endl; //print data current = current->next; //move to next link } } //////////////////////////////////////////////////////////////// int main() { linklist li; //make linked list li.additem(25); //add four items to list li.additem(36); li.additem(49); li.additem(64); Chapter 10470 11 3087 CH10 11/29/01 2:21 PM Page 470li.display(); //display entire list return 0; } The linklist class has only one member data item: the pointer to the start of the list. When the list is first created, the constructor initializes this pointer, which is called first,to NULL. The NULL constant is defined to be 0. This value serves as a signal that a pointer does not hold a valid address. In our program a link whose next member has a value of NULL is assumed to be at the end of the list. Adding an Item to the List The additem() member function adds an item to the linked list. A new link is inserted at the beginning of the list. (We could write the additem() function to insert items at the end of the list, but that is a little more complex to program.) Let’s look at the steps involved in inserting a new link. First, a new structure of type link is created by the line link* newlink = new link; This creates memory for the new link structure with new and saves the pointer to it in the newlink variable. Next we want to set the members of the newly created structure to appropriate values. A struc- ture is similar to a class in that, when it is referred to by pointer rather than by name, its mem- bers are accessed using the -> member-access operator. The following two lines set the data variable to the value passed as an argument to additem(), and the next pointer to point to whatever address was in first, which holds the pointer to the start of the list. newlink->data = d; newlink->next = first; Finally, we want the first variable to point to the new link: first = newlink; The effect is to uncouple the connection between first and the old first link, insert the new link, and move the old first link into the second position. Figure 10.16 shows this process. Pointers 10 P OINTERS 471 11 3087 CH10 11/29/01 2:21 PM Page 471FIGURE 10.16 Adding to a linked list. Displaying the List Contents Once the list is created it’s easy to step through all the members, displaying them (or perform- ing other operations). All we need to do is follow from one next pointer to another until we find a next that is NULL, signaling the end of the list. In the function display(),the line cout << endl << current->data; prints the value of the data, and current = current->next; moves us along from one link to another, until current != NULL in the while expression becomes false. Here’s the output of LINKLIST: 64 49 36 25 Chapter 10472 11 3087 CH10 11/29/01 2:21 PM Page 472Linked lists are perhaps the most commonly used data storage arrangements after arrays. As we noted, they avoid the wasting of memory space engendered by arrays. The disadvantage is that finding a particular item on a linked list requires following the chain of links from the head of the list until the desired link is reached. This can be time-consuming. An array ele- ment, on the other hand, can be accessed quickly, provided its index is known in advance. We’ll have more to say about linked lists and other data-storage techniques in Chapter 15, “The Standard Template Library.” Self-Containing Classes We should note a possible pitfall in the use of self-referential classes and structures. The link structure in LINKLIST contained a pointer to the same kind of structure. You can do the same with classes: class sampleclass { sampleclass* ptr; // this is fine }; However, while a class can contain a pointer to an object of its own type, it cannot contain an object of its own type: class sampleclass { sampleclass obj; // can’t do this }; This is true of structures as well as classes. Augmenting LINKLIST The general organization of LINKLIST can serve for a more complex situation than that shown. There could be more data in each link. Instead of an integer, a link could hold a number of data items or it could hold a pointer to a structure or object. Additional member functions could perform such activities as adding and removing links from an arbitrary part of the chain. Another important member function is a destructor. As we men- tioned, it’s important to delete blocks of memory that are no longer in use. A destructor that performs this task would be a highly desirable addition to the linklist class. It could go through the list using delete to free the memory occupied by each link. Pointers 10 P OINTERS 473 11 3087 CH10 11/29/01 2:21 PM Page 473Pointers to Pointers Our next example demonstrates an array of pointers to objects, and shows how to sort these pointers based on data in the object. This involves the idea of pointers to pointers, and may help demonstrate why people lose sleep over pointers. The idea in the next program is to create an array of pointers to objects of the person class. This is similar to the PTROBJS example, but we go further and add variations of the order() and bsort() functions from the PTRSORT example so that we can sort a group of person objects based on the alphabetical order of their names. Here’s the listing for PERSORT: // persort.cpp // sorts person objects using array of pointers #include #include //for string class using namespace std; //////////////////////////////////////////////////////////////// class person //class of persons { protected: string name; //person’s name public: void setName() //set the name { cout << “Enter name: “; cin >> name; } void printName() //display the name { cout << endl << name; } string getName() //return the name { return name; } }; //////////////////////////////////////////////////////////////// int main() { void bsort(person**, int); //prototype person* persPtr[100]; //array of pointers to persons int n = 0; //number of persons in array char choice; //input char do { //put persons in array persPtr[n] = new person; //make new object persPtr[n]->setName(); //set person’s name n++; //count new person cout << “Enter another (y/n)? “; //enter another cin >> choice; // person? } Chapter 10474 11 3087 CH10 11/29/01 2:21 PM Page 474while( choice==’y’ ); //quit on ‘n’ cout << “\nUnsorted list:”; for(int j=0; jprintName(); } bsort(persPtr, n); //sort pointers cout << “\nSorted list:”; for(j=0; jprintName(); } cout << endl; return 0; } //end main() //-------------------------------------------------------------- void bsort(person** pp, int n) //sort pointers to persons { void order(person**, person**); //prototype int j, k; //indexes to array for(j=0; jgetName() > (*pp2)->getName() ) { person* tempptr = *pp1; //swap the pointers *pp1 = *pp2; *pp2 = tempptr; } } When the program is first executed it asks for a name. When the user gives it one, it creates an object of type person and sets the name data in this object to the name entered by the user. The program also stores a pointer to the object in the persPtr array. When the user types n to indicate that no more names will be entered, the program calls the bsort() function to sort the person objects based on their name member variables. Here’s some sample interaction with the program: Enter name: Washington Enter another (y/n)? y Enter name: Adams Pointers 10 P OINTERS 475 11 3087 CH10 11/29/01 2:21 PM Page 475Enter another (y/n)? y Enter name: Jefferson Enter another (y/n)? y Enter name: Madison Enter another (y/n)? n Unsorted list: Washington Adams Jefferson Madison Sorted list: Adams Jefferson Madison Washington Sorting Pointers Actually, when we sort person objects, we don’t move the objects themselves; we move the pointers to the objects. This eliminates the need to shuffle the objects around in memory, which can be very time-consuming if the objects are large. It could also, if we wanted, allow us to keep multiple sorts—one by name and another by phone number, for example—in memory at the same time without storing the objects multiple times. The process is shown in Figure 10.17. To facilitate the sorting activity, we’ve added a getName() member function to the person class so we can access the names from order() to decide when to swap pointers. The person** Data Type You will notice that the first argument to the bsort() function, and both arguments to order(),have the type person**. What do the two asterisks mean? These arguments are used to pass the address of the array persPtr,or—in the case of order()—the addresses of ele- ments of the array. If this were an array of type person,the address of the array would be type person*. However, the array is of type pointers to person,or person*,so its address is type person**. The address of a pointer is a pointer to a pointer. Figure 10.18 shows how this looks. Chapter 10476 11 3087 CH10 11/29/01 2:21 PM Page 476FIGURE 10.17 Sorting an array of pointers. Pointers 10 P OINTERS 477 11 3087 CH10 11/29/01 2:21 PM Page 477FIGURE 10.18 Pointer to an array of pointers. Compare this program with PTRSORT, which sorted an array of type int. You’ll find that the data types passed to functions in PERSORT all have one more asterisk than they did in PTRSORT, because the array is an array of pointers. Since the persPtr array contains pointers, the construction persPtr[j]->printName() executes the printName() function in the object pointed to by element j of persPtr. Comparing Strings The order() function in PERSORT has been modified to order two strings lexigraphically—that is, by putting them in alphabetical order. To do this it compares the strings using the C++ library function strcmp(). This function takes the two strings s1 and s2 as arguments, as in strcmp(s1, s2), and returns one of the following values. Value Condition <0 s1 comes before s2 0 s1 is the same as s2 >0 s1 comes after s2 The strings are accessed using the syntax (*pp1)->getname() Chapter 10478 11 3087 CH10 11/29/01 2:21 PM Page 478The argument pp1 is a pointer to a pointer, and we want the name pointed to by the pointer it points to. The member-access operator -> dereferences one level, but we need to dereference another level, hence the asterisk preceding pp1. Just as there can be pointers to pointers, there can be pointers to pointers to pointers, and so on. Fortunately such complexities are seldom encountered. A Parsing Example Programmers are frequently faced with the problem of unravelling or parsing a string of sym- bols. Examples are commands typed by a user at the keyboard, sentences in natural languages (such as English), statements in a programming language, and algebraic expressions. Now that we’ve learned about pointers and strings, we can handle this sort of problem. Our next (somewhat longer) example will show how to parse arithmetic expressions like 6/3+2*3-1 The user enters the expression, the program works its way through it, character by character, figures out what it means in arithmetic terms, and displays the resulting value (7 in the exam- ple). Our expressions will use the four arithmetic operators: +, -, *, and /. We’ll simplify the numbers we use to make the programming easier by restricting them to a single digit. Also, we won’t allow parentheses. This program makes use of our old friend the Stack class (see the STAKARAY program in Chapter 7). We’ve modified this class so that it stores data of type char. We use the stack to store both numbers and operators (both as characters). The stack is a useful storage mechanism because, when parsing expressions, we frequently need to access the last item stored, and a stack is a last-in-first-out (LIFO) container. Besides the Stack class, we’ll use a class called express (short for expression), representing an entire arithmetic expression. Member functions for this class allow us to initialize an object with an expression in the form of a string (entered by the user), parse the expression, and return the resulting arithmetic value. Parsing Arithmetic Expressions Here’s how we parse an arithmetic expression. We start at the left, and look at each character in turn. It can be either a number (always a single digit—a character between 0 and 9), or an operator (the characters +, -, *, and /). If the character is a number, we always push it onto the stack. We also push the first operator we encounter. The trick is how we handle subsequent operators. Note that we can’t execute the current operator, because we haven’t yet read the number that follows it. Finding an operator is Pointers 10 P OINTERS 479 11 3087 CH10 11/29/01 2:21 PM Page 479merely the signal that we can execute the previous operator, which is stored on the stack. That is, if the sequence 2+3 is on the stack, we wait until we find another operator before carrying out the addition. Thus whenever we find that the current character is an operator (except the first), we pop the previous number (3 in the preceding example) and the previous operator (+) off the stack, plac- ing them in the variables lastval and lastop. Finally we pop the first number (2) and carry out the arithmetic operation on the two numbers (obtaining 5). Can we always execute the pre- vious operator? No. Remember that * and / have a higher precedence than + and -. In the expression 3+4/2, we can’t execute the + until we’ve done the division. So when we get to the / in this expression, we must put the 2 and the + back on the stack until we’ve carried out the division. On the other hand, if the current operator is a + or -, we know we can always execute the pre- vious operator. That is, when we see the + in the expression 4-5+6, we know it’s all right to execute the -, and when we see the - in 6/2-3 , we know it’s okay to do the division. Table 10.1 shows the four possibilities. TABLE 10.1 Operators and Parsing Actions Previous Current Operator Operator Example Action + or -*or / 3+4/ Push previous operator and previous number (+, 4) * or /*or / 9/3* Execute previous operator, push result (3) + or -+or - 6+3+ Execute previous operator, push result (9) * or /+or - 8/2- Execute previous operator, push result (4) The parse() member function carries out this process of going through the input expression and performing those operations it can. However, there is more work to do. The stack still con- tains either a single number or several sequences of number-operator-number. Working down through the stack, we can execute these sequences. Finally, a single number is left on the stack; this is the value of the original expression. The solve() member function carries out this task, working its way down through the stack until only a single number is left. In general, parse() puts things on the stack, and solve() takes them off. Chapter 10480 11 3087 CH10 11/29/01 2:21 PM Page 480The PARSE Program Some typical interaction with PARSE might look like this: Enter an arithmetic expression of the form 2+3*4/3-2. No number may have more than one digit. Don’t use any spaces or parentheses. Expression: 9+6/3 The numerical value is: 11 Do another (Enter y or n)? Note that it’s all right if the results of arithmetic operations contain more than one digit. They are limited only by the numerical size of type char,from –128 to +127. Only the input string is limited to numbers from 0 to 9. Here’s the listing for the program: // parse.cpp // evaluates arithmetic expressions composed of 1-digit numbers #include #include //for strlen(), etc using namespace std; const int LEN = 80; //length of expressions, in characters const int MAX = 40; //size of stack //////////////////////////////////////////////////////////////// class Stack { private: char st[MAX]; //stack: array of chars int top; //number of top of stack public: Stack() //constructor { top = 0; } void push(char var) //put char on stack { st[++top] = var; } char pop() //take char off stack { return st[top--]; } int gettop() //get top of stack { return top; } }; //////////////////////////////////////////////////////////////// class express //expression class { private: Stack s; //stack for analysis char* pStr; //pointer to input string int len; //length of input string Pointers 10 P OINTERS 481 11 3087 CH10 11/29/01 2:21 PM Page 481public: express(char* ptr) //constructor { pStr = ptr; //set pointer to string len = strlen(pStr); //set length } void parse(); //parse the input string int solve(); //evaluate the stack }; //-------------------------------------------------------------- void express::parse() //add items to stack { char ch; //char from input string char lastval; //last value char lastop; //last operator for(int j=0; j=’0’ && ch<=’9’) //if it’s a digit, s.push(ch-’0’); //save numerical value //if it’s operator else if(ch==’+’ || ch==’-’ || ch==’*’ || ch==’/’) { if(s.gettop()==1) //if it’s first operator s.push(ch); //put on stack else //not first operator { lastval = s.pop(); //get previous digit lastop = s.pop(); //get previous operator //if this is * or / AND last operator was + or - if( (ch==’*’ || ch==’/’) && (lastop==’+’ || lastop==’-’) ) { s.push(lastop); //restore last two pops s.push(lastval); } else //in all other cases { switch(lastop) //do last operation { //push result on stack case ‘+’: s.push(s.pop() + lastval); break; case ‘-’: s.push(s.pop() - lastval); break; case ‘*’: s.push(s.pop() * lastval); break; case ‘/’: s.push(s.pop() / lastval); break; default: cout << “\nUnknown oper”; exit(1); Chapter 10482 11 3087 CH10 11/29/01 2:21 PM Page 482} //end switch } //end else, in all other cases s.push(ch); //put current op on stack } //end else, not first operator } //end else if, it’s an operator else //not a known character { cout << “\nUnknown input character”; exit(1); } } //end for } //end parse() //-------------------------------------------------------------- int express::solve() //remove items from stack { char lastval; //previous value while(s.gettop() > 1) { lastval = s.pop(); //get previous value switch( s.pop() ) //get previous operator { //do operation, push answer case ‘+’: s.push(s.pop() + lastval); break; case ‘-’: s.push(s.pop() - lastval); break; case ‘*’: s.push(s.pop() * lastval); break; case ‘/’: s.push(s.pop() / lastval); break; default: cout << “\nUnknown operator”; exit(1); } //end switch } //end while return int( s.pop() ); //last item on stack is ans } //end solve() //////////////////////////////////////////////////////////////// int main() { char ans; //’y’ or ‘n’ char string[LEN]; //input string from user cout << “\nEnter an arithmetic expression” “\nof the form 2+3*4/3-2.” “\nNo number may have more than one digit.” “\nDon’t use any spaces or parentheses.”; do { cout << “\nEnter expresssion: “; cin >> string; //input from user express* eptr = new express(string); //make expression eptr->parse(); //parse it cout << “\nThe numerical value is: “ << eptr->solve(); //solve it delete eptr; //delete expression Pointers 10 P OINTERS 483 11 3087 CH10 11/29/01 2:21 PM Page 483cout << “\nDo another (Enter y or n)? “; cin >> ans; } while(ans == ‘y’); return 0; } This is a longish program, but it shows how a previously designed class, Stack, can come in handy in a new situation; it demonstrates the use of pointers in a variety of ways; and it shows how useful it can be to treat a string as an array of characters. Simulation: A Horse Race As our final example in this chapter we’ll show a horse-racing game. In this game a number of horses appear on the screen, and, starting from the left, race to a finish line on the right. This program will demonstrate pointers in a new situation, and also a little bit about object-oriented design. Each horse’s speed is determined randomly, so there is no way to figure out in advance which one will win. The program uses console graphics, so the horses are easily, although somewhat crudely, displayed. You’ll need to compile the program with the MSOFTCON.H or BORLACON.H header file (depending on your compiler), and the MSOFTCON.CPP or BORLACON.CPP source file. (See Appendix E, “Console Graphics Lite,” for more information.) When our program, HORSE, is started, it asks the user to supply the race’s distance and the number of horses that will run in it. The classic unit of distance for horse racing (at least in English-speaking countries) is the furlong, which is 1/8 of a mile. Typical races are 6, 8, 10, or 12 furlongs. You can enter from 1 to 7 horses. The program draws vertical lines corresponding to each furlong, along with start and finish lines. Each horse is represented by a rectangle with a number in the middle. Figure 10.19 shows the screen with a race in progress. Chapter 10484 FIGURE 10.19 Output of the HORSE program. 11 3087 CH10 11/29/01 2:21 PM Page 484Designing the Horse Race How do we approach an OOP design for our horse race? Our first question might be, is there a group of similar entities that we’re trying to model? The answer is yes, the horses. So it seems reasonable to make each horse an object. There will be a class called horse, which will contain data specific to each horse, such as its number and the distance it has run so far (which is used to display the horse in the correct screen position). However, there is also data that applies to the entire race track, rather than to individual horses. This includes the track length, the elapsed time in minutes and seconds (0:00 at the start of the race), and the total number of horses. It makes sense then to have a track object, which will be a single member of the track class. You can think of other real-world objects associated with horse racing, such as riders and saddles, but they aren’t relevant to this program. Are there other ways to design the program? For example, what about using inheritance to make the horses descendants of the track? This doesn’t make much sense, because the horses aren’t a “kind of” race track; they’re a completely different thing. Another option is to make the track data into static data of the horse class. However, it’s generally better to make each different kind of thing in the problem domain (the real world) a separate class in the program. One advantage of this is that it’s easier to use the classes in other contexts, such as using the track to race cars instead of horses. How will the horse objects and the track object communicate? (Or in UML terms, what will their association consist of?) An array of pointers to horse objects can be a member of the track class, so the track can access the horses through these pointers. The track will create the horses when it’s created. As it does so, it will pass a pointer to itself to each horse, so the horse can access the track. Here’s the listing for HORSE: // horse.cpp // models a horse race #include “msoftcon.h” //for console graphics #include //for I/O #include //for random() #include //for time() using namespace std; const int CPF = 5; //columns per furlong const int maxHorses = 7; //maximum number of horses class track; //for forward references //////////////////////////////////////////////////////////////// class horse { private: const track* ptrTrack; //pointer to track const int horse_number; //this horse’s number Pointers 10 P OINTERS 485 11 3087 CH10 11/29/01 2:21 PM Page 485float finish_time; //this horse’s finish time float distance_run; //distance run so far public: //create the horse horse(const int n, const track* ptrT) : horse_number(n), ptrTrack(ptrT), distance_run(0.0) //haven’t moved yet { } ~horse() //destroy the horse { /*empty*/ } //display the horse void display_horse(const float elapsed_time); }; //end class horse //////////////////////////////////////////////////////////////// class track { private: horse* hArray[maxHorses]; //array of ptrs-to-horses int total_horses; //total number of horses int horse_count; //horses created so far const float track_length; //track length in furlongs float elapsed_time; //time since start of race public: track(float lenT, int nH); //2-arg constructor ~track(); //destructor void display_track(); //display track void run(); //run the race float get_track_len() const; //return total track length }; //end class track //--------------------------------------------------------------- void horse::display_horse(float elapsed_time) //for each horse { //display horse & number set_cursor_pos( 1 + int(distance_run * CPF), 2 + horse_number*2 ); //horse 0 is blue set_color(static_cast(cBLUE+horse_number)); //draw horse char horse_char = ‘0’ + static_cast(horse_number); putch(‘ ‘); putch(‘\xDB’); putch(horse_char); putch(‘\xDB’); //until finish, if( distance_run < ptrTrack->get_track_len() + 1.0 / CPF ) { if( rand() % 3 ) //skip about 1 of 3 ticks distance_run += 0.2F; //advance 0.2 furlongs finish_time = elapsed_time; //update finish time } Chapter 10486 11 3087 CH10 11/29/01 2:21 PM Page 486else { //display finish time int mins = int(finish_time)/60; int secs = int(finish_time) - mins*60; cout << “ Time=” << mins << “:” << secs; } } //end display_horse() //--------------------------------------------------------------- track::track(float lenT, int nH) : //track constructor track_length(lenT), total_horses(nH), horse_count(0), elapsed_time(0.0) { init_graphics(); //start graphics total_horses = //not more than 7 horses (total_horses > maxHorses) ? maxHorses : total_horses; for(int j=0; j(time(&aTime)) ); display_track(); } //end track constructor //--------------------------------------------------------------- track::~track() //track destructor { for(int j=0; jdisplay_horse(elapsed_time); wait(500); } getch(); //eat the keystroke cout << endl; } //--------------------------------------------------------------- float track::get_track_len() const { return track_length; } ///////////////////////////////////////////////////////////////// int main() { float length; int total; //get data from user cout << “\nEnter track length (furlongs; 1 to 12): “; cin >> length; cout << “\nEnter number of horses (1 to 7): “; cin >> total; track theTrack(length, total); //create the track theTrack.run(); //run the race return 0; } //end main() Keeping Time Simulation programs usually involve an activity taking place over a period of time. To model the passage of time, such programs typically energize themselves at fixed intervals. In the HORSE program, the main() program calls the track’s run() function. This function makes a series of calls within a while loop, one for each horse, to a function display_horse(). This function redraws each horse in its new position. The while loop then pauses 500 milliseconds, using the console graphics wait() function. Then it does the same thing again, until the race is over or the user presses a key. Deleting an Array of Pointers to Objects At the end of the program the destructor for the track must delete the horse objects, which it obtained with new in its constructor. Notice that we can’t just say delete[] hArray; //deletes pointers, but not horses Chapter 10488 11 3087 CH10 11/29/01 2:21 PM Page 488This deletes the array of pointers, but not what the pointers point to. Instead we must go through the array element by element, and delete each horse individually: for(int j=0; j = track length] after (500 ms) [distance < track length] created deleted Finished 11 3087 CH10 11/29/01 2:21 PM Page 491The event that triggers the other two transitions is the timing out of a 500 millisecond timer. The keyword after is used to name these transitions, with the time as a parameter. Transitions can also be labeled with what the UML calls a guard:a condition that must be sat- isfied if the transition is to occur. Guards are written in brackets. The two after() transitions have guards as well as event names. Because the events are the same, the guards determine which transition will occur. Note that one of these transitions is a self transition: it returns to the same state where it began. Racing from State to State Each time it enters the Running state, the horse object carries out an activity that consists of increasing the distance it has run by 0.2 furlongs. As long as it has not yet reached the finish line, the [distance < track length] guard is true and the Running state transitions back to itself. When the horse reaches the finish line, [distance >= track length] becomes true, and the horse transitions to the Finished state, where it displays its total time for the race. It then waits to be deleted. We’ve shown enough to give you an idea what state diagrams do. There is of course much more to learn about them. We’ll see an example of a more complex state diagram that describes an elevator object in Chapter 13, “Multifile Programs.” Debugging Pointers Pointers can be the source of mysterious and catastrophic program bugs. The most common problem is that the programmer has failed to place a valid address in a pointer variable. When this happens the pointer can end up pointing anywhere in memory. It could be pointing to the program code, or to the operating system. If the programmer then inserts a value into memory using the pointer, the value will write over the program or operating instructions, and the com- puter will crash or evince other uncharming behavior. A particular version of this scenario takes place when the pointer points to address 0, which is called NULL. This happens, for example, if the pointer variable is defined as a global variable, since global variables are automatically initialized to 0. Here’s a miniprogram that demon- strates the situation: int* intptr; //global variable, initialized to 0 void main() { //failure to put valid address in intptr *intptr = 37; //attempts to put 37 in address at 0 } //result is error Chapter 10492 11 3087 CH10 11/29/01 2:21 PM Page 492When intptr is defined, it is given the value 0, since it is global. The single program state- ment will attempt to insert the value 37 into the address at 0. Fortunately, however, the runtime error-checking unit built into the program by the compiler is waiting for attempts to access address 0, and will display an error message (perhaps an access violation, null pointer assignment,orpage fault) and terminate the program. If you see such a message, one possibility is that you have failed to properly initialize a pointer. Summary This has been a whirlwind tour through the land of pointers. There is far more to learn, but the topics we’ve covered here will provide a basis for the examples in the balance of the book and for further study of pointers. We’ve learned that everything in the computer’s memory has an address, and that addresses are pointer constants. We can find the addresses of variables using the address-of operator &. Pointers are variables that hold address values. Pointers are defined using an asterisk (*) to mean pointer to. A data type is always included in pointer definitions (except void*), since the compiler must know what is being pointed to, so that it can perform arithmetic correctly on the pointer. We access the thing pointed to using the asterisk in a different way, as the dereference operator, meaning contents of the variable pointed to by. The special type void* means a pointer to any type. It’s used in certain difficult situations where the same pointer must hold addresses of different types. Array elements can be accessed using array notation with brackets or pointer notation with an asterisk. Like other addresses, the address of an array is a constant, but it can be assigned to a variable, which can be incremented and changed in other ways. When the address of a variable is passed to a function, the function can work with the original variable. (This is not true when arguments are passed by value.) In this respect passing by pointer offers the same benefits as passing by reference, although pointer arguments must be dereferenced or accessed using the dereference operator. However, pointers offer more flexibil- ity in some cases. A string constant can be defined as an array or as a pointer. The pointer approach may be more flexible, but there is a danger that the pointer value will be corrupted. Strings, being arrays of type char,are commonly passed to functions and accessed using pointers. The new operator obtains a specified amount of memory from the system and returns a pointer to the memory. This operator is used to create variables and data structures during program execution. The delete operator releases memory obtained with new. Pointers 10 P OINTERS 493 11 3087 CH10 11/29/01 2:21 PM Page 493When a pointer points to an object, members of the object’s class can be accessed using the access operator ->. The same syntax is used to access structure members. Classes and structures may contain data members that are pointers to their own type. This per- mits the creation of complex data structures such as linked lists. There can be pointers to pointers. These variables are defined using the double asterisk; for example, int** pptr. Multiplicity in UML class diagrams shows the number of objects involved in an association. UML state diagrams show how a particular object’s situation changes over time. States are rep- resented by rectangles with rounded corners, and transitions between states are represented by directed lines. Questions Answers to these questions can be found in Appendix G. 1. Write a statement that displays the address of the variable testvar. 2. The contents of two pointers that point to adjacent variables of type float differ by _____________. 3. A pointer is a. the address of a variable. b. an indication of the variable to be accessed next. c. a variable for storing addresses. d. the data type of an address variable. 4. Write expressions for the following: a. The address of var b. The contents of the variable pointed to by var c. The variable var used as a reference argument d. The data type pointer-to-char 5. An address is a _____________, while a pointer is a ____________. 6. Write a definition for a variable of type pointer-to-float. 7. Pointers are useful for referring to a memory address that has no _______________. 8. If a pointer testptr points to a variable testvar,write a statement that represents the contents of testvar but does not use its name. Chapter 10494 11 3087 CH10 11/29/01 2:21 PM Page 4949. An asterisk placed after a data type means _________. An asterisk placed in front of a variable name means __________. 10. The expression *test can be said to a. be a pointer to test. b. refer to the contents of test. c. dereference test. d. refer to the value of the variable pointed to by test. 11. Is the following code correct? int intvar = 333; int* intptr; cout << *intptr; 12. A pointer to void can hold pointers to _________. 13. What is the difference between intarr[3] and *(intarr+3)? 14. Write some code that uses pointer notation to display every value in the array intarr, which has 77 elements. 15. If intarr is an array of integers, why is the expression intarr++ not legal? 16. Of the three ways to pass arguments to functions, only passing by __________ and pass- ing by __________ allow the function to modify the argument in the calling program. 17. The type of variable a pointer points to must be part of the pointer’s definition so that a. data types don’t get mixed up when arithmetic is performed on them. b. pointers can be added to one another to access structure members. c. no one’s religious conviction will be attacked. d. the compiler can perform arithmetic correctly to access array elements. 18. Using pointer notation, write a prototype (declaration) for a function called func() that returns type void and takes a single argument that is an array of type char. 19. Using pointer notation, write some code that will transfer 80 characters from the string s1 to the string s2. 20. The first element in a string is a. the name of the string. b. the first character in the string. c. the length of the string. d. the name of the array holding the string. 21. Using pointer notation, write the prototype for a function called revstr() that returns a string value and takes one argument that represents a string. Pointers 10 P OINTERS 495 11 3087 CH10 11/29/01 2:21 PM Page 49522. Write a definition for an array numptrs of pointers to the strings One, Two, and Three. 23. The new operator a. returns a pointer to a variable. b. creates a variable called new. c. obtains memory for a new variable. d. tells how much memory is available. 24. Using new may result in less __________ memory than using an array. 25. The delete operator returns ____________ to the operating system. 26. Given a pointer p that points to an object of type upperclass,write an expression that executes the exclu() member function in this object. 27. Given an object with index number 7 in array objarr,write an expression that executes the exclu() member function in this object. 28. In a linked list a. each link contains a pointer to the next link. b. an array of pointers points to the links. c. each link contains data or a pointer to data. d. the links are stored in an array. 29. Write a definition for an array arr of 8 pointers that point to variables of type float. 30. If you wanted to sort many large objects or structures, it would be most efficient to a. place them in an array and sort the array. b. place pointers to them in an array and sort the array. c. place them in a linked list and sort the linked list. d. place references to them in an array and sort the array. 31. Express the multiplicities of an association that has fewer than 10 objects at one end and more than 2 objects at the other. 32. The states in a state diagram correspond to a. messages between objects. b. circumstances in which an object finds itself. c. objects in the program. d. changes in an object’s situation. Chapter 10496 11 3087 CH10 11/29/01 2:21 PM Page 49633. True or false: a transition between states exists for the duration of the program. 34. A guard in a state diagram is a. a constraint on when a transition can occur. b. a name for certain kinds of transitions. c. a name for certain kinds of states. d. a restriction on the creation of certain states. Exercises Answers to starred exercises can be found in Appendix G. *1. Write a program that reads a group of numbers from the user and places them in an array of type float. Once the numbers are stored in the array, the program should average them and print the result. Use pointer notation wherever possible. *2. Start with the String class from the NEWSTR example in this chapter. Add a member function called upit() that converts the string to all uppercase. You can use the toupper() library function, which takes a single character as an argument and returns a character that has been converted (if necessary) to uppercase. This function uses the CCTYPE header file. Write some code in main() to test upit(). *3. Start with an array of pointers to strings representing the days of the week, as found in the PTRTOSTR program in this chapter. Provide functions to sort the strings into alphabeti- cal order, using variations of the bsort() and order() functions from the PTRSORT pro- gram in this chapter. Sort the pointers to the strings, not the actual strings. *4. Add a destructor to the LINKLIST program. It should delete all the links when a linklist object is destroyed. It can do this by following along the chain, deleting each link as it goes. You can test the destructor by having it display a message each time it deletes a link; it should delete the same number of links that were added to the list. (A destructor is called automatically by the system for any existing objects when the program exits.) 5. Suppose you have a main() with three local arrays, all the same size and type (say float). The first two are already initialized to values. Write a function called addarrays() that accepts the addresses of the three arrays as arguments; adds the con- tents of the first two arrays together, element by element; and places the results in the third array before returning. A fourth argument to this function can carry the size of the arrays. Use pointer notation throughout; the only place you need brackets is in defining the arrays. Pointers 10 P OINTERS 497 11 3087 CH10 11/29/01 2:21 PM Page 4976. Make your own version of the library function strcmp(s1, s2), which compares two strings and returns –1 if s1 comes first alphabetically, 0 if s1 and s2 are the same, and 1 if s2 comes first alphabetically. Call your function compstr(). It should take two char* strings as arguments, compare them character by character, and return an int. Write a main() program to test the function with different combinations of strings. Use pointer notation throughout. 7. Modify the person class in the PERSORT program in this chapter so that it includes not only a name, but also a salary item of type float representing the person’s salary. You’ll need to change the setName() and printName() member functions to setData() and printData(), and include in them the ability to set and display the salary as well as the name. You’ll also need a getSalary() function. Using pointer notation, write a salsort() function that sorts the pointers in the persPtr array by salary rather than by name. Try doing all the sorting in salsort(),rather than calling another function as PERSORT does. If you do this, don’t forget that -> takes precedence over *,so you’ll need to say if( (*(pp+j))->getSalary() > (*(pp+k))->getSalary() ) { /* swap the pointers */ } 8. Revise the additem() member function from the LINKLIST program so that it adds the item at the end of the list, rather than the beginning. This will cause the first item inserted to be the first item displayed, so the output of the program will be 25 36 49 64 To add the item, you’ll need to follow the chain of pointers to the end of the list, then change the last link to point to the new link. 9. Let’s say that you need to store 100 integers so that they’re easily accessible. However, let’s further assume that there’s a problem: The memory in your computer is so frag- mented that the largest array that you can use holds only 10 integers. (Such problems actually arise, although usually with larger memory objects.) You can solve this problem by defining 10 separate int arrays of 10 integers each, and an array of 10 pointers to these arrays. The int arrays can have names like a0, a1, a2, and so on. The address of each of these arrays can be stored in the pointer array of type int*, which can have a name like ap (for array of pointers). You can then access individual integers using expressions like ap[j][k], where j steps through the pointers in ap and k steps through individual integers in each array. This looks as if you’re accessing a two-dimensional array, but it’s really a group of one-dimensional arrays. Fill such a group of arrays with test data (say the numbers 0, 10, 20, and so on up to 990). Then display the data to make sure it’s correct. Chapter 10498 11 3087 CH10 11/29/01 2:21 PM Page 49810. As presented, Exercise 9 is rather inelegant because each of the 10 int arrays is declared in a different program statement, using a different name. Each of their addresses must also be obtained using a separate statement. You can simplify things by using new, which allows you to allocate the arrays in a loop and assign pointers to them at the same time: for(j=0; jdraw(); This is an amazing capability: Completely different functions are executed by the same func- tion call. If the pointer in ptrarr points to a ball, the function that draws a ball is called; if it points to a triangle, the triangle-drawing function is called. This is called polymorphism, which means different forms. The functions have the same appearance, the draw() expression, but dif- ferent actual functions are called, depending on the contents of ptrarr[j]. Polymorphism is one of the key features of object-oriented programming, after classes and inheritance. For the polymorphic approach to work, several conditions must be met. First, all the different classes of shapes, such as balls and triangles, must be descended from a single base class (called shape in MULTSHAP). Second, the draw() function must be declared to be virtual in the base class. 12 3087 CH11 11/29/01 2:19 PM Page 504This is all rather abstract, so let’s start with some short programs that show parts of the situa- tion, and put everything together later. Normal Member Functions Accessed with Pointers Our first example shows what happens when a base class and derived classes all have functions with the same name, and you access these functions using pointers but without using virtual functions. Here’s the listing for NOTVIRT: // notvirt.cpp // normal functions accessed from pointer #include using namespace std; //////////////////////////////////////////////////////////////// class Base //base class { public: void show() //normal function { cout << “Base\n”; } }; //////////////////////////////////////////////////////////////// class Derv1 : public Base //derived class 1 { public: void show() { cout << “Derv1\n”; } }; //////////////////////////////////////////////////////////////// class Derv2 : public Base //derived class 2 { public: void show() { cout << “Derv2\n”; } }; //////////////////////////////////////////////////////////////// int main() { Derv1 dv1; //object of derived class 1 Derv2 dv2; //object of derived class 2 Base* ptr; //pointer to base class ptr = &dv1; //put address of dv1 in pointer ptr->show(); //execute show() Virtual Functions 11 V IRTUAL F UNCTIONS 505 12 3087 CH11 11/29/01 2:19 PM Page 505ptr = &dv2; //put address of dv2 in pointer ptr->show(); //execute show() return 0; } The Derv1 and Derv2 classes are derived from class Base. Each of these three classes has a member function show(). In main() we create objects of class Derv1 and Derv2, and a pointer to class Base. Then we put the address of a derived class object in the base class pointer in the line ptr = &dv1; // derived class address in base class pointer But wait—how can we get away with this? Doesn’t the compiler complain that we’re assigning an address of one type (Derv1) to a pointer of another (Base)? On the contrary, the compiler is perfectly happy, because type checking has been relaxed in this situation, for reasons that will become apparent soon. The rule is that pointers to objects of a derived class are type- compatible with pointers to objects of the base class. Now the question is, when you execute the line ptr->show(); what function is called? Is it Base::show() or Derv1::show()? Again, in the last two lines of NOTVIRT we put the address of an object of class Derv2 in the pointer, and again execute ptr->show(); Which of the show() functions is called here? The output from the program answers these questions: Base Base As you can see, the function in the base class is always executed. The compiler ignores the contents of the pointer ptr and chooses the member function that matches the type of the pointer, as shown in Figure 11.1. Sometimes this is what we want, but it doesn’t solve the problem posed at the beginning of this section: accessing objects of different classes using the same statement. Chapter 11506 12 3087 CH11 11/29/01 2:19 PM Page 506FIGURE 11.1 Nonvirtual pointer access. Virtual Member Functions Accessed with Pointers Let’s make a single change in our program: We’ll place the keyword virtual in front of the declarator for the show() function in the base class. Here’s the listing for the resulting pro- gram, VIRT: // virt.cpp // virtual functions accessed from pointer #include using namespace std; //////////////////////////////////////////////////////////////// class Base //base class { public: virtual void show() //virtual function { cout << “Base\n”; } }; //////////////////////////////////////////////////////////////// class Derv1 : public Base //derived class 1 { Virtual Functions 11 V IRTUAL F UNCTIONS 507 12 3087 CH11 11/29/01 2:19 PM Page 507public: void show() { cout << “Derv1\n”; } }; //////////////////////////////////////////////////////////////// class Derv2 : public Base //derived class 2 { public: void show() { cout << “Derv2\n”; } }; //////////////////////////////////////////////////////////////// int main() { Derv1 dv1; //object of derived class 1 Derv2 dv2; //object of derived class 2 Base* ptr; //pointer to base class ptr = &dv1; //put address of dv1 in pointer ptr->show(); //execute show() ptr = &dv2; //put address of dv2 in pointer ptr->show(); //execute show() return 0; } The output of this program is Derv1 Derv2 Now, as you can see, the member functions of the derived classes, not the base class, are exe- cuted. We change the contents of ptr from the address of Derv1 to that of Derv2, and the par- ticular instance of show() that is executed also changes. So the same function call ptr->show(); executes different functions, depending on the contents of ptr. The rule is that the compiler selects the function based on the contents of the pointer ptr, not on the type of the pointer, as in NOTVIRT. This is shown in Figure 11.2. Chapter 11508 12 3087 CH11 11/29/01 2:19 PM Page 508FIGURE 11.2 Virtual pointer access. Late Binding The astute reader may wonder how the compiler knows what function to compile. In NOTVIRT the compiler has no problem with the expression ptr->show(); It always compiles a call to the show() function in the base class. But in VIRT the compiler doesn’t know what class the contents of ptr may contain. It could be the address of an object of the Derv1 class or of the Derv2 class. Which version of draw() does the compiler call? In fact the compiler doesn’t know what to do, so it arranges for the decision to be deferred until the program is running. At runtime, when it is known what class is pointed to by ptr,the appropriate version of draw will be called. This is called late binding or dynamic binding. (Choosing functions in the normal way, during compilation, is called early binding or static binding.) Late binding requires some overhead but provides increased power and flexibility. We’ll put these ideas to use in a moment, but first let’s consider a refinement to the idea of virtual functions. Virtual Functions 11 V IRTUAL F UNCTIONS 509 12 3087 CH11 11/29/01 2:19 PM Page 509Abstract Classes and Pure Virtual Functions Think of the shape class in the multshap program in Chapter 9. We’ll never make an object of the shape class; we’ll only make specific shapes such as circles and triangles. When we will never want to instantiate objects of a base class, we call it an abstract class. Such a class exists only to act as a parent of derived classes that will be used to instantiate objects. It may also provide an interface for the class hierarchy. How can we make it clear to someone using our family of classes that we don’t want anyone to instantiate objects of the base class? We could just say this in the documentation, and count on the users of the class to remember it, but of course it’s much better to write our classes so that such instantiation is impossible. How can we can do that? By placing at least one pure virtual function in the base class. A pure virtual function is one with the expression =0 added to the declaration. This is shown in the VIRTPURE example. // virtpure.cpp // pure virtual function #include using namespace std; //////////////////////////////////////////////////////////////// class Base //base class { public: virtual void show() = 0; //pure virtual function }; //////////////////////////////////////////////////////////////// class Derv1 : public Base //derived class 1 { public: void show() { cout << “Derv1\n”; } }; //////////////////////////////////////////////////////////////// class Derv2 : public Base //derived class 2 { public: void show() { cout << “Derv2\n”; } }; //////////////////////////////////////////////////////////////// int main() { // Base bad; //can’t make object from abstract class Base* arr[2]; //array of pointers to base class Derv1 dv1; //object of derived class 1 Chapter 11510 12 3087 CH11 11/29/01 2:19 PM Page 510Derv2 dv2; //object of derived class 2 arr[0] = &dv1; //put address of dv1 in array arr[1] = &dv2; //put address of dv2 in array arr[0]->show(); //execute show() in both objects arr[1]->show(); return 0; } Here the virtual function show() is declared as virtual void show() = 0; // pure virtual function The equal sign here has nothing to do with assignment; the value 0 is not assigned to anything. The =0 syntax is simply how we tell the compiler that a virtual function will be pure. Now if in main() you attempt to create objects of class Base,the compiler will complain that you’re try- ing to instantiate an object of an abstract class. It will also tell you the name of the pure virtual function that makes it an abstract class. Notice that, although this is only a declaration, you never need to write a definition of the base class show(),although you can if you need to. Once you’ve placed a pure virtual function in the base class, you must override it in all the derived classes from which you want to instantiate objects. If a class doesn’t override the pure virtual function, it becomes an abstract class itself, and you can’t instantiate objects from it (although you might from classes derived from it). For consistency, you may want to make all the virtual functions in the base class pure. As you can see, we’ve made another, unrelated, change in VIRTPURE:The addresses of the member functions are stored in an array of pointers and accessed using array elements. This works in just the same way as using a single pointer. The output of VIRTPURE is the same as VIRT: Derv1 Derv2 Virtual Functions and the person Class Now that we understand some of the mechanics of virtual functions, let’s look at a situation where it makes sense to use them. Our example is an extension of the PTROBJ and PERSORT examples from Chapter 10, “Pointers.” It uses the same person class, but adds two derived classes, student and professor. These derived classes each contain a function called isOutstanding(). This function makes it easy for the school administrators to create a list of outstanding students and professors for the venerable Awards Day ceremony. Here’s the listing for VIRTPERS: Virtual Functions 11 V IRTUAL F UNCTIONS 511 12 3087 CH11 11/29/01 2:19 PM Page 511// virtpers.cpp // virtual functions with person class #include using namespace std; //////////////////////////////////////////////////////////////// class person //person class { protected: char name[40]; public: void getName() { cout << “ Enter name: “; cin >> name; } void putName() { cout << “Name is: “ << name << endl; } virtual void getData() = 0; //pure virtual func virtual bool isOutstanding() = 0; //pure virtual func }; //////////////////////////////////////////////////////////////// class student : public person //student class { private: float gpa; //grade point average public: void getData() //get student data from user { person::getName(); cout << “ Enter student’s GPA: “; cin >> gpa; } bool isOutstanding() { return (gpa > 3.5) ? true : false; } }; //////////////////////////////////////////////////////////////// class professor : public person //professor class { private: int numPubs; //number of papers published public: void getData() //get professor data from user { person::getName(); cout << “ Enter number of professor’s publications: “; cin >> numPubs; } bool isOutstanding() { return (numPubs > 100) ? true : false; } }; Chapter 11512 12 3087 CH11 11/29/01 2:19 PM Page 512//////////////////////////////////////////////////////////////// int main() { person* persPtr[100]; //array of pointers to persons int n = 0; //number of persons on list char choice; do { cout << “Enter student or professor (s/p): “; cin >> choice; if(choice==’s’) //put new student persPtr[n] = new student; // in array else //put new professor persPtr[n] = new professor; // in array persPtr[n++]->getData(); //get data for person cout << “ Enter another (y/n)? “; //do another person? cin >> choice; } while( choice==’y’ ); //cycle until not ‘y’ for(int j=0; jputName(); //say if outstanding if( persPtr[j]->isOutstanding() ) cout << “ This person is outstanding\n”; } return 0; } //end main() The Classes The person class is an abstract class because it contains the pure virtual functions getData() and isOutstanding(). No person objects can ever be created. This class exists only to be the base class for the student and professor classes. The student and professor classes add new data items to the base class. The student class contains a variable gpa of type float, which represents the student’s grade point average (GPA). The professor class contains a variable numPubs, of type int, which represents the number of scholarly publications the pro- fessor has published. Students with a GPA of over 3.5 and professors who have published more than 100 papers are considered outstanding. (We’ll refrain from comment on the desir- ability of these criteria for judging educational excellence.) The isOutstanding() Function The isOutstanding() function is declared as a pure virtual function in person. In the student class this function returns a bool true if the student’s GPA is greater than 3.5, and false oth- erwise. In professor it returns true if the professor’s numPubs variable is greater than 100. Virtual Functions 11 V IRTUAL F UNCTIONS 513 12 3087 CH11 11/29/01 2:19 PM Page 513The getData() function asks the user for the GPA for a student,but for the number of publi- cations for a professor. The main() Program In main() we first let the user enter a number of student and teacher names. For students, the program also asks for the GPA, and for professors it asks for the number of publications. When the user is finished, the program prints out the names of all the students and professors, noting those who are outstanding. Here’s some sample interaction: Enter student or professor (s/p): s Enter name: Timmy Enter student’s GPA: 1.2 Enter another (y/n)? y Enter student or professor (s/p): s Enter name: Brenda Enter student’s GPA: 3.9 Enter another (y/n)? y Enter student or professor (s/p): s Enter name: Sandy Enter student’s GPA: 2.4 Enter another (y/n)? y Enter student or professor (s/p): p Enter name: Shipley Enter number of professor’s publications: 714 Enter another (y/n)? y Enter student or professor (s/p): p Enter name: Wainright Enter number of professor’s publications: 13 Enter another (y/n)? n Name is: Timmy Name is: Brenda This person is outstanding Name is: Sandy Name is: Shipley This person is outstanding Name is: Wainright Virtual Functions in a Graphics Example Let’s try another example of virtual functions. This one is a graphics example derived from the MULTSHAP program in Chapter 9, “Inheritance.” As we noted at the beginning of this section, you may want to draw a number of shapes using the same statement. The VIRTSHAP program does this. Remember that you must build this program with the appropriate console graphics file, as described in Appendix E, “Console Graphics Lite.” Chapter 11514 12 3087 CH11 11/29/01 2:19 PM Page 514// virtshap.cpp // virtual functions with shapes #include using namespace std; #include “msoftcon.h” //for graphics functions //////////////////////////////////////////////////////////////// class shape //base class { protected: int xCo, yCo; //coordinates of center color fillcolor; //color fstyle fillstyle; //fill pattern public: //no-arg constructor shape() : xCo(0), yCo(0), fillcolor(cWHITE), fillstyle(SOLID_FILL) { } //4-arg constructor shape(int x, int y, color fc, fstyle fs) : xCo(x), yCo(y), fillcolor(fc), fillstyle(fs) { } virtual void draw()=0 //pure virtual draw function { set_color(fillcolor); set_fill_style(fillstyle); } }; //////////////////////////////////////////////////////////////// class ball : public shape { private: int radius; //(xCo, yCo) is center public: ball() : shape() //no-arg constr { } //5-arg constructor ball(int x, int y, int r, color fc, fstyle fs) : shape(x, y, fc, fs), radius(r) { } void draw() //draw the ball { shape::draw(); draw_circle(xCo, yCo, radius); } }; //////////////////////////////////////////////////////////////// class rect : public shape { Virtual Functions 11 V IRTUAL F UNCTIONS 515 12 3087 CH11 11/29/01 2:19 PM Page 515private: int width, height; //(xCo, yCo) is upper left corner public: rect() : shape(), height(0), width(0) //no-arg ctor { } //6-arg ctor rect(int x, int y, int h, int w, color fc, fstyle fs) : shape(x, y, fc, fs), height(h), width(w) { } void draw() //draw the rectangle { shape::draw(); draw_rectangle(xCo, yCo, xCo+width, yCo+height); set_color(cWHITE); //draw diagonal draw_line(xCo, yCo, xCo+width, yCo+height); } }; //////////////////////////////////////////////////////////////// class tria : public shape { private: int height; //(xCo, yCo) is tip of pyramid public: tria() : shape(), height(0) //no-arg constructor { } //5-arg constructor tria(int x, int y, int h, color fc, fstyle fs) : shape(x, y, fc, fs), height(h) { } void draw() //draw the triangle { shape::draw(); draw_pyramid(xCo, yCo, height); } }; //////////////////////////////////////////////////////////////// int main() { int j; init_graphics(); //initialize graphics system shape* pShapes[3]; //array of pointers to shapes //define three shapes pShapes[0] = new ball(40, 12, 5, cBLUE, X_FILL); pShapes[1] = new rect(12, 7, 10, 15, cRED, SOLID_FILL); pShapes[2] = new tria(60, 7, 11, cGREEN, MEDIUM_FILL); Chapter 11516 12 3087 CH11 11/29/01 2:19 PM Page 516for(j=0; j<3; j++) //draw all shapes pShapes[j]->draw(); for(j=0; j<3; j++) //delete all shapes delete pShapes[j]; set_cursor_pos(1, 25); return 0; } The class specifiers in VIRTSHAP are similar to those in MULTSHAP,except that the draw() func- tion in the shape class has been made into a pure virtual function. In main(), we set up an array, ptrarr,of pointers to shapes. Next we create three objects, one of each class, and place their addresses in an array. Now it’s easy to draw all three shapes. The statement ptrarr[j]->draw(); does this as the loop variable j changes. This is a powerful approach to combining graphics elements, especially when a large number of objects need to be grouped together and drawn as a unit. Virtual Destructors Base class destructors should always be virtual. Suppose you use delete with a base class pointer to a derived class object to destroy the derived-class object. If the base-class destructor is not virtual then delete, like a normal member function, calls the destructor for the base class, not the destructor for the derived class. This will cause only the base part of the object to be destroyed. The VIRTDEST program shows how this looks. //vertdest.cpp //tests non-virtual and virtual destructors #include using namespace std; //////////////////////////////////////////////////////////////// class Base { public: ~Base() //non-virtual destructor // virtual ~Base() //virtual destructor { cout << “Base destroyed\n”; } }; //////////////////////////////////////////////////////////////// class Derv : public Base { Virtual Functions 11 V IRTUAL F UNCTIONS 517 12 3087 CH11 11/29/01 2:19 PM Page 517public: ~Derv() { cout << “Derv destroyed\n”; } }; //////////////////////////////////////////////////////////////// int main() { Base* pBase = new Derv; delete pBase; return 0; } The output for this program as written is Base destroyed This shows that the destructor for the Derv part of the object isn’t called. In the listing the base class destructor is not virtual, but you can make it so by commenting out the first definition for the destructor and substituting the second. Now the output is Derv destroyed Base destroyed Now both parts of the derived class object are destroyed properly. Of course, if none of the destructors has anything important to do (like deleting memory obtained with new) then virtual destructors aren’t important. But in general, to ensure that derived-class objects are destroyed properly, you should make destructors in all base classes virtual. Most class libraries have a base class that includes a virtual destructor, which ensures that all derived classes have virtual destructors. Virtual Base Classes Before leaving the subject of virtual programming elements, we should mention virtual base classes as they relate to multiple inheritance. Consider the situation shown in Figure 11.3, with a base class, Parent; two derived classes, Child1 and Child2; and a fourth class, Grandchild, derived from both Child1 and Child2. In this arrangement a problem can arise if a member function in the Grandchild class wants to access data or functions in the Parent class. The NORMBASE program shows what happens. Chapter 11518 12 3087 CH11 11/29/01 2:19 PM Page 518FIGURE 11.3 Virtual base classes. // normbase.cpp // ambiguous reference to base class class Parent { protected: int basedata; }; class Child1 : public Parent { }; class Child2 : public Parent { }; class Grandchild : public Child1, public Child2 { public: int getdata() { return basedata; } // ERROR: ambiguous }; A compiler error occurs when the getdata() member function in Grandchild attempts to access basedata in Parent. Why? When the Child1 and Child2 classes are derived from Parent, each inherits a copy of Parent; this copy is called a subobject. Each of the two subob- jects contains its own copy of Parent’s data, including basedata. Now, when Grandchild refers to basedata, which of the two copies will it access? The situation is ambiguous, and that’s what the compiler reports. Virtual Functions 11 V IRTUAL F UNCTIONS 519 Parent Child2Child1 Grandchild 12 3087 CH11 11/29/01 2:19 PM Page 519To eliminate the ambiguity, we make Child1 and Child2 into virtual base classes, as shown by the example VIRTBASE. // virtbase.cpp // virtual base classes class Parent { protected: int basedata; }; class Child1 : virtual public Parent // shares copy of Parent { }; class Child2 : virtual public Parent // shares copy of Parent { }; class Grandchild : public Child1, public Child2 { public: int getdata() { return basedata; } // OK: only one copy of Parent }; The use of the keyword virtual in these two classes causes them to share a single common subobject of their base class Parent. Since there is only one copy of basedata,there is no ambiguity when it is referred to in Grandchild. The need for virtual base classes may indicate a conceptual problem with your use of multiple inheritance, so they should be used with caution. Friend Functions The concepts of encapsulation and data hiding dictate that nonmember functions should not be able to access an object’s private or protected data. The policy is, if you’re not a member, you can’t get in. However, there are situations where such rigid discrimination leads to considerable inconvenience. Friends as Bridges Imagine that you want a function to operate on objects of two different classes. Perhaps the function will take objects of the two classes as arguments, and operate on their private data. In this situation there’s nothing like a friend function. Here’s a simple example, FRIEND,that shows how friend functions can act as a bridge between two classes: // friend.cpp // friend functions Chapter 11520 12 3087 CH11 11/29/01 2:19 PM Page 520#include using namespace std; //////////////////////////////////////////////////////////////// class beta; //needed for frifunc declaration class alpha { private: int data; public: alpha() : data(3) { } //no-arg constructor friend int frifunc(alpha, beta); //friend function }; //////////////////////////////////////////////////////////////// class beta { private: int data; public: beta() : data(7) { } //no-arg constructor friend int frifunc(alpha, beta); //friend function }; //////////////////////////////////////////////////////////////// int frifunc(alpha a, beta b) //function definition { return( a.data + b.data ); } //-------------------------------------------------------------- int main() { alpha aa; beta bb; cout << frifunc(aa, bb) << endl; //call the function return 0; } In this program, the two classes are alpha and beta. The constructors in these classes initialize their single data items to fixed values (3 in alpha and 7 in beta). We want the function frifunc() to have access to both of these private data members, so we make it a friend function. It’s declared with the friend keyword in both classes: friend int frifunc(alpha, beta); This declaration can be placed anywhere in the class; it doesn’t matter whether it goes in the public or the private section. Virtual Functions 11 V IRTUAL F UNCTIONS 521 12 3087 CH11 11/29/01 2:19 PM Page 521An object of each class is passed as an argument to the function frifunc(), and it accesses the private data member of both classes through these arguments. The function doesn’t do much: It adds the data items and returns the sum. The main() program calls this function and prints the result. A minor point: Remember that a class can’t be referred to until it has been declared. Class beta is referred to in the declaration of the function frifunc() in class alpha,so beta must be declared before alpha. Hence the declaration class beta; at the beginning of the program. Breaching the Walls We should note that friend functions are controversial. During the development of C++, argu- ments raged over the desirability of including this feature. On the one hand, it adds flexibility to the language; on the other, it is not in keeping with data hiding, the philosophy that only member functions can access a class’s private data. How serious is the breach of data integrity when friend functions are used? A friend function must be declared as such within the class whose data it will access. Thus a programmer who does not have access to the source code for the class cannot make a function into a friend. In this respect, the integrity of the class is still protected. Even so, friend functions are conceptu- ally messy, and potentially lead to a spaghetti-code situation if numerous friends muddy the clear boundaries between classes. For this reason friend functions should be used sparingly. If you find yourself using many friends, you may need to rethink the design of the program. English Distance Example However, sometimes friend functions are too convenient to avoid. Perhaps the most common example is when friends are used to increase the versatility of overloaded operators. The fol- lowing program shows a limitation in the use of such operators when friends are not used. This example is a variation on the ENGLPLUS and ENGLCONV programs in Chapter 8, “Operator Overloading.” It’s called NOFRI. // nofri.cpp // limitation to overloaded + operator #include using namespace std; //////////////////////////////////////////////////////////////// class Distance //English Distance class { private: int feet; float inches; Chapter 11522 12 3087 CH11 11/29/01 2:19 PM Page 522public: Distance() : feet(0), inches(0.0) //constructor (no args) { } //constructor (one arg) Distance(float fltfeet) //convert float to Distance { //feet is integer part feet = static_cast(fltfeet); inches = 12*(fltfeet-feet); //inches is what’s left } Distance(int ft, float in) //constructor (two args) { feet = ft; inches = in; } void showdist() //display distance { cout << feet << “\’-” << inches << ‘\”’; } Distance operator + (Distance); }; //-------------------------------------------------------------- //add this distance to d2 Distance Distance::operator + (Distance d2) //return the sum { int f = feet + d2.feet; //add the feet float i = inches + d2.inches; //add the inches if(i >= 12.0) //if total exceeds 12.0, { i -= 12.0; f++; } //less 12 inches, plus 1 foot return Distance(f,i); //return new Distance with sum } //////////////////////////////////////////////////////////////// int main() { Distance d1 = 2.5; //constructor converts Distance d2 = 1.25; //float feet to Distance Distance d3; cout << “\nd1 = “; d1.showdist(); cout << “\nd2 = “; d2.showdist(); d3 = d1 + 10.0; //distance + float: OK cout << “\nd3 = “; d3.showdist(); // d3 = 10.0 + d1; //float + Distance: ERROR // cout << “\nd3 = “; d3.showdist(); cout << endl; return 0; } In this program, the + operator is overloaded to add two objects of type Distance. Also, there is a one-argument constructor that converts a value of type float,representing feet and deci- mal fractions of feet, into a Distance value. (That is, it converts 10.25' into 10'–3''.) Virtual Functions 11 V IRTUAL F UNCTIONS 523 12 3087 CH11 11/29/01 2:19 PM Page 523When such a constructor exists, you can make statements like this in main(): d3 = d1 + 10.0; The overloaded + is looking for objects of type Distance both on its left and on its right, but if the argument on the right is type float,the compiler will use the one-argument constructor to convert this float to a Distance value, and then carry out the addition. Here is what appears to be a subtle variation on this statement: d3 = 10.0 + d1; Does this work? No, because the object of which the overloaded + operator is a member must be the variable to the left of the operator. When we place a variable of a different type there, or a constant, then the compiler uses the + operator that adds that type (float in this case), not the one that adds Distance objects. Unfortunately, this operator does not know how to convert float to Distance,so it can’t handle this situation. Here’s the output from NOFRI: d1 = 2’-6” d2 = 1’-3” d3 = 12’-6” The second addition won’t compile, so these statements are commented out. We could get around this problem by creating a new object of type Distance: d3 = Distance(10, 0) + d1; but this is nonintuitive and inelegant. How can we write natural-looking statements that have nonmember data types to the left of the operator? As you may have guessed, a friend can help you out of this dilemma. The FRENGL program shows how. // frengl.cpp // friend overloaded + operator #include using namespace std; //////////////////////////////////////////////////////////////// class Distance //English Distance class { private: int feet; float inches; public: Distance() //constructor (no args) { feet = 0; inches = 0.0; } Distance( float fltfeet ) //constructor (one arg) { //convert float to Distance feet = int(fltfeet); //feet is integer part inches = 12*(fltfeet-feet); //inches is what’s left } Chapter 11524 12 3087 CH11 11/29/01 2:19 PM Page 524Distance(int ft, float in) //constructor (two args) { feet = ft; inches = in; } void showdist() //display distance { cout << feet << “\’-” << inches << ‘\”’; } friend Distance operator + (Distance, Distance); //friend }; // Distance operator + (Distance d1, Distance d2) //add d1 to d2 { int f = d1.feet + d2.feet; //add the feet float i = d1.inches + d2.inches; //add the inches if(i >= 12.0) //if inches exceeds 12.0, { i -= 12.0; f++; } //less 12 inches, plus 1 foot return Distance(f,i); //return new Distance with sum } //-------------------------------------------------------------- int main() { Distance d1 = 2.5; //constructor converts Distance d2 = 1.25; //float-feet to Distance Distance d3; cout << “\nd1 = “; d1.showdist(); cout << “\nd2 = “; d2.showdist(); d3 = d1 + 10.0; //distance + float: OK cout << “\nd3 = “; d3.showdist(); d3 = 10.0 + d1; //float + Distance: OK cout << “\nd3 = “; d3.showdist(); cout << endl; return 0; } The overloaded + operator is made into a friend: friend Distance operator + (Distance, Distance); Notice that, while the overloaded + operator took one argument as a member function, it takes two as a friend function. In a member function, one of the objects on which the + operates is the object of which it was a member, and the second is an argument. In a friend, both objects must be arguments. The only change to the body of the overloaded + function is that the variables feet and inches,used in NOFRI for direct access to the object’s data, have been replaced in FRENGL by d1.feet and d1.inches, since this object is supplied as an argument. Virtual Functions 11 V IRTUAL F UNCTIONS 525 12 3087 CH11 11/29/01 2:19 PM Page 525Remember that, to make a function a friend, only the function declaration within the class is preceded by the keyword friend. The class definition is written normally, as are calls to the function. friends for Functional Notation Sometimes a friend allows a more obvious syntax for calling a function than does a member function. For example, suppose we want a function that will square (multiply by itself) an object of the English Distance class and return the result in square feet, as a type float. The MISQ example shows how this might be done with a member function. // misq.cpp // member square() function for Distance #include using namespace std; //////////////////////////////////////////////////////////////// class Distance //English Distance class { private: int feet; float inches; public: //constructor (no args) Distance() : feet(0), inches(0.0) { } //constructor (two args) Distance(int ft, float in) : feet(ft), inches(in) { } void showdist() //display distance { cout << feet << “\’-” << inches << ‘\”’; } float square(); //member function }; //-------------------------------------------------------------- float Distance::square() //return square of { //this Distance float fltfeet = feet + inches/12; //convert to float float feetsqrd = fltfeet * fltfeet; //find the square return feetsqrd; //return square feet } //////////////////////////////////////////////////////////////// int main() { Distance dist(3, 6.0); //two-arg constructor (3’-6”) float sqft; sqft = dist.square(); //return square of dist //display distance and square Chapter 11526 12 3087 CH11 11/29/01 2:19 PM Page 526cout << “\nDistance = “; dist.showdist(); cout << “\nSquare = “ << sqft << “ square feet\n”; return 0; } The main() part of the program creates a Distance value, squares it, and prints out the result. The output shows the original distance and the square: Distance = 3’-6” Square = 12.25 square feet In main() we use the statement sqft = dist.square(); to find the square of dist and assign it to sqft. This works all right, but if we want to work with Distance objects using the same syntax that we use with ordinary numbers, we would probably prefer a functional notation: sqft = square(dist); We can achieve this effect by making square() a friend of the Distance class, as shown in FRISQ: // frisq.cpp // friend square() function for Distance #include using namespace std; //////////////////////////////////////////////////////////////// class Distance //English Distance class { private: int feet; float inches; public: Distance() : feet(0), inches(0.0) //constructor (no args) { } //constructor (two args) Distance(int ft, float in) : feet(ft), inches(in) { } void showdist() //display distance { cout << feet << “\’-” << inches << ‘\”’; } friend float square(Distance); //friend function }; //-------------------------------------------------------------- float square(Distance d) //return square of { //this Distance Virtual Functions 11 V IRTUAL F UNCTIONS 527 12 3087 CH11 11/29/01 2:19 PM Page 527float fltfeet = d.feet + d.inches/12; //convert to float float feetsqrd = fltfeet * fltfeet; //find the square return feetsqrd; //return square feet } //////////////////////////////////////////////////////////////// int main() { Distance dist(3, 6.0); //two-arg constructor (3’-6”) float sqft; sqft = square(dist); //return square of dist //display distance and square cout << “\nDistance = “; dist.showdist(); cout << “\nSquare = “ << sqft << “ square feet\n”; return 0; } Whereas square() takes no arguments as a member function in MISQ, it takes one as a friend in FRISQ. In general, the friend version of a function requires one more argument than when the function is a member. The square() function in FRISQ is similar to that in MISQ,but it refers to the data in the source Distance object as d.feet and d.inches,instead of as feet and inches. friend Classes The member functions of a class can all be made friends at the same time when you make the entire class a friend. The program FRICLASS shows how this looks. // friclass.cpp // friend classes #include using namespace std; //////////////////////////////////////////////////////////////// class alpha { private: int data1; public: alpha() : data1(99) { } //constructor friend class beta; //beta is a friend class }; //////////////////////////////////////////////////////////////// class beta { //all member functions can Chapter 11528 12 3087 CH11 11/29/01 2:19 PM Page 528public: //access private alpha data void func1(alpha a) { cout << “\ndata1=” << a.data1; } void func2(alpha a) { cout << “\ndata1=” << a.data1; } }; //////////////////////////////////////////////////////////////// int main() { alpha a; beta b; b.func1(a); b.func2(a); cout << endl; return 0; } In class alpha the entire class beta is proclaimed a friend. Now all the member functions of beta can access the private data of alpha (in this program, the single data item data1). Note that in the friend declaration we specify that beta is a class using the class keyword: friend class beta; We could have also declared beta to be a class before the alpha class specifier, as in previous examples class beta; and then, within alpha,referred to beta without the class keyword: friend beta; Static Functions In the STATIC example in Chapter 6, “Objects and Classes,” we introduced static data mem- bers. As you may recall, a static data member is not duplicated for each object; rather a single data item is shared by all objects of a class. The STATIC example showed a class that kept track of how many objects of itself there were. Let’s extend this concept by showing how functions as well as data may be static. Besides showing static functions, our example will model a class that provides an ID number for each of its objects. This allows you to query an object to find out which object it is—a capability that is sometimes useful in debugging a program, among other situations. The program also casts some light on the operation of destructors. Here’s the listing for STATFUNC: // statfunc.cpp // static functions and ID numbers for objects #include using namespace std; Virtual Functions 11 V IRTUAL F UNCTIONS 529 12 3087 CH11 11/29/01 2:19 PM Page 529//////////////////////////////////////////////////////////////// class gamma { private: static int total; //total objects of this class // (declaration only) int id; //ID number of this object public: gamma() //no-argument constructor { total++; //add another object id = total; //id equals current total } ~gamma() //destructor { total--; cout << “Destroying ID number “ << id << endl; } static void showtotal() //static function { cout << “Total is “ << total << endl; } void showid() //non-static function { cout << “ID number is “ << id << endl; } }; //-------------------------------------------------------------- int gamma::total = 0; //definition of total //////////////////////////////////////////////////////////////// int main() { gamma g1; gamma::showtotal(); gamma g2, g3; gamma::showtotal(); g1.showid(); g2.showid(); g3.showid(); cout << “----------end of program----------\n”; return 0; } Chapter 11530 12 3087 CH11 11/29/01 2:19 PM Page 530Accessing static Functions In this program there is a static data member, total,in the class gamma. This data keeps track of how many objects of the class there are. It is incremented by the constructor and decre- mented by the destructor. Suppose we want to access total from outside the class. We construct a function, showtotal(),that prints the total’s value. But how do we access this function? When a data member is declared static,there is only one such data value for the entire class, no matter how many objects of the class are created. In fact, there may be no such objects at all, but we still want to be able to learn this fact. We could create a dummy object to use in calling a member function, as in gamma dummyObj; // make an object so we can call function dummyObj.showtotal(); // call function But this is rather inelegant. We shouldn’t need to refer to a specific object when we’re doing something that relates to the entire class. It’s more reasonable to use the name of the class itself with the scope-resolution operator. gamma::showtotal(); // more reasonable However, this won’t work if showtotal() is a normal member function; an object and the dot member-access operator are required in such cases. To access showtotal() using only the class name, we must declare it to be a static member function. This is what we do in STATFUNC,in the declarator static void showtotal() Now the function can be accessed using only the class name. Here’s the output: Total is 1 Total is 3 ID number is 1 ID number is 2 ID number is 3 ----------end of program-------- Destroying ID number 3 Destroying ID number 2 Destroying ID number 1 We define one object, g1, and then print out the value of total, which is 1. Then we define two more objects, g2 and g3, and again print out the total, which is now 3. Virtual Functions 11 V IRTUAL F UNCTIONS 531 12 3087 CH11 11/29/01 2:19 PM Page 531Numbering the Objects We’ve placed another function in gamma() to print out the ID number of individual members. This ID number is set equal to total when an object is created, so each object has a unique number. The showid() function prints out the ID of its object. We call it three times in main(), in the statements g1.showid(); g2.showid(); g3.showid(); As the output shows, each object has a unique number. The g1 object is numbered 1, g2 is 2, and g3 is 3. Investigating Destructors Now that we know how to number objects, we can investigate an interesting fact about destruc- tors. STATFUNC prints an end of program message in its last statement, but it’s not done yet, as the output shows. The three objects created in the program must be destroyed before the pro- gram terminates, so that memory is not left in an inaccessible state. The compiler takes care of this by invoking the destructor. We can see that this happens by inserting a statement in the destructor that prints a message. Since we’ve numbered the objects, we can also find out the order in which the objects are destroyed. As the output shows, the last object created, g3, is destroyed first. One can infer from this last-in-first-out approach that local objects are stored on the stack. Assignment and Copy Initialization The C++ compiler is always busy on your behalf, doing things you can’t be bothered to do. If you take charge, it will defer to your judgment; otherwise it will do things its own way. Two important examples of this process are the assignment operator and the copy constructor. You’ve used the assignment operator many times, probably without thinking too much about it. Suppose a1 and a2 are objects. Unless you tell the compiler otherwise, the statement a2 = a1; // set a2 to the value of a1 will cause the compiler to copy the data from a1, member by member, into a2. This is the default action of the assignment operator, =. You’re also familiar with initializing variables. Initializing an object with another object, as in alpha a2(a1); // initialize a2 to the value of a1 Chapter 11532 12 3087 CH11 11/29/01 2:19 PM Page 532causes a similar action. The compiler creates a new object, a2, and copies the data from a1, member by member, into a2. This is the default action of the copy constructor. Both of these default activities are provided, free of charge, by the compiler. If member-by- member copying is what you want, you need take no further action. However, if you want assignment or initialization to do something more complex, you can override the default func- tions. We’ll discuss the techniques for overloading the assignment operator and the copy con- structor separately, and then put them together in an example that gives a String class a more efficient way to manage memory. We’ll also introduce a new UML feature: the object diagram. Overloading the Assignment Operator Let’s look at a short example that demonstrates the technique of overloading the assignment operator. Here’s the listing for ASSIGN: // assign.cpp // overloads assignment operator (=) #include using namespace std; //////////////////////////////////////////////////////////////// class alpha { private: int data; public: alpha() //no-arg constructor { } alpha(int d) //one-arg constructor { data = d; } void display() //display data { cout << data; } alpha operator = (alpha& a) //overloaded = operator { data = a.data; //not done automatically cout << “\nAssignment operator invoked”; return alpha(data); //return copy of this alpha } }; //////////////////////////////////////////////////////////////// int main() { alpha a1(37); alpha a2; Virtual Functions 11 V IRTUAL F UNCTIONS 533 12 3087 CH11 11/29/01 2:19 PM Page 533a2 = a1; //invoke overloaded = cout << “\na2=”; a2.display(); //display a2 alpha a3 = a2; //does NOT invoke = cout << “\na3=”; a3.display(); //display a3 cout << endl; return 0; } The alpha class is very simple; it contains only one data member. Constructors initialize the data, and a member function can print out its value. The new aspect of ASSIGN is the function operator=(), which overloads the = operator. In main(), we define a1 and give it the value 37, and define a2 but give it no value. Then we use the assignment operator to set a2 to the value of a1: a2 = a1; // assignment statement This causes our overloaded operator=() function to be invoked. Here’s the output from ASSIGN: Assignment operator invoked a2=37 a3=37 Initialization Is Not Assignment In the last two lines of ASSIGN, we initialize the object a3 to the value a2 and display it. Don’t be confused by the syntax here. The equal sign in alpha a3 = a2; // copy initialization, not an assignment is not an assignment but an initialization, with the same effect as alpha a3(a2); // alternative form of copy initialization This is why the assignment operator is executed only once, as shown by the single invocation of the line Assignment operator invoked in the output of ASSIGN. Taking Responsibility When you overload the = operator you assume responsibility for doing whatever the default assignment operator did. Often this involves copying data members from one object to another. The alpha class in ASSIGN has only one data item, data,so the operator=() function copies its value with the statement data = a.data; Chapter 11534 12 3087 CH11 11/29/01 2:20 PM Page 534The function also prints the Assignment operator invoked message so that we can tell when it executes. Passing by Reference Notice that the argument to operator=() is passed by reference. It is not absolutely necessary to do this, but it’s usually a good idea. Why? As you know, an argument passed by value gen- erates a copy of itself in the function to which it is passed. The argument passed to the operator=() function is no exception. If such objects are large, the copies can waste a lot of memory. Values passed by reference don’t generate copies, and thus help to conserve memory. Also, there are certain situations in which you want to keep track of the number of objects (as in the STATFUNC example, where we assigned numbers to the objects). If the compiler is gener- ating extra objects every time you use the assignment operator, you may wind up with more objects than you expected. Passing by reference helps avoid such spurious object creation. Returning a Value As we’ve seen, a function can return information to the calling program by value or by refer- ence. When an object is returned by value, a new object is created and returned to the calling program. In the calling program, the value of this object can be assigned to a new object or it can be used in other ways. When an object is returned by reference, no new object is created. A reference to the original object in the function is all that’s returned to the calling program. The operator=() function in ASSIGN returns a value by creating a temporary alpha object and initializing it using the one-argument constructor in the statement return alpha(data); The value returned is a copy of, but not the same object as, the object of which the overloaded = operator is a member. Returning a value makes it possible to chain = operators: a3 = a2 = a1; However, returning by value has the same disadvantages as passing an argument by value: It creates an extra copy that wastes memory and can cause confusion. Can we return this value with a reference, using the declarator shown here for the overloaded = operator? alpha& operator = (alpha& a) // bad idea in this case Unfortunately, we can’t use reference returns on variables that are local to a function. Remember that local (automatic) variables—that is, those created within a function (and not designated static)—are destroyed when the function returns. A return by reference returns only the address of the data being returned, and, for local data, this address points to data within the function. When the function is terminated and this data is destroyed, the pointer is left with a meaningless value. Your compiler may flag this usage with a warning. (We’ll see one way to solve this problem in the section “The this Pointer” later in this chapter.) Virtual Functions 11 V IRTUAL F UNCTIONS 535 12 3087 CH11 11/29/01 2:20 PM Page 535Not Inherited The assignment operator is unique among operators in that it is not inherited. If you overload the assignment operator in a base class, you can’t use this same function in any derived classes. The Copy Constructor As we discussed, you can define and at the same time initialize an object to the value of another object with two kinds of statements: alpha a3(a2); // copy initialization alpha a3 = a2; // copy initialization, alternate syntax Both styles of definition invoke a copy constructor: a constructor that creates a new object and copies its argument into it. The default copy constructor, which is provided automatically by the compiler for every object, performs a member-by-member copy. This is similar to what the assignment operator does; the difference is that the copy constructor also creates a new object. Like the assignment operator, the copy constructor can be overloaded by the user. The XOFXREF example shows how it’s done. // xofxref.cpp // copy constructor: X(X&) #include using namespace std; //////////////////////////////////////////////////////////////// class alpha { private: int data; public: alpha() //no-arg constructor { } alpha(int d) //one-arg constructor { data = d; } alpha(alpha& a) //copy constructor { data = a.data; cout << “\nCopy constructor invoked”; } void display() //display { cout << data; } void operator = (alpha& a) //overloaded = operator { data = a.data; cout << “\nAssignment operator invoked”; } }; Chapter 11536 12 3087 CH11 11/29/01 2:20 PM Page 536//////////////////////////////////////////////////////////////// int main() { alpha a1(37); alpha a2; a2 = a1; //invoke overloaded = cout << “\na2=”; a2.display(); //display a2 alpha a3(a1); //invoke copy constructor // alpha a3 = a1; //equivalent definition of a3 cout << “\na3=”; a3.display(); //display a3 cout << endl; return 0; } This program overloads both the assignment operator and the copy constructor. The overloaded assignment operator is similar to that in the ASSIGN example. The copy constructor takes one argument: an object of type alpha, passed by reference. Here’s its declarator: alpha(alpha&) This declarator has the form X(X&) (pronounced “X of X ref”). Here’s the output of XOFXREF: Assignment operator invoked a2=37 Copy constructor invoked a3=37 The statement a2 = a1; invokes the assignment operator, while alpha a3(a1); invokes the copy constructor. The equivalent statement alpha a3 = a1; could also be used to invoke the copy constructor. We’ve seen that the copy constructor may be invoked when an object is defined. It is also invoked when arguments are passed by value to functions and when values are returned from functions. Let’s discuss these situations briefly. Virtual Functions 11 V IRTUAL F UNCTIONS 537 12 3087 CH11 11/29/01 2:20 PM Page 537Function Arguments The copy constructor is invoked when an object is passed by value to a function. It creates the copy that the function operates on. Thus if the function void func(alpha); were declared in XOFXREF, and this function were called by the statement func(a1); then the copy constructor would be invoked to create a copy of the a1 object for use by func(). (Of course, the copy constructor is not invoked if the argument is passed by reference or if a pointer to it is passed. In these cases no copy is created; the function operates on the original variable.) Function Return Values The copy constructor also creates a temporary object when a value is returned from a function. Suppose there was a function like this in XOFXREF alpha func(); and this function was called by the statement a2 = func(); The copy constructor would be invoked to create a copy of the value returned by func(), and this value would be assigned (invoking the assignment operator) to a2. Why Not an X(X) Constructor? Do we need to use a reference in the argument to the copy constructor? Could we pass by value instead? No, the compiler complains that it is out of memory if we try to compile alpha(alpha a) Why? Because when an argument is passed by value, a copy of it is constructed. What makes the copy? The copy constructor. But this is the copy constructor, so it calls itself. In fact, it calls itself over and over until the compiler runs out of memory. So, in the copy constructor, the argument must be passed by reference, which creates no copies. Watch Out for Destructors In the sections “Passing by Reference” and “Returning a Value,” we discussed passing argu- ments to a function by value and returning by value. These situations cause the destructor to be called as well, when the temporary objects created by the function are destroyed when the function returns. This can cause considerable consternation if you’re not expecting it. The moral is, when working with objects that require more than member-by-member copying, pass and return by reference—not by value—whenever possible. Chapter 11538 12 3087 CH11 11/29/01 2:20 PM Page 538Define Both Copy Constructor and Assignment Operator When you overload the assignment operator, you almost always want to overload the copy constructor as well (and vice versa). You don’t want your custom copying routine used in some situations, and the default member-by-member scheme used in others. Even if you don’t think you’ll use one or the other, you may find the compiler using them in nonobvious situations, such as passing an argument to a function by value, and returning from a function by value. In fact, if the constructor to a class involves the use of system resources such as memory or disk files, you should almost always overload both the assignment operator and the copy con- structor, and make sure they do what you want. How to Prohibit Copying We’ve discussed how to customize the copying of objects using the assignment operator and the copy constructor. Sometimes, however, you may want to prohibit the copying of an object using these operations. For example, it might be essential that each member of a class be cre- ated with a unique value for some member, which is provided as an argument to the construc- tor. If an object is copied, the copy will be given the same value. To avoid copying, overload the assignment operator and the copy constructor as private members. class alpha { private: alpha& operator = (alpha&); // private assignment operator alpha(alpha&); // private copy constructor }; As soon as you attempt a copying operation, such as alpha a1, a2; a1 = a2; // assignment alpha a3(a1); // copy constructor the compiler will tell you that the function is not accessible. You don’t need to define the func- tions, since they will never be called. UML Object Diagrams We’ve seen examples of class diagrams in previous chapters. It will probably not surprise you to know that the UML supports object diagrams as well. Object diagrams depict specific objects (for instance, the Mike_Gonzalez object of the Professor class). Because the relation- ships among objects change during the course of a program’s operation (indeed, objects may even be created and destroyed) an object diagram is like a snapshot, representing objects at a particular moment in time. It’s said to be a static UML diagram. You use an object diagram to model a particular thing your program does. You freeze the pro- gram at a moment in time and look at the objects that participate in the behavior you’re inter- ested in, and the communications among these objects at that point in time. Virtual Functions 11 V IRTUAL F UNCTIONS 539 12 3087 CH11 11/29/01 2:20 PM Page 539In object diagrams, objects are represented by rectangles, just as classes are in class diagrams. The object’s name, attributes, and operations are shown in a similar way. However, objects are distinguished from classes by having their names underlined. Both the object name and the class name can be used, separated by a colon: anObj:aClass If you don’t know the name of the object (because it’s only known through a pointer, for exam- ple) you can use just the class name preceded by the colon: :aClass Lines between the objects are called links,and represent one object communicating with another. Navigability can be shown. The value of an attribute can be shown using an equal sign: count = 0 Notice there’s no semicolon at the end; this is the UML, not C++. Another UML feature we’ll encounter is the note. Notes are shown as rectangles with a dog- eared (turned down) corner. They hold comments or explanations. A dotted line connects a note to the relevent element in the diagram. Unlike associations and links, a note can refer to an element inside a class or object rectangle. Notes can be used in any kind of UML diagram. We’ll see a number of object diagrams in the balance of this chapter. A Memory-Efficient String Class The ASSIGN and XOFXREF examples don’t really need to have overloaded assignment operators and copy constructors. They use straightforward classes with only one data item, so the default assignment operator and copy constructor would work just as well. Let’s look at an example where it is essential for the user to overload these operators. Defects with the String Class We’ve seen various versions of our homemade String class in previous chapters. However, these versions are not very sophisticated. It would be nice to overload the = operator so that we could assign the value of one String object to another with the statement s2 = s1; If we overload the = operator, the question arises of how we will handle the actual string (the array of type char), which is the principal data item in the String class. One possibility is for each String object to have a place to store a string. If we assign one String object to another (from s1 into s2 in the previous statement), we simply copy the string from the source into the destination object. If you’re concerned with conserving memory, the Chapter 11540 12 3087 CH11 11/29/01 2:20 PM Page 540problem with this is that the same string now exists in two (or more) places in memory. This is not very efficient, especially if the strings are long. Figure 11.4 shows how this looks. Virtual Functions 11 V IRTUAL F UNCTIONS 541 s1:String str “This is a long string in memory.” s2:String str “This is a long string in memory.” s1:String str s2:String str “This is a long string in memory.” FIGURE 11.4 UML object diagram: replicating strings. Instead of having each String object contain its own char* string, we could arrange for it to contain only a pointer to a string. Now, if we assign one String object to another, we need only copy the pointer from one object to another; both pointers will point to the same string. This is efficient, since only a single copy of the string itself needs to be stored in memory. Figure 11.5 shows how this looks. FIGURE 11.5 UML object diagram: replicating pointers to strings. However, if we use this system we need to be careful when we destroy a String object. If a String’s destructor uses delete to free the memory occupied by the string, and if there are several objects with pointers pointing to the string, these other objects will be left with pointers pointing to memory that may no longer hold the string they think it does; they become dan- gling pointers. 12 3087 CH11 11/29/01 2:20 PM Page 541To use pointers to strings in String objects, we need a way to keep track of how many String objects point to a particular string, so that we can avoid using delete on the string until the last String that points to it is itself deleted. Our next example, STRIMEM, does just this. A String-Counter Class Suppose we have several String objects pointing to the same string and we want to keep a count of how many Strings point to the string. Where will we store this count? It would be cumbersome for every String object to maintain a count of how many of its fellow Strings were pointing to a particular string, so we don’t want to use a member variable in String for the count. Could we use a static variable? This is a possibility; we could create a static array and use it to store a list of string addresses and counts. However, this requires con- siderable overhead. It’s more efficient to create a new class to store the count. Each object of this class, which we call strCount, contains a count and also a pointer to the string itself. Each String object contains a pointer to the appropriate strCount object. Figure 11.6 shows how this looks. Chapter 11542 s1:String psc s2:String psc s3:String psc s4:String psc :strCount :strCount count = 2 str count = 3 str “This is a different long string.”s5:String psc “This is a long string in memory.” FIGURE 11.6 String and strCount objects. 12 3087 CH11 11/29/01 2:20 PM Page 542To ensure that String objects have access to strCount objects, we make String a friend of strCount. Also, we want to ensure that the strCount class is used only by the String class. To prevent access to any of its functions, we make all member functions of strCount private. Because String is a friend, it can nevertheless access any part of strCount. Here’s the listing for STRIMEM: // strimem.cpp // memory-saving String class // overloaded assignment and copy constructor #include #include //for strcpy(), etc. using namespace std; //////////////////////////////////////////////////////////////// class strCount //keep track of number { //of unique strings private: int count; //number of instances char* str; //pointer to string friend class String; //make ourselves available //member functions are private //-------------------------------------------------------------- strCount(char* s) //one-arg constructor { int length = strlen(s); //length of string argument str = new char[length+1]; //get memory for string strcpy(str, s); //copy argument to it count=1; //start count at 1 } //-------------------------------------------------------------- ~strCount() //destructor { delete[] str; } //delete the string }; //////////////////////////////////////////////////////////////// class String //String class { private: strCount* psc; //pointer to strCount public: String() //no-arg constructor { psc = new strCount(“NULL”); } //-------------------------------------------------------------- String(char* s) //1-arg constructor { psc = new strCount(s); } //-------------------------------------------------------------- String(String& S) //copy constructor { Virtual Functions 11 V IRTUAL F UNCTIONS 543 12 3087 CH11 11/29/01 2:20 PM Page 543psc = S.psc; (psc->count)++; } //-------------------------------------------------------------- ~String() //destructor { if(psc->count==1) //if we are its last user, delete psc; // delete our strCount else // otherwise, (psc->count)--; // decrement its count } //-------------------------------------------------------------- void display() //display the String { cout << psc->str; //print string cout << “ (addr=” << psc << “)”; //print address } //-------------------------------------------------------------- void operator = (String& S) //assign the string { if(psc->count==1) //if we are its last user, delete psc; // delete our strCount else // otherwise, (psc->count)--; // decrement its count psc = S.psc; //use argument’s strCount (psc->count)++; //increment its count } }; //////////////////////////////////////////////////////////////// int main() { String s3 = “When the fox preaches, look to your geese.”; cout << “\ns3=”; s3.display(); //display s3 String s1; //define String s1 = s3; //assign it another String cout << “\ns1=”; s1.display(); //display it String s2(s3); //initialize with String cout << “\ns2=”; s2.display(); //display it cout << endl; return 0; } Chapter 11544 12 3087 CH11 11/29/01 2:20 PM Page 544In the main() part of STRIMEM we define a String object, s3,to contain the proverb “When the fox preaches, look to your geese.” We define another String s1 and set it equal to s3; then we define s2 and initialize it to s3. Setting s1 equal to s3 invokes the overloaded assignment oper- ator; initializing s2 to s3 invokes the overloaded copy constructor. We print out all three strings, and also the address of the strCount object pointed to by each object’s psc pointer, to show that these objects are all the same. Here’s the output from STRIMEM: s3=When the fox preaches, look to your geese. (addr=0x8f510e00) s1=When the fox preaches, look to your geese. (addr=0x8f510e00) s2=When the fox preaches, look to your geese. (addr=0x8f510e00) The other duties of the String class are divided between the String and strCount classes. Let’s see what they do. The strCount Class The strCount class contains the pointer to the actual string and the count of how many String class objects point to this string. Its single constructor takes a pointer to a string as an argu- ment and creates a new memory area for the string. It copies the string into this area and sets the count to 1, since just one String points to it when it is created. The destructor in strCount frees the memory used by the string. (We use delete[] with brackets because a string is an array.) The String Class The String class uses three constructors. If a new string is being created, as in the zero- argument and C-string-argument constructors, a new strCount object is created to hold the string, and the psc pointer is set to point to this object. If an existing String object is being copied, as in the copy constructor and the overloaded assignment operator, the pointer psc is set to point to the old strCount object, and the count in this object is incremented. The overloaded assignment operator, as well as the destructor, must also delete the old strCount object pointed to by psc if the count is 1. (We don’t need brackets on delete because we’re deleting only a single strCount object.) Why must the assignment operator worry about deletion? Remember that the String object on the left of the equal sign (call it s1) was pointing at some strCount object (call it oldStrCnt) before the assignment. After the assignment s1 will be pointing to the object on the right of the equal sign. If there are now no String objects pointing to oldStrCnt, it should be deleted. If there are other objects pointing to it, its count must be decremented. Figure 11.7 shows the action of the overloaded assign- ment operator, and Figure 11.8 shows the copy constructor. Virtual Functions 11 V IRTUAL F UNCTIONS 545 12 3087 CH11 11/29/01 2:20 PM Page 545FIGURE 11.7 Assignment operator in STRIMEM. Chapter 11546 s3:String psc s1:String psc :strCount Before execution of s1 = s2; After execution of s1 = s2; count = 1 str “When the fox preaches…” :strCount count = 1 str “NULL” s3:String psc s1:String psc :strCount count = 2 str “When the fox preaches…” s3:String psc :strCount Before execution of String s2 (s3); After execution of String s2 (s3); count = 1 str “When the fox preaches…” s3:String psc s2:String psc :strCount count = 2 str “When the fox preaches…” FIGURE 11.8 Copy constructor in STRIMEM. 12 3087 CH11 11/29/01 2:20 PM Page 546The this Pointer The member functions of every object have access to a sort of magic pointer named this, which points to the object itself. Thus any member function can find out the address of the object of which it is a member. Here’s a short example, WHERE,that shows the mechanism: // where.cpp // the this pointer #include using namespace std; //////////////////////////////////////////////////////////////// class where { private: char charray[10]; //occupies 10 bytes public: void reveal() { cout << “\nMy object’s address is “ << this; } }; //////////////////////////////////////////////////////////////// int main() { where w1, w2, w3; //make three objects w1.reveal(); //see where they are w2.reveal(); w3.reveal(); cout << endl; return 0; } The main() program in this example creates three objects of type where. It then asks each object to print its address, using the reveal() member function. This function prints out the value of the this pointer. Here’s the output: My object’s address is 0x8f4effec My object’s address is 0x8f4effe2 My object’s address is 0x8f4effd8 Since the data in each object consists of an array of 10 bytes, the objects are spaced 10 bytes apart in memory. (EC minus E2 is 10 decimal, as is E2 minus D8.) Some compilers may place extra bytes in objects, making them slightly larger than 10 bytes. Accessing Member Data with this When you call a member function, it comes into existence with the value of this set to the address of the object for which it was called. The this pointer can be treated like any other pointer to an object, and can thus be used to access the data in the object it points to, as shown in the DOTHIS program: Virtual Functions 11 V IRTUAL F UNCTIONS 547 12 3087 CH11 11/29/01 2:20 PM Page 547// dothis.cpp // the this pointer referring to data #include using namespace std; //////////////////////////////////////////////////////////////// class what { private: int alpha; public: void tester() { this->alpha = 11; //same as alpha = 11; cout << this->alpha; //same as cout << alpha; } }; //////////////////////////////////////////////////////////////// int main() { what w; w.tester(); cout << endl; return 0; } This program simply prints out the value 11. Notice that the tester() member function accesses the variable alpha as this->alpha This is exactly the same as referring to alpha directly. This syntax works, but there is no rea- son for it except to show that this does indeed point to the object. Using this for Returning Values A more practical use for this is in returning values from member functions and overloaded operators. Recall that in the ASSIGN program we could not return an object by reference, because the object was local to the function returning it and thus was destroyed when the function returned. We need a more permanent object if we’re going to return it by reference. The object of which a function is a member is more permanent than its individual member functions. An object’s member functions are created and destroyed every time they’re called, but the object itself endures until it is destroyed by some outside agency (for example, when it is deleted). Thus returning by reference the object of which a function is a member is a better bet than returning a temporary object created in a member function. The this pointer makes this easy. Chapter 11548 12 3087 CH11 11/29/01 2:20 PM Page 548Here’s the listing for ASSIGN2, in which the operator=() function returns by reference the object that invoked it: //assign2.cpp // returns contents of the this pointer #include using namespace std; //////////////////////////////////////////////////////////////// class alpha { private: int data; public: alpha() //no-arg constructor { } alpha(int d) //one-arg constructor { data = d; } void display() //display data { cout << data; } alpha& operator = (alpha& a) //overloaded = operator { data = a.data; //not done automatically cout << “\nAssignment operator invoked”; return *this; //return copy of this alpha } }; //////////////////////////////////////////////////////////////// int main() { alpha a1(37); alpha a2, a3; a3 = a2 = a1; //invoke overloaded =, twice cout << “\na2=”; a2.display(); //display a2 cout << “\na3=”; a3.display(); //display a3 cout << endl; return 0; } In this program we can use the declaration alpha& operator = (alpha& a) which returns by reference, instead of alpha operator = (alpha& a) which returns by value. The last statement in this function is return *this; Virtual Functions 11 V IRTUAL F UNCTIONS 549 12 3087 CH11 11/29/01 2:20 PM Page 549Since this is a pointer to the object of which the function is a member, *this is that object itself, and the statement returns it by reference. Here’s the output of ASSIGN2: Assignment operator invoked Assignment operator invoked a2=37 a3=37 Each time the equal sign is encountered in a3 = a2 = a1; the overloaded operator=() function is called, which prints the messages. The three objects all end up with the same value. You usually want to return by reference from overloaded assignment operators, using *this,to avoid the creation of extra objects. Revised STRIMEM Program Using the this pointer we can revise the operator=() function in STRIMEM to return a value by reference, thus making possible multiple assignment operators for String objects, such as s1 = s2 = s3; At the same time, we can avoid the creation of spurious objects, such as those that are created when objects are returned by value. Here’s the listing for STRIMEM2: // strimem2.cpp // memory-saving String class // the this pointer in overloaded assignment #include #include //for strcpy(), etc using namespace std; //////////////////////////////////////////////////////////////// class strCount //keep track of number { //of unique strings private: int count; //number of instances char* str; //pointer to string friend class String; //make ourselves available //member functions are private strCount(char* s) //one-arg constructor { int length = strlen(s); //length of string argument str = new char[length+1]; //get memory for string strcpy(str, s); //copy argument to it count=1; //start count at 1 } Chapter 11550 12 3087 CH11 11/29/01 2:20 PM Page 550//-------------------------------------------------------------- ~strCount() //destructor { delete[] str; } //delete the string }; //////////////////////////////////////////////////////////////// class String //String class { private: strCount* psc; //pointer to strCount public: String() //no-arg constructor { psc = new strCount(“NULL”); } //-------------------------------------------------------------- String(char* s) //1-arg constructor { psc = new strCount(s); } //-------------------------------------------------------------- String(String& S) //copy constructor { cout << “\nCOPY CONSTRUCTOR”; psc = S.psc; (psc->count)++; } //-------------------------------------------------------------- ~String() //destructor { if(psc->count==1) //if we are its last user, delete psc; // delete our strCount else // otherwise, (psc->count)--; // decrement its count } //-------------------------------------------------------------- void display() //display the String { cout << psc->str; //print string cout << “ (addr=” << psc << “)”; //print address } //-------------------------------------------------------------- String& operator = (String& S) //assign the string { cout << “\nASSIGNMENT”; if(psc->count==1) //if we are its last user, delete psc; //delete our strCount else // otherwise, (psc->count)--; // decrement its count psc = S.psc; //use argument’s strCount (psc->count)++; //increment count Virtual Functions 11 V IRTUAL F UNCTIONS 551 12 3087 CH11 11/29/01 2:20 PM Page 551return *this; //return this object } }; int main() { String s3 = “When the fox preaches, look to your geese.”; cout << “\ns3=”; s3.display(); //display s3 String s1, s2; //define Strings s1 = s2 = s3; //assign them cout << “\ns1=”; s1.display(); //display it cout << “\ns2=”; s2.display(); //display it cout << endl; //wait for keypress return 0; } Now the declarator for the = operator is String& operator = (String& S) // return by reference And, as in ASSIGN2, this function returns a pointer to this. Here’s the output: s3=When the fox preaches, look to your geese. (addr=0x8f640d3a) ASSIGNMENT ASSIGNMENT s1=When the fox preaches, look to your geese. (addr=0x8f640d3a) s2=When the fox preaches, look to your geese. (addr=0x8f640d3a) The output shows that, following the assignment statement, all three String objects point to the same strCount object. We should note that the this pointer is not available in static member functions, since they are not associated with a particular object. Beware of Self-Assignment A corollary of Murphy’s Law states that whatever is possible, someone will eventually do. This is certainly true in programming, so you can expect that if you have overloaded the = operator, someone will use it to set an object equal to itself: alpha = alpha; Your overloaded assignment operator should be prepared to handle such self-assignment. Otherwise, bad things may happen. For example, in the main() part of the STRIMEM2program, if you set a String object equal to itself, the program will crash (unless there are other String objects using the same strCount object). The problem is that the code for the assignment oper- ator deletes the strCount object if it thinks the object that called it is the only object using the strCount. Self-assignment will cause it to believe this, even though nothing should be deleted. Chapter 11552 12 3087 CH11 11/29/01 2:20 PM Page 552To fix this, you should check for self-assignment at the start of any overloaded assignment operator. You can do this in most cases by comparing the address of the object for which the operator was called with the address of its argument. If the addresses are the same, the objects are identical and you should return immediately. (You don’t need to assign one to the other; they’re already the same.) For example, in STRIMEM2, you can insert the lines if(this == &S) return *this; at the start of operator=(). That should solve the problem. Dynamic Type Information It’s possible to find out information about an object’s class and even change the class of an object at runtime. We’ll look briefly at two mechanisms: the dynamic_cast operator, and the typeid operator. These are advanced capabilities, but you may find them useful someday. These capabilities are usually used in situations where a variety of classes are descended (sometimes in complicated ways) from a base class. For dynamic casts to work, the base class must be polymorphic; that is, it must have at least one virtual function. For both dynamic_cast and typeid to work, your compiler must enable Run-Time Type Information (RTTI). Borland C++Builder has this capability enabled by default, but in Microsoft Visual C++ you’ll need to turn it on overtly. See Appendix C, “Microsoft Visual C++,” for details on how this is done. You’ll also need to include the header file TYPEINFO. Checking the Type of a Class with dynamic_cast Suppose some other program sends your program an object (as the operating system might do with a callback function). It’s supposed to be a certain type of object, but you want to check it to be sure. How can you tell if an object is a certain type? The dynamic_cast operator pro- vides a way, assuming that the classes whose objects you want to check are all descended from a common ancestor. The DYNCAST1program shows how this looks. //dyncast1.cpp //dynamic cast used to test type of object //RTTI must be enabled in compiler #include #include //for dynamic_cast using namespace std; //////////////////////////////////////////////////////////////// class Base { virtual void vertFunc() //needed for dynamic cast { } }; Virtual Functions 11 V IRTUAL F UNCTIONS 553 12 3087 CH11 11/29/01 2:20 PM Page 553class Derv1 : public Base { }; class Derv2 : public Base { }; //////////////////////////////////////////////////////////////// //checks if pUnknown points to a Derv1 bool isDerv1(Base* pUnknown) //unknown subclass of Base { Derv1* pDerv1; if( pDerv1 = dynamic_cast(pUnknown) ) return true; else return false; } //-------------------------------------------------------------- int main() { Derv1* d1 = new Derv1; Derv2* d2 = new Derv2; if( isDerv1(d1) ) cout << “d1 is a member of the Derv1 class\n”; else cout << “d1 is not a member of the Derv1 class\n”; if( isDerv1(d2) ) cout << “d2 is a member of the Derv1 class\n”; else cout << “d2 is not a member of the Derv1 class\n”; return 0; } Here we have a base class Base and two derived classes Derv1 and Derv2. There’s also a func- tion, isDerv1(), which returns true if the pointer it received as an argument points to an object of class Derv1. This argument is of class Base,so the object passed can be either Derv1 or Derv2. The dynamic_cast operator attempts to convert this unknown pointer pUnknown to type Derv1. If the result is not zero, pUnknown did point to a Derv1 object. If the result is zero, it pointed to something else. Changing Pointer Types with dynamic_cast The dynamic_cast operator allows you to cast upward and downward in the inheritance tree. However, it allows such casting only in limited ways. The DYNCAST2program shows examples of such casts. Chapter 11554 12 3087 CH11 11/29/01 2:20 PM Page 554//dyncast2.cpp //tests dynamic casts //RTTI must be enabled in compiler #include #include //for dynamic_cast using namespace std; //////////////////////////////////////////////////////////////// class Base { protected: int ba; public: Base() : ba(0) { } Base(int b) : ba(b) { } virtual void vertFunc() //needed for dynamic_cast { } void show() { cout << “Base: ba=” << ba << endl; } }; //////////////////////////////////////////////////////////////// class Derv : public Base { private: int da; public: Derv(int b, int d) : da(d) { ba = b; } void show() { cout << “Derv: ba=” << ba << “, da=” << da << endl; } }; //////////////////////////////////////////////////////////////// int main() { Base* pBase = new Base(10); //pointer to Base Derv* pDerv = new Derv(21, 22); //pointer to Derv //derived-to-base: upcast -- points to Base subobject of Derv pBase = dynamic_cast(pDerv); pBase->show(); //”Base: ba=21” pBase = new Derv(31, 32); //normal //base-to-derived: downcast -- (pBase must point to a Derv) pDerv = dynamic_cast(pBase); pDerv->show(); //”Derv: ba=31, da=32” return 0; } Virtual Functions 11 V IRTUAL F UNCTIONS 555 12 3087 CH11 11/29/01 2:20 PM Page 555Here we have a base and a derived class. We’ve given each of these classes a data item to bet- ter demonstrate the effects of dynamic casts. In an upcast you attempt to change a derived-class object into a base-class object. What you get is the base part of the derived class object. In the example we make an object of class Derv. The base class part of this object holds member data ba, which has a value of 21, and the derived part holds data member da, which has the value 22. After the cast, pBase points to the base-class part of this Derv class object, so when called upon to display itself, it prints Base: ba=21. Upcasts are fine if all you want is the base part of the object. In a downcast we put a derived class object, which is pointed to by a base-class pointer, into a derived-class pointer. The typeid Operator Sometimes you want more information about an object than simple verification that it’s of a certain class. You can obtain information about the type of an unknown object, such as its class name, using the typeid operator. The TYPEID program demonstrates how it works. // typeid.cpp // demonstrates typeid() function // RTTI must be enabled in compiler #include #include //for typeid() using namespace std; //////////////////////////////////////////////////////////////// class Base { virtual void virtFunc() //needed for typeid { } }; class Derv1 : public Base { }; class Derv2 : public Base { }; //////////////////////////////////////////////////////////////// void displayName(Base* pB) { cout << “pointer to an object of “; //display name of class cout << typeid(*pB).name() << endl; //pointed to by pB } //-------------------------------------------------------------- int main() { Base* pBase = new Derv1; Chapter 11556 12 3087 CH11 11/29/01 2:20 PM Page 556displayName(pBase); //”pointer to an object of class Derv1” pBase = new Derv2; displayName(pBase); //”pointer to an object of class Derv2” return 0; } In this example the displayName() function displays the name of the class of the object passed to it. To do this, it uses the name member of the type_info class, along with the typeid opera- tor. In main() we pass this function two objects of class Derv1 and Derv2 respectively, and the program’s output is pointer to an object of class Derv1 pointer to an object of class Derv2 Besides its name, other information about a class is available using typeid. For example, you can check for equality of classes using an overloaded == operator. We’ll show an example of this in the EMPL_IO program in Chapter 12, “Streams and Files.” Although the examples in this section have used pointers, dynamic_cast and typeid work equally well with references. Summary Virtual functions provide a way for a program to decide while it is running what function to call. Ordinarily such decisions are made at compile time. Virtual functions make possible greater flexibility in performing the same kind of action on different kinds of objects. In partic- ular, they allow the use of functions called from an array of type pointer-to-base that actually holds pointers (or references) to a variety of derived types. This is an example of polymor- phism. Typically a function is declared virtual in the base class, and other functions with the same name are declared in derived classes. The use of one or more pure virtual functions in a class makes the class abstract, which means that no objects can be instantiated from it. A friend function can access a class’s private data, even though it is not a member function of the class. This is useful when one function must have access to two or more unrelated classes and when an overloaded operator must use, on its left side, a value of a class other than the one of which it is a member. friends are also used to facilitate functional notation. A static function is one that operates on the class in general, rather than on objects of the class. In particular it can operate on static variables. It can be called with the class name and scope-resolution operator. Virtual Functions 11 V IRTUAL F UNCTIONS 557 12 3087 CH11 11/29/01 2:20 PM Page 557The assignment operator = can be overloaded. This is necessary when it must do more than merely copy one object’s contents into another. The copy constructor, which creates copies during initialization, and also when arguments are passed and returned by value, can also be overloaded. This is necessary when the copy constructor must do more than simply copy an object. The this pointer is predefined in member functions to point to the object of which the func- tion is a member. The this pointer is useful in returning the object of which the function is a member. The dynamic_cast operator plays several roles. It can be used to determine what type of object a pointer points to, and, in certain situations, it can change the type of a pointer. The typeid operator can discover certain information about an object’s class, such as its name. The UML object diagram shows the relationship of a group of objects at a specific point in a program’s operation. Questions Answers to these questions can be found in Appendix G. 1. Virtual functions allow you to a. create an array of type pointer-to-base class that can hold pointers to derived classes. b. create functions that can never be accessed. c. group objects of different classes so they can all be accessed by the same function code. d. use the same function call to execute member functions of objects from different classes. 2. True or false: A pointer to a base class can point to objects of a derived class. 3. If there is a pointer p to objects of a base class, and it contains the address of an object of a derived class, and both classes contain a nonvirtual member function, ding(),then the statement p->ding(); will cause the version of ding() in the ________ class to be exe- cuted. 4. Write a declarator for a virtual function called dang() that returns type void and takes one argument of type int. 5. Deciding—after a program starts to execute—what function will be executed by a partic- ular function call statement is called ________. 6. If there is a pointer, p,to objects of a base class, and it contains the address of an object of a derived class, and both classes contain a virtual member function, ding(),the state- ment p->ding(); will cause the version of ding() in the ________ class to be executed. Chapter 11558 12 3087 CH11 11/29/01 2:20 PM Page 5587. Write the declaration for a pure virtual function called aragorn that returns no value and takes no arguments. 8. A pure virtual function is a virtual function that a. causes its class to be abstract. b. returns nothing. c. is used in a base class. d. takes no arguments. 9. Write the definition of an array called parr of 10 pointers to objects of class dong. 10. An abstract class is useful when a. no classes should be derived from it. b. there are multiple paths from one derived class to another. c. no objects should be instantiated from it. d. you want to defer the declaration of the class. 11. True or false: A friend function can access a class’s private data without being a mem- ber of the class. 12. A friend function can be used to a. mediate arguments between classes. b. allow access to classes whose source code is unavailable. c. allow access to an unrelated class. d. increase the versatility of an overloaded operator. 13. Write the declaration for a friend function called harry() that returns type void and takes one argument of class george. 14. The keyword friend appears in a. the class allowing access to another class. b. the class desiring access to another class. c. the private section of a class. d. the public section of a class. 15. Write a declaration that, in the class in which it appears, will make every member of the class harry a friend function. 16. A static function a. should be called when an object is destroyed. b. is closely connected to an individual object of a class. c. can be called using the class name and function name. d. is used when a dummy object must be created. Virtual Functions 11 V IRTUAL F UNCTIONS 559 12 3087 CH11 11/29/01 2:20 PM Page 55917. Explain what the default assignment operator = does when applied to objects. 18. Write a declaration for an overloaded assignment operator in class zeta. 19. An assignment operator might be overloaded to a. help keep track of the number of identical objects. b. assign a separate ID number to each object. c. ensure that all member data is copied exactly. d. signal when assignment takes place. 20. True or false: The user must always define the operation of the copy constructor. 21. The operation of the assignment operator and that of the copy constructor are a. similar, except that the copy constructor creates a new object. b. similar, except that the assignment operator copies member data. c. different, except that they both create a new object. d. different, except that they both copy member data. 22. Write the declaration of a copy constructor for a class called Bertha. 23. True or false: A copy constructor could be defined to copy only part of an object’s data. 24. The lifetime of a variable that is a. local to a member function coincides with the lifetime of the function. b. global coincides with the lifetime of a class. c. nonstatic member data of an object coincides with the lifetime of the object. d. static in a member function coincides with the lifetime of the function. 25. True or false: There is no problem with returning the value of a variable defined as local within a member function so long as it is returned by value. 26. Explain the difference in operation between these two statements. person p1(p0); person p1 = p0; 27. A copy constructor is invoked when a. a function returns by value. b. an argument is passed by value. c. a function returns by reference. d. an argument is passed by reference. 28. What does the this pointer point to? 29. If, within a class, da is a member variable, will the statement this.da=37; assign 37 to da? Chapter 11560 12 3087 CH11 11/29/01 2:20 PM Page 56030. Write a statement that a member function can use to return the entire object of which it is a member, without creating any temporary object