COEN244: Operator overloading - Encsusers.encs.concordia.ca/~amer/teach/coen244/notes/opOverloading.pdfc A. Amer Operator Overloading 26. Overloading object assignment = The = operator

COEN244: Operator overloading

Aishy Amer

Electrical & Computer Engineering

OutlineFriend functions

Operator notation in C++

Why operator overloading?

What can/cannot be overloaded?

Two ways to overload: member andnon-member operators

Unary operator overload

Binary operator overload

Multi-Overloading

Assignment operator versusCopy constructor

Some rules to overloading

Chain statements

Special operators:[]() + + −− <<>>

A complete example with member functions

A complete example with friend functions

c© A. Amer Operator Overloading 1

Introduction

Compiler identifies a function by its interface consisting ofits namenumber of its parameterstype of its parametersorder of its parameters

Recall:private members can be accessed/modified only whileINSIDE *a* member function


Friend functions

A friend function is a function that is not a member function ofa particular class but still can access its non-public members

For an external function to access the private and protectedmembers of a class:

it needs to be a friend of the class that defines thesemembersThe keyword friend is used


Friend functions

Exampleclass C i r c l e {

private :i n t rad ius , x , y ;

public :. . .i n t Area ( ) ; / / Member f u n c t i o n to Compute areaf r iend C i r c l e copy ( const C i r c l e &) ; / / Fr iend f u n c t i o n

} ;

i n t C i r c l e : : Area ( ) { return PI∗ rad ius∗ rad ius ; }

C i r c l e copy ( const C i r c l e& src ) { return C i r c l e ( src . rad ius , s rc . x , s rc . y ) ; }

/ / UsageC i r c l e a (10 ,3 ,6 ) , b ; cout << a . area ( ) ;

b = copy ( a ) ; cout << b . Area ( ) << ’ \ n ’ ;


Friend functions

Use member functions instead of friend functions whenpossible

In some case, friend functions can be very usefule.g., operator overloading

Friend functions are also used to specify friend classesTo declare all member functions of a class A as friends of another class B

use the following declaration in the definition of the class A: friend class B;

The friend relationship is neither symmetric nor transitiveIf a class A is a friend of class B it does not imply that B is a friend of A

Note that friend functions break encapsulationTherefore, pick your friend classes carefully!

We can also specify only the member functions that we want them to be friends insteadof all the member functionsfriend void B::print(A &);


Notation

C++ operators: + ∗ / + + >><< ...

Binary operator z=x+y; cout<<z;Unary operator z++;

x += y;x is left-hand (LH) objecty is right-hand (RH) objectx is target objecty is the source object

Recall: an object is implicitly accessed via the this pointer

⇒ Use this pointer to implicitly access the target object



C++ goal:

treat programmer-defined and built-in types equally

provide an intuitive interface to USERS of your class

Solution: Extending built-in operators so that the same

expression syntax with operator can be applied to objects



C++ goal:

treat programmer-defined and built-in types equally

provide an intuitive interface to USERS of your class

Solution: Extending built-in operators so that the same

expression syntax with operator can be applied to objects

Examples:i n t a =5 ,b =3; a +=b ;time3 = time1 + time2 ;tim3 ++; cout<< time3 ;Complex x ( 2 0 , 4 0 ) , y ( 3 0 , 5 0 ) ;x +=y ;cout<<x ; / / => x =(20+30) + i (40+50)


Two ways to overload

You may overloadmember & non-member (e.g., friend) functions of a class

Member function:Prototype Syntax :< re turn_type > operator <op> ( <RHobject>& ) ;Example : void operator +=( t ime &) ;

Non-member function: (normally declared as a friend)Prototype Syntax :f r iend <re turn_type > operator <op> ( < LHobject >&, <RHobject >&);Example : f r iend void operator << ( iost ream &, t ime &) ;


Overloading using member functions

Uses this pointer to implicitly access the source objectOnly object for unary operatorsLeft-hand object for binary operators

Syntax:<re turn_type > operator <op> ( <RHobject>& ) ;

Example:void Complex : : operator += ( const Complex &b ) { / / t a r g e t ob jec t changes

r e a l = r e a l + b . r e a l ; / / add to the r e a l component o f the t a r g e timag = imag + b . imag ; / / add to the imag component o f the t a r g e t

}/ / Test ingComplex x (20 ,40) , y (30 ,50 ) ;x +=y ; / / x i s the t a r g e t ob jec t , y i s the source ob jec t


Recall: Friend functions

A friend function is a a NON-member function thata class permits to have access to an object’s private datamembers

Friend functions are NOT member functions of the class butfree functions which have been given access to a class’s datamember

Friend functions are implemented without the scoperesolution operator


Overloading as non-member (friend) function

Should only be done when necessary

Accepts all object parameters through argument list

Syntax:f r iend <re turn_type > operator <op> ( < LHobject >&, <RHobject >&);

Example:/ / Dec la ra t i onf r iend ostream& operator <<(ostream&, const Complex&) ;

/ / Implementat ionostream& operator <<(ostream& os , const Complex& Z) {

os << ( (Z . r e a l ) ? Z . r e a l : ’ ’ ) ;i f (Z . imag ) os << " + i " << Z . imag ;return os ;

}


Why to use friend functions

1. To overload operators that have another class as the leftoperand, e.g., <<

2. To permit operators to be commutative such as inComplex a , c ; c= 5+a ; / / c =?. opera tor +(a ) ;

3. To make some function calls more attractive such as inTime t1 , t2 ;i f ( t1 . compareTo ( t2 ) == 0) . . .

==> i f ( compareTo ( t1 , t2 ) == 0) . . .

4. Friend function can be more efficient:they can avoid some of the costs of going through memberfunctions


Operator overloading: Example

class Time {f r iend ostream& operator <<(ostream&, const Time &) ;public :

. . . / / cons t ruc to r svoid set ( i n t h=0 , i n t m=0 , i n t s =0) ;. . .void operator +=(Time &t ) ; / / Overload the b inary += opera torvoid operator + + ( ) ; / / Overload the unary ++ p r e f i x opera tor

. . .private : i n t hour , minute , second ;

} ;

/ / D e f i n i t i o n o f the overloaded b inary += opera torvoid Time : : operator +=(Time &t ) {

/ / Use the set f u n c t i o n to take care o f the a d d i t i o nset ( hour + t . hour , minute + t . minute , second + t . second ) ;

}



/ / D e f i n i t i o n o f the overloaded unary ++ p r e f i x opera torvoid Time : : operator ++() {

/ / Use the set f u n c t i o n to take care o f the incrementset ( hour , minute , second +1) ;

}. . ./ / Use of the overloaded opsTime t , p (13 ,0 ,10 ) ;for ( i n t i =0; i <N; i ++) {

t ++;t . d i sp lay ( ) ;

}t +=p ;


What can/cannot be overloaded?

Operator overloading:the ability to re-define built-in C++ operators to manipulateuser-defined objects

Can be:+ − ∗ / % ^ & | ~ != < > += −= ∗= /= %=^= &= |= << >> >>= <<= == != <= >=&& | | ++ −− −>∗ , −>[ ] ( )new delete new [ ] delete [ ]

Cannot be:. .∗ : : ? : sizeof/ / .∗ i s p o i n t e r to member

Why would you need to overload the “cannot be” operators?


Two types of operators to overload

C++ operators: + ∗ / + + >><< ...

Binary operator z=x+y; cout<<z;Unary operator z++;



Member unary Ops:Syntax : < re turn_type > operator <op symbol> ( ) ;

Example : void operator + + ( ) ; / / Overload the unary ++ p r e f i x opera tor

Number o f parameters accepted : none

Friend unary Ops:Syntax : f r iend <re turn_type > operator <op symbol> ( < RHobject >&);

Example : f r iend void operator ++( Rat iona l &x ) ;

Number o f parameters accepted : one for unary opera tors



Member binary Ops:Syntax : < re turn_type > operator <op symbol >(<RHobject >&);

Example : void operator +=(Time &t ) ; / / Overload the b inary += opera tor

Number o f parameters accepted : one

Friend binary Ops:Syntax : f r iend <re turn_type > operator <op symbol >(< LHobject >&, <RHobject >

Example : f r iend void operator += ( Rat iona l &x , const Rat iona l &y ) ;

Number o f parameters accepted : two for b inary opera tors


Overloading: Multi-Overloading

Can we do the following?/ / Test ingComplex Z1 ( 1 , 1 ) , Z2 ( 0 . 5 , 0 .707) , r e s u l t ( 0 , 0 ) ;double x = 1.264 ;r e s u l t = Z1 + Z2 ; / / yesr e s u l t = Z1 + x ; / / yesr e s u l t = x + Z1 ; / / ???? wi th member f u n c t i o n s ?


Overloading: Multi-Overloading

class Complex {f r iend Complex operator +( double , Complex& ) ;public :

void Complex ( double = 0 , double = 0 ) ;Complex operator +( Complex& ) ;Complex operator +( double ) ;

private : double rea l , imag ;} ;

Complex Complex : : operator +( Complex& RHZ ){ return Complex ( r e a l + RHZ. rea l , I + RHZ. imag ) ; }

Complex Complex : : operator +( double x ){ return Complex ( r e a l + x , imag ) ; }

Complex operator +( double x , Complex& RHZ ){ return Complex ( RHZ. r e a l + x , RHZ. imag ) ;



/ / L i s t i n g 10.4 Overloaded opera tor f u n c t i o n sclass Complex { / / programmer−def ined data type

private :double rea l , imag ; / / p r i v a t e data

public :. . .Complex ( double r , double i ) ; / / genera l cons t ruc to rComplex operator +( const Complex &b ) const ; / / no change to t a r g e tvoid operator += ( const Complex &b ) ; / / t a r g e t ob jec t changesvoid operator += ( double b ) ; / / t a r g e t ob jec t changesvoid operator + ( ) const ; / / no change to t a r g e t

} ; / / end of c lass Complex



Complex : : Complex ( double r , double i ) / / genera l cons t ruc to r{ r e a l = r ; imag = i ; }

Complex Complex : : operator +( const Complex &b ) const / / over load +{

return Complex ( r e a l + b . rea l , imag + b . imag ) ; / / no change to t a r g e t}

Note: "Complex (real + b.real, imag + b.imag);" creates anunnamed object by calling the general constructor Complex(...)



/ / Note b . r e a l : a p r i v a t e data member can only be accessed/ / from w i t h i n the class , i . e . , the data can be accessed or/ / modi f ied only wh i le INSIDE ∗a∗ member f u n c t i o n o f the c lass .

/ / over load +=void Complex : : operator += ( const Complex &b ) { / / t a r g e t ob jec t changes

r e a l = r e a l + b . r e a l ; / / add to the r e a l component o f the t a r g e timag = imag + b . imag ; / / add to the imag component o f the t a r g e t

}

/ / over load += againvoid Complex : : operator += ( double b ) / / t a r g e t ob jec t changes

{ r e a l += b ; } / / add to r e a l component o f the t a r g e t

/ / over load + again −− p r e f i x ( see l a t e r )/ / opera tor + ( ) means p r e f i x => +z ; opera tor +( i n t ) means p o s t f i x => z+void Complex : : operator + ( ) const / / no change to t a r g e t

{ cout << " ( " << r e a l << " , " << imag << " ) " << endl ; } / / t a r g e t ’ s f i e l d s



/ / Test ingComplex x (20 ,40) , y (30 ,50) , z1 ( 0 , 0 ) , z2 ( 0 , 0 ) ; / / def ined , i n i t i a l i z e dcout << " Value of x : " ; +x ; / / same as x . opera tor + ( ) ;cout << " Value of y : " ; y . operator + ( ) ; / / anyth ing goesz1 = x . operator +( y ) ; / / use i n the f u n c t i o n c a l lcout << " z1 = x + y : " ; +z1 ;z2 = x + y ; / / same as z2=x . opera tor +( y ) ;cout << " z2 = x + y : " ; +z2 ;z1 += x ; / / same as z1 . opera tor +=(x ) ;cout << " Add x to z1 : " ; +z1 ;z2 += 30 .0 ; / / same as z2 . opera tor += (30 .0 ) ;cout << " Add 30 to z2 : " ; +z2 ;


Overloading: really needed?

class Complex {private : . . .public : . . .

} ;

/ / Without opera tor over load ing :Complex Complex add ( const Complex& x , const Complex& y ) ;Complex mul ( const Complex& x , const Complex& y ) ;Complex foo ( const Complex& a , const Complex& b , const Complex& c ){ / / a∗b + b∗c + c∗a

return add ( add ( mul ( a , b ) , mul ( b , c ) ) , mul ( c , a ) ) ;}

/ / With opera tor over load ing :Complex operator+ ( const Complex& x , const Complex& y ) ;Complex operator∗ ( const Complex& x , const Complex& y ) ;Complex f ( const Complex& a , const Complex& b , const Complex& c ){ return a∗b + b∗c + c∗a ; }


Overloading: really needed?

This class looks more complex..

But remember: the goal is to make the use of your classeasierit makes life easier for the users of your class, not for thedeveloper of the class!

Remember: in a reuse-oriented ( or OO) language, there willusually be many people who use your class

Your design should favor the many users


Overloading object assignment =

The = operator is a method related to a class

If an object is assigned to another object

a member-wise copy is performed

Example:Time t1 (12 , 01 , 30) , t2 ;t2 = t1 ; / / assignment method =/ / t2 gets the values 12 , 01 , 30

Be careful with member-wise copyIf member data is a pointer, the pointer address is copiedthis could be disastrous


Overloading object assignment =

If your object has a pointer to memory that was dynamicallyallocated previously, e.g., in the constructor, you will need anoverloaded assignment operator

Typical steps to overload the assignment operator:Make sure you aren’t assigning an object to itselfe.g. x = xDelete the associated memoryCopy all the membersReturn *this which is necessary to allow multipleassignment,e.g., x = y = z; → x = (y = z);


Overloading =: Example

class Person {private :char∗ name ;i n t i d ;

public :. . .

} ;

/ / Would l i k e toPerson a , b , c ; . . . b=a ; / / a : source ob jec t ; b : t a r g e t ob jec t


Overloading =: Example

Person& Person : : operator =( const Person& p ) {i f ( th is != &p ) { / / make sure not same ob jec t

delete [ ] name ; / / f r ee o ld name ’ s memory .

name = new char [ s t r l e n ( p . name ) + 1 ] ; / / get new spacei f ( p == NULL) { / / t e s t f o r success , q u i t i f no memory

cout <<" Person : : opera tor = ( ) : no memory" ;e x i t ( 1 ) ;

}

s t r cpy (name, p . name ) ; / / copy new namei d = p . i d ; / / copy i d

}

return ∗ th is ; / / r e t u r n r e f f o r m u l t i p l e assignment}


Overloading =: More Example

/ / an ob jec t i s i m p l i c i t l y accessed v ia the t h i s p o i n t e rClass t r i a n g l e {

/ / assume pt2d ( a po in t i n 2D) i s a c lass def ined p rev ious l yprivate :

pt2d∗ v e r t i c e s ;public :

. . ./ / cons t r uc to r & d e s t r u c t o rt r i a n g l e ( const t r i a n g l e& t ) ; / / copy cons t ruc to rt r i a n g l e operator =( const t r i a n g l e& t ) ; / / assignment opera tor. . . .

} ;


Overloading =: More Example

/ / copy assignment opera tort r i a n g l e t r i a n g l e : : operator =( const t r i a n g l e& t ) {

i f ( th is != & t ) / / s e l f−assignment l i k e t = t ?for ( i n t i =0; i <3; i ++)

v e r t i c e s [ i ] = t . v e r t i c e s [ i ] ;return ????? / / what do you t h i n k ? r e t u r n t h i s ; or r e t u r n ∗ t h i s

}

/ / Test ingpt2d p1 (20 ,40) , p2 (30 ,90) , p3 (40 ,10 ) ;t r i a n g l e a ( p1 , p2 , p3 ) , b ;b = a ; / / OK?


Copy constructor vs. =

What is the difference between a copy constructor and anoverloaded assignment operator?

A copy constructor:constructs a new object by using the content of the argumentobject

Complex a , c ; / / dec lare a & c c a l l d e f a u l t cons t ruc to r sComplex b=a ; / / dec lare b and copy a to i t by c a l l i n g copy cons t ruc to r

An overloaded assignment operator: assigns the contents ofan existing object to another existing object of the same class

c=a ; / / somewhere i n your program : an overloaded assignment opera tor


Outline



What can/cannot be

overloaded?

Two ways to overload:

member and non-member

operators



Multi-Overloading

Assignment operator versus

Copy constructor


Chain statements

Special operators:

[]() + + −− <<>>

A complete example with

member functions


friend functions


Operator overloading: Rules

You can only overload existing C++ built-in operatorscan only redefine existing operators - no making up new ones

You cannot change the meaning of existing operators forbuilt-in types by overloading them for built-in types

You can only overload operators for programmer definedtypes⇒ You can define the meaning of the operator for theprogrammer-defined type

You cannot change the number of operands that the operatortakes

You cannot change the precedence of the operator or itsassociativity



New behaviors for operators should be consistent with defaultbehavior

At least one operand of any overloaded operator must be ofsome user-defined type (i.e., class)

Make sure the class design support its own methods (e.g.,define constructor)

Use the const keyword for parameters of overloaded operatorfunctions that do not change the value of their parameters

Mark the function as constant if the target object does notchange

The absence of the const keyword is the evidence that thefunction changes the target object



An overloaded unary operator declared as member functiontakes no parameters, i.e., works on target object

An overloaded unary operator declared as friend functiontakes one parameter which must be object of the same classor a pointer to an object of the class

An overloaded binary operator declared as member functiontakes one parameter:

Target object is always used on left side of operator

Object on right side of operator taken from parameter passed

An overloaded binary operator declared as friend functiontakes two parameters:

1st is on left side of operator & 2nd is on right side of operator


Overloading: Chain statements

What does object.method1().method2() means?

First object.method1() is executed

This returns some object which might bea reference to object, i.e., method1() might end with return*this; ORsome other object

Let’s call the returned object objectB

Then objectB becomes the this object of method2()

Examples of method chaining:in the iostream library, e.g., cout << x << y which worksbecause cout << x is a function that returns cout

in assignments x = y = z;


Overloading: Chain statements

Recall: After a function return statement, a temporary copy ofthe return value is returned

Complex& Complex : : operator = ( . . . . ) {Complex temp ;/ / r e t u r n temp ;return ∗ th is ;

} / / temp i s destroyed/ / a copy of the r e t u r n ob jec t value i s made/ / t h i s copy i s a v a i l a b l e to the c a l l e r o f the f u n c t i o n

x = y = z; means x = (y = z); //precedence rules of the = Op

Sometime you want: (x = y) = z;Thus (x = y) should return an object if you overload =

cout << x << y; means (cout << x) << y;Recall: Input/Output statements executes from left to right


Outline



What can/cannot be

overloaded?



operators



Multi-Overloading


Copy constructor


Chain statements

Special operators: ⇐

[]() + + −− <<>>


member functions


friend functions


Overloading: Insertion/Extraction

The iostream library is an object-oriented library that providesinput and output functionality using streams

A stream is an abstraction that represents a device in whichinput and output operations are performed

A stream can basically be represented as an indefinite sourceor destination of characters

The number of character to be in/output is not needed to beknown in advance

We can simply write characters to the stream or get them untilwe find it convenient or until the source of characters exhausts



The streams are generally associated to a physical source ordestination of characters:

Examples: a disk file, the keyboard or a text console, ..

The characters we get or write to/from our abstraction calledstream are physically input/output to the physical device

For example, file streams are C++ objects to manipulate andinteract with filesOnce a file stream is used to open a file, any input or outputoperation performed on that stream is physically reflected inthe file.



C++ provides a standard iostream library to work with streams

The iostream library provides us with standard IO-objectssuch as cin, cout, & cerr.

These objects are used to perform input/output operations onthe standard input and output devices



The Stream Insertion/Extraction Operators are binaryoperators:

cout << i; with cout as an object of type ostream where<< is an output operatorcin >> j; with cin as object of type istream where >> isinput operatorQuestion: should << or >> be overloaded as a memberfunction of ostream class (e.g., cout.operator << (i);) oras a friend function of ostream (e.g.,operator << (cout, i);?



Recall: LH Obj. has to be of the same type as the class weare writing the op for

The LH side is an iostream object

Therefore, we cannot have access to the this pointer if weimplemented the overload using a member function

Thus: stream Insertion/Extraction operators MUST beoverload with non-member functions (i.e. friends)


Stream Insertion >> Overload

class Time {f r iend i s t ream& operator >>( is t ream &, Time &) ;public :

void Time ( i n t = 0 , i n t = 0 , i n t = 0 ) ;private :

i n t hour ; i n t min ; i n t sec ;} ;

Time : : Time ( i n t h , i n t m, i n t s ) { hour = h ; min = m ; sec = s ; }


Stream Insertion >> Overload

i s t ream& operator >>( is t ream& is , Time& T) {i s >> setw ( 2 ) >> T . hour ; i s . ignore ( ) ;i s >> setw ( 2 ) >> T . min ; i s . ignore ( ) ;i s >> setw ( 2 ) >> T . sec ;return i s ;

}

/ / Use : what the user enters goes to T . hour T . min & T . secTime noon ;cout <<" enter data f o r t ime " ; c in >>noon ;

Note: “cin” and “noon” are objects of different types


Stream Extraction << Overload

class Complex {f r iend ostream& operator <<(ostream&, const Complex&) ;public :

void Complex ( double = 0 , double = 0 ) ;private :double rea l , imag ;

} ;

Complex : : Complex ( double re , double im ) { r e a l = re ; imag = im ; }


Stream Extraction << Overload

ostream& operator <<(ostream& os , const Complex& Z) {os << ( (Z . r e a l ) ? Z . r e a l : ’ ’ ) ;i f (Z . imag )os << " + i " << Z . imag ;

return os ;}

/ / Test ing :complex a ( 1 , 5 ) ;cout << a ; / / p r i n t s 1 + i 5

Note: “cout” and “a” are objects of different types


Overloading: Subscript op. []

Does f() = 7 makes sense? What about a[i] = 7 ? What is []?



Does f() = 7 makes sense? What about a[i] = 7 ? What is []?

Subscript operator [] is used for class Array

It is unique: has an object as its LH operand and a subscriptindex as a RH operand

Prototype Syntax:<re turn type> & operator [ ] ( <index type> ) ;

The operator function takes the implied object and a subscriptindex parameter

Usually returns a reference to the indexed value to allowassignment, e.g. object[2] = 10;



class Array {private :

f l o a t ∗ data ;public :

i n t s ize ( ) const ;f l o a t& operator [ ] ( i n t index ) ;. . .

} ;

f l o a t& operator [ ] ( i n t index ) { return data [ index ] ; }

/ / Test ingArray a ;for ( i n t i = 0 ; i < a . s ize ( ) ; ++ i )

a [ i ] = 7 ; / / This l i n e invokes Array : : opera tor [ ] ( i n t )


Overloading: Function call Op. ()

Also called “Parenthesis Op”

The Parenthesis operator gives the user a mean to pass anyparameters to an object and return anything

It is like a function call with no function name

Examples:Complex Z ( 1 , 1 ) ; / / Const ruc tor \ \Complex Z( ’ i ’ ) ; / / Parenthes is Operator/ / pass the charac te r i to the ob jec t Z and do something/ / e . g . , ou tput the values o f t h i s ob jec ts


Overloading: Function call Op. ()

class Poin t {private : i n t x , y ;public :

Po in t ( i n t i , i n t j ) { x= i ; y= j ; }Po in t operator ( ) ( i n t a , i n t b )

{ Po in t tmp ; tmp . x = x + a ; tmp . y = y + b ; return tmp ; }Po in t operator ( ) ( i n t a )

{ x = x + a ; y = y + a ; return ∗ th is ; }} ;/ / Usage :

Poin t p (1 , 2 ) ; / / genera l cons t ruc to rPoin t q ( p ) ; / / copy cons t ruc to r

p = q ( 5 , 3 ) ; / / invoke opera tor ( )cout << p << q ; / / p =6 ,5; q =1 ,2

q ( 6 ) ; / / invoke opera tor ( )cout << p << q ; / / p=6 ,5: q=7 ,8


Increment/Decrement Op. ++/−−

a++; // post-fix ++a; // pre-fix

When used on their own (as above) the prefix and postfixhave the same effect BUT within an expression there is a bigdifference

Prefix notation will increment the variable BEFORE theexpression is evaluated

Postfix notation will increment AFTER the expressionevaluation

main ( ) { main ( ) {i n t a=1; i n t a=1;cout << " a i s "<< ++a ; cout << " a i s " << a++;

/ / ou tput : 2 / / Output 1cout << " a f t e r : a i s "<< a ; cout << " a f t e r : a i s " << a ;/ / Output : 2 / / Output : 2

} }


Increment/Decrement Op. ++/−−

There are two types of increment/decrement: Pre and Post

The syntax for pre-increment/decrement is the same as theusual operator overload

<re turn_type > operator + + ( ) ;

The syntax for post-increment/decrement is:

<re turn_type > operator ++( i n t ) ;

int is used to only indicate that the operator is postfixOtherwise how would the compiler distinguish the twofunctions?

Complex Complex : : operator + + ( ) ;Complex Complex : : operator ++( i n t i ) ;


Increment/Decrement Op ++/−−

Examples:Complex Complex : : operator ++() { / / p r e f i x as i n ++z

re ++; im++ / / increment both componentsreturn ∗ th is ;

}

Complex Complex : : operator ++( i n t i ) { / / p o s t f i x as i n z++Complex temp=∗ th is ;re ++; im++ / / increment both components

return temp ;}

/ / UseComplex aa =5.0 ; / / aa=Complex ( 5 . 0 ) ;Complex bb=++aa ; / / bb=Complex ( 6 , 1 ) , aa=Complex (6 ,1 )Complex cc=aa++; / / cc=Complex (6 ,1 ) aa=Complex (7 ,2 )


Operator overloading: Destructor?

Quiz 1: Can I overload the destructor of a class?


Operator overloading: Destructor?

Quiz 1: Can I overload the destructor of a class?

NO. A class can have only one destructor

You never explicitly call a destructor (almost never)

Quiz 2: In which order are objects destroyed?

Objects are destroyed in the reverse order they are declared


Outline



What can/cannot be

overloaded?



operators



Multi-Overloading


Copy constructor


Chain statements

Special operators:

[]() + + −− <<>>


member functions


friend functions


Overloading: Complete example with member functions

/ / L i s t i n g 10.6 Class Rat iona l t h a t supports mixed types i n expressions .class Rat iona l {

private :long nmr , dnm; / / p r i v a t e datavoid normal ize ( ) ; / / p r i v a t e member f u n c t i o n

public :Ra t iona l ( long n=0 , long d=1) / / general , conversion , d e f a u l t cons t r uc to r{ nmr = n ; dnm = d ;

this−>normal ize ( ) ; }Ra t iona l operator + ( const Rat iona l &x ) const ; / / constant t a r g e tRat iona l operator − ( const Rat iona l &x ) const ;Ra t iona l operator ∗ ( const Rat iona l &x ) const ;Ra t iona l operator / ( const Rat iona l &x ) const ;void operator += ( const Rat iona l &x ) ; / / t a r g e t changesvoid operator −= ( const Rat iona l &x ) ;void operator ∗= ( const Rat iona l &x ) ;void operator /= ( const Rat iona l &x ) ;


Overloading: Example with member functions

bool operator == ( const Rat iona l &other ) const ; / / constant t a r g e tbool operator < ( const Rat iona l &other ) const ;bool operator > ( const Rat iona l &other ) const ;void show ( ) const ;

} ; / / end of c lass s p e c i f i c a t i o n

Rat iona l Ra t iona l : : operator + ( const Rat iona l &x ) const{ return Rat iona l ( nmr∗x .dnm + x . nmr∗dnm, dnm∗x .dnm ) ; } / / e . g . , 1/4+3/2=14/8

Rat iona l Ra t iona l : : operator − ( const Rat iona l &x ) const{ return Rat iona l ( nmr∗x .dnm − x . nmr∗dnm, dnm∗x .dnm ) ; } / / e . g . , 3/2−1/4=10/8

Rat iona l Ra t iona l : : operator ∗ ( const Rat iona l &x ) const{ return Rat iona l ( nmr ∗ x . nmr , dnm ∗ x .dnm ) ; }

Ra t iona l Ra t iona l : : operator / ( const Rat iona l &x ) const{ return Rat iona l ( nmr ∗ x .dnm, dnm ∗ x . nmr ) ; }



void Rat iona l : : operator += ( const Rat iona l &x ) {nmr = nmr ∗ x .dnm + x . nmr ∗ dnm; / / 3 /8+3/2=(6+24)/16=15/8dnm = dnm ∗ x .dnm; / / n1 / d1+n2 / d2 = ( n1∗d2+n2∗d1 ) / ( d1∗d2 )this−>normal ize ( ) ;

}

void Rat iona l : : operator −= ( const Rat iona l &x ) {nmr = nmr ∗ x .dnm − x . nmr ∗ dnm; / / 3 /8+3/2=(6+24)/16=15/8dnm = dnm ∗ x .dnm; / / n1 / d1+n2 / d2 = ( n1∗d2−n2∗d1 ) / ( d1∗d2 )this−>normal ize ( ) ;

}

void Rat iona l : : operator ∗= ( const Rat iona l &x ) {nmr = nmr ∗ x . nmr ; dnm = dnm ∗ x .dnm;this−>normal ize ( ) ;

}



void Rat iona l : : operator /= ( const Rat iona l &x ) {nmr = nmr ∗ x .dnm; dnm = dnm ∗ x . nmr ;this−>normal ize ( ) ;

}

bool Rat iona l : : operator == ( const Rat iona l &other ) const{ return ( nmr ∗ other .dnm == dnm ∗ other . nmr ) ; }

bool Rat iona l : : operator < ( const Rat iona l &other ) const{ return ( nmr ∗ other .dnm < dnm ∗ other . nmr ) ; }

bool Rat iona l : : operator > ( const Rat iona l &other ) const{ return ( nmr ∗ other .dnm > dnm ∗ other . nmr ) ; }

void Rat iona l : : show ( ) const{ cout << " " << nmr << " / " << dnm; }



void Rat iona l : : normal ize ( ) {/ / p r i v a t e member f u n c t i o n

i f ( nmr == 0) { dnm = 1; return ; }i n t s ign = 1;i f ( nmr < 0) { s ign = −1; nmr = −nmr ; }i f (dnm < 0) { s ign = −s ign ; dnm = −dnm; }

long gcd = nmr , value = dnm; / / s t a r t l ook ing f o r g rea tes t common d i v i s o rwhile ( value != gcd ) / / s top when the GCD i s found

{i f ( gcd > value )

gcd = gcd − value ; / / sub t r ac t smal le r number from the grea te relse value = value − gcd ;

}nmr = s ign ∗ ( nmr / gcd ) ; dnm = dnm/ gcd ; / / denominator i s always p o s i t i v e

}



−−>Test ing : now we can work w i th r a t i o n a l numbers as wi th o ther numbersRat iona l a ( 1 , 4 ) , b ( 3 , 2 ) , c , d ;c = a + 5; / / c = 5 + a? l a t e ra . show ( ) ; cout << " + " << 5 << " =" ; c . show ( ) ; cout << endl ;d = b − 1;b . show ( ) ; cout << " − " << 1 << " =" ; d . show ( ) ; cout << endl ;c = a ∗ 7;a . show ( ) ; cout << " ∗ " << 7 << " =" ; c . show ( ) ; cout << endl ;d = b / 2 ;b . show ( ) ; cout << " / " << 2 << " =" ; d . show ( ) ; cout << endl ;c . show ( ) ;c += 3;cout << " += " << 3 << " =" ; c . show ( ) ; cout << endl ;d . show ( ) ;d ∗= 2;cout << " ∗= " << 2 << " =" ; d . show ( ) ; cout << endl ;i f ( b < 2){ b . show ( ) ; cout << " < " << 2 << endl ; }


Overloading: Complete example with friend functions

/ / L i s t i n g 10 .7 : Class Rat iona l using f r i e n d f u n c t i o n s/ / to suppor t mixed type expressionsclass Rat iona l {

private :long nmr , dnm; / / p r i v a t e datavoid normal ize ( ) ; / / p r i v a t e member f u n c t i o n

public :Ra t iona l ( long n=0 , long d=1) / / general , conversion , d e f a u l t cons t r uc to r{ nmr = n ; dnm = d ;

this−>normal ize ( ) ; }f r iend Rat iona l operator + ( const Rat iona l &x , const Rat iona l &y ) ;

/ / does t h i s + opera tor handle a l l cases R+W, R+6 , 6+R?/ / Maybe convers ion takes place?

f r iend Rat iona l operator − ( const Rat iona l &x , const Rat iona l &y ) ;f r iend Rat iona l operator ∗ ( const Rat iona l &x , const Rat iona l &y ) ;f r iend Rat iona l operator / ( const Rat iona l &x , const Rat iona l &y ) ;f r iend void operator += ( Rat iona l &x , const Rat iona l &y ) ;


Overloading: Example with friend functions

fr iend void operator −= ( Rat iona l &x , const Rat iona l &y ) ;f r iend void operator ∗= ( Rat iona l &x , const Rat iona l &y ) ;f r iend void operator /= ( Ra t iona l &x , const Rat iona l &y ) ;f r iend bool operator == ( const Rat iona l &x , const Rat iona l &y ) ;f r iend bool operator < ( const Rat iona l &x , const Rat iona l &y ) ;f r iend bool operator > ( const Rat iona l &x , const Rat iona l &y ) ;void show ( ) const ;

} ; / / end of c lass s p e c i f i c a t i o n

void Rat iona l : : show ( ) const{ cout << " " << nmr << " / " << dnm; }

void Rat iona l : : normal ize ( ) / / p r i v a t e member f u n c t i o n{ . . . . }



Rat iona l operator + ( const Rat iona l &x , const Rat iona l &y ){ return Rat iona l ( y . nmr∗x .dnm + x . nmr∗y .dnm, y .dnm∗x .dnm ) ; }

Ra t iona l operator − ( const Rat iona l &x , const Rat iona l &y ){ return Rat iona l ( x . nmr∗y .dnm − y . nmr∗x .dnm, x .dnm∗y .dnm ) ; }

Ra t iona l operator ∗ ( const Rat iona l &x , const Rat iona l &y ){ return Rat iona l ( x . nmr ∗ y . nmr , x .dnm ∗ y .dnm ) ; }

Ra t iona l operator / ( const Rat iona l &x , const Rat iona l &y ){ return Rat iona l ( x . nmr ∗ y .dnm, x .dnm ∗ y . nmr ) ; }

void operator += ( Rat iona l &x , const Rat iona l &y ){x . nmr = x . nmr ∗ y .dnm + y . nmr ∗ x .dnm; x .dnm ∗= y .dnm;x . normal ize ( ) ;

}



void operator −= ( Rat iona l &x , const Rat iona l &y ){x . nmr = x . nmr∗y .dnm + y . nmr∗x .dnm; x .dnm ∗= y .dnm;x . normal ize ( ) ;

}void operator ∗= ( Rat iona l &x , const Rat iona l &y )

{x . nmr ∗= y . nmr ; x .dnm ∗= y .dnm;x . normal ize ( ) ;

}void operator /= ( Ra t iona l &x , const Rat iona l &y )

{x . nmr = x . nmr ∗ y .dnm; x .dnm = x .dnm ∗ y . nmr ;x . normal ize ( ) ;

}



bool operator == ( const Rat iona l &x , const Rat iona l &y ){ return ( x . nmr ∗ y .dnm == x .dnm ∗ y . nmr ) ; }

bool operator < ( const Rat iona l &x , const Rat iona l &y ){ return ( x . nmr ∗ y .dnm < x .dnm ∗ y . nmr ) ; }

bool operator > ( const Rat iona l &x , const Rat iona l &y ){ return ( x . nmr ∗ y .dnm > x .dnm ∗ y . nmr ) ; }



/ Test ingRat iona l a ( 1 , 4 ) , b ( 3 , 2 ) , c , d ;c = 5 + a ;cout << " " << 5 << " +" ; a . show ( ) ; cout << " =" ; c . show ( ) ; cout << endl ;d = 1 − b ; / / operator −(Ra t iona l ( 1 ) , b ) ;cout << " 1 −" ; b . show ( ) ; cout << " = " ; d . show ( ) ; cout << endl ;c = 7 ∗ a ; / / opera tor ∗ ( Ra t iona l ( 7 ) , a ) ;cout << " 7 ∗ " ; a . show ( ) ; cout << " = " ; c . show ( ) ; cout << endl ;d = 2 / b ; / / opera tor / ( Ra t iona l ( 2 ) , b ) ;cout << " 2 / " ; b . show ( ) ; cout << " = " ; d . show ( ) ; cout << endl ;c . show ( ) ;c += 3; / / opera tor +=(c , Ra t iona l ( 3 ) ) ;cout << " += " << 3 << " =" ; c . show ( ) ; cout << endl ;d . show ( ) ;d ∗= 2; / / opera tor ∗=(d , Ra t iona l ( 2 ) )cout << " ∗= " << 2 << " =" ; d . show ( ) ; cout << endl ;



i f ( a < 5) cout << " a < 5\ n " ; / / operator <(a , Ra t iona l ( 5 ) ) ;i f (1 < b ) cout << " 1 < b \ n " ; / / operator <( Rat iona l ( 1 ) , b ) ;i f (1 < 5) cout << " 1 < 5\ n " ; / / b u i l t −i n i n e q u a l i t y opera tori f ( d ∗ b − a == c − 1) cout << " d∗b−a == c−1 ==" ;( c − 1 ) . show ( ) ; cout << endl ;


Outline



What can/cannot be

overloaded?



operators



Multi-Overloading


Copy constructor


Chain statements

Special operators:

[]() + + −− <<>>


member functions


friend functions


Overloading and class design

First think about what the object logically represents,then about how to physically build it

Design your classes from the outsideinterfaces first, i.e., set of public methods

A good interface provides a simplified view that is expressedin the vocabulary of a user

Example, a linked list:A linked list is sequence of elements – what is it?It is built as a chain of nodes – How it is built?The value in the nodes is solely the responsibility of thelinked list userHow would you design a class LinkedList?Start with what a LinkedList is not how it is physicallybuilt!



Logical interface/abstraction from the user’s point of view:A → B → C → Doperator++ to go to the next elementa get()/set() to access its value stored in the Nodebegin/end method to go to beginning or end or a listappend method to adds an element after the end

Physical structureThe Heap

4

A DCB

prepointer

head

Length



A linked list of objects of type Tclass Node {

f r iend class L inkedL i s t ;

private :Node ( const T& elem , Node∗ next = NULL ) ;

T elem ; / / T i s the type of elem such as i n t or f l o a tNode∗ next ;

} ;



class L inkedL i s t {private :Node∗ const head ; / / dummy head node ; p o i n t e r never changes .i n t l eng th ; / / how many elem are i n l i s t ? (dummy node not counted )Node∗ precPo in te r ; / / po in t s 1 node before l o g i c a l p o s i t i o n o f cpo in te r

public :L i nkedL i s t ( ) ;~ L inkedL i s t ( ) ;

/ / movementbool operator ++() ;bool operator−−() ;void toBegin ( ) ;void toEnd ( ) ;



/ / s t a t ebool atEnd ( ) const ; / / pre p o i n t e r i s one beyond l a s t element i n l i s t

/ / mutat ionbool set ( const T& elem ) ; / / change elem at p o i n t e rbool i n s e r t ( const T& elem ) ; / / put new elem i n f r o n t o f o ld elembool remove ( ) ; / / remove elem at p o i n t e r ; p o i n t e r moves on to next

/ / i nspec t i oni n t l eng th ( ) const ; / / # elems ( not #nodes ) i n l i s tbool get ( T& elem ) const ; / / r e t u r n elem at p o i n t e r/ / and more . . .

} ;


Documents

COEN244: Operator overloading - Encsusers.encs.concordia.ca/~amer/teach/coen244/notes/opOverloading.pdfc A. Amer Operator Overloading 26. Overloading object assignment = The = operator