Assignment Operator - Object Oriented Programming - Lecture Slides, Exercises of Object Oriented Programming

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Object-Oriented Programming

(OOP)

Lecture No. 18

• Modifying:

class String{

public:

void operator =(const String &);

int main(){

String str1(“ABC");

String str2(“DE”), str3(“FG”);

str1 = str2; // Valid…

str1 = str2 = str3; // Error…

return 0;

}

• str1=str2=str3 is resolved as:

str1.operator=(str2.operator=

(str3))

Return type is

void. Parameter

can’t be void

String & String :: operator = (const String & rhs){

size = rhs.size;

delete [] bufferPtr;

if(rhs.size != 0){

bufferPtr = new char[rhs.size+1];

strcpy(bufferPtr,rhs.bufferPtr);

}

else bufferPtr = NULL;

return *this;

}

void main(){

String str1(“AB"); String str2(“CD”), str3(“EF”); str1 = str2; str1 = str2 = str3; // Now valid…

}

int main(){

String str1("Fakhir");

// Self Assignment problem…

str1 = str1;

return 0;

}

… // size = rhs.size; // delete [] bufferPtr; …

• Result of str1 = str

str

Fakhir

• Now self-assignment is properly handled:

int main(){

String str1("Fakhir");

str1 = str1;

return 0;

}

• Solution: modify the operator= function

as follows:

class String{

public:

const String & operator=

(const String &);

But we can do the following with

primitive types:

int main(){

int a, b, c;

(a = b) = c;

return 0;

}

Other Binary operators

• Overloading += operator:

class Complex{

double real, img;

public:

Complex & operator+=(const Complex &

rhs);

Complex & operator+=(count double &

rhs);

Other Binary operators

Complex & Complex::operator +=

(const double & rhs){

real = real + rhs;

return * this;

Other Binary operators

int main(){

Complex c1, c2, c3;

c1 += c2;

c3 += 0.087;

return 0;