Increment ++ and Decrement -- Operator Overloading in C++.

Increment ++ and Decrement -- Operator Overloading

Increment Operator Overloading

/* C++ program to demonstrate the overloading of ++ operator. */

#include <iostream.h>

#include <conio.h>

class Check

{

    private:

       int i;

    public:

       Check(): i(0) {  }

       void operator ++()

          { ++i; }

       void Display()

          { cout<<"i="<<i<<endl; }

};

int main()

{

    Check obj;

 /* Displays the value of data member i for object obj */

    obj.Display();

/* Invokes operator function void operator ++( ) */

    ++obj;

 

/* Displays the value of data member i for object obj */

    obj.Display(); 

    return 0;

}

Expected Output:-

i=0

i=1

Explanation

• Initially when the object obj is declared the data member I value for the the object is 0(constructor initializes I to 0). 
• The operator function void operator++) (is invoked when++ operator is worked on obj, which increases the data member value I to 1. 
• This program is not complete in the sense that you can not use code:

obj1=++obj;

• It is because the return type of operator function in above program is void. Here is the little modification of above program so that you can use code obj1=++obj.

/* C++ program to demonstrate the working of ++ operator overloading. */

#include <iostream.h>

#include <conio.h>

class Check

{

  private:

    int i;

  public:

    Check(): i(0) {  }

    Check operator ++() /* Notice, return type Check*/

    {

       Check temp;  /* Temporary object check created */

       ++i;         /* i increased by 1. */

       temp.i=i;   /* i of object temp is given same value as i */

       return temp; /* Returning object temp */

    }

   void Display()

      { cout<<"i="<<i<<endl; }

};

int main()

{

    Check obj, obj1;

    obj.Display();

    obj1.Display();

    obj1=++obj;

    obj.Display();

    obj1.Display();

    return 0;

}

Expected Output:-

i=0

i=0

i=1

i=1

• This program is similar to above program. 
• The only difference is that, the return type of operator function is Check in this case which allows to use both codes ++obj; obj1=++obj;. 
• It is because, temp returned from operator function is stored in object obj. Since, the return type of operator function is Check, you can also assign the value of obj to another object. 
• Notice that, = (assignment operator) does not need to be overloaded because this operator is already overloaded in C++ library.

Operator Overloading of Post-fix Operator

• Up to this point, increment operator overloading is valid only if it is used in prefix form. 
• It is the software update above to make it function with both prefix form as well as post-fix form.

/* C++ program to demonstrate the working of ++ operator overloading. *

#include <iostream.h>

#include <conio.h>

class Check

{

  private:

    int i;

  public:

    Check(): i(0) {  }

    Check operator ++ ()

    {

        Check temp;

        temp.i=++i;

        return temp;

    }

/* Notice int inside barcket which indicates postfix increment. */

    Check operator ++ (int)

    {

        Check temp;

        temp.i=i++;

        return temp;

    }

   void Display()

      { cout<<"i="<<i<<endl; }

};

int main()

{

    Check obj, obj1;   

    obj.Display();

    obj1.Display();

    obj1=++obj;     /* Operator function is called then only value of obj is assigned to obj1. */

    obj.Display();

    obj1.Display();

    obj1=obj++;     /* Assigns value of obj to obj1++ then only operator function is called. */

    obj.Display();

    obj1.Display();

    return 0;

}

Expected Output:-

i=0

i=0

i=1

i=1

i=2

i=1

• When increment operator is overloaded in prefix form; Check operator ++ () is called but, when increment operator is overloaded in post-fix form; Check operator ++ (int) is invoked. 
• Notice, the int inside bracket. This int gives information to the compiler that it is the post-fix version of operator. Don't confuse this int doesn't indicate integer.

Operator Overloading of Decrement -- Operator

#include <iostream.h>

#include <conio.h>

class Check

{

  private:

    int i;

  public:

    Check(): i(3) {  }

    Check operator -- ()

    {

        Check temp;

        temp.i = --i;

        return temp;

    }

    // Notice int inside barcket which indicates postfix decrement.

    Check operator -- (int)

    {

        Check temp;

        temp.i = i--;

        return temp;

    }

    void Display()

    { cout << "i = "<< i <<endl; }

};

int main()

{

    Check obj, obj1;    

    obj.Display(); 

    obj1.Display();

    // Operator function is called, only then value of obj is assigned to obj1

    obj1 = --obj;

    obj.Display();

    obj1.Display();

    // Assigns value of obj to obj1, only then operator function is called.

    obj1 = obj--;

    obj.Display();

    obj1.Display();

    return 0;

}

Expected Output:-

i = 3

i = 3

i = 2

i = 2

i = 1

i = 2