Increment dan Decrement Operator Overloading di C++

In this tutorial, increment ++ and decrements -- operator are overloaded in best possible way, i.e., increase the value of a data member by 1 if ++ operator operates on an object and decrease value of data member by 1 if -- operator is used.

Increment Operator Overloading

Output

i=0
i=1

Explanation
Initially when the object obj is declared, the value of data member i for object obj is 0( constructor initializes i to 0). When ++ operator is operated on obj, operator function void operator++( ) is invoked which increases the value of data member i to 1.
This program is not complete in the sense that, you cannot used 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 overlading. */
#include <iostream>
using namespace std;
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;
}

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 Postfix Operator

Overloading of increment operator up to this point is only true if it is used in prefix form. This is the modification of above program to make this work both for prefix form and postfix form.

/* C++ program to demonstrate the working of ++ operator overlading. */
#include <iostream>
using namespace std;
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();

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

/* Assigns value of obj to obj1++ then only operator function is called. */ 
    obj1=obj++;     
    obj.Display();
    obj1.Display();
    return 0;
}

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 postfix form; Check operator ++ (int) is invoked. Notice, the int inside bracket. This int gives information to the compiler that it is the postfix version of operator. Don't confuse this int doesn't indicate integer.

 

Operator Overloading of Decrement -- Operator

Decrement operator can be overloaded in similar way as increment operator. Also, unary operators like: !, ~ etc can be overloaded in similar manner.