Unary operator overloading : What's the use of int argument in postfix form

[subith86]

Here the value of int x is always zero. I don't know if there is any case where it is non-zero. If so pls reply.

int OperOverld::operator--(int x)
{
    cout<<x<<endl; //always zero
    //some code goes here
    //...
    //...
    //...
    //atlast return some int
}

[Narue]

It's simply a way to distinguish between the prefix and postfix overloads. If the two didn't differ in the parameter list, they would be ambiguous.

[subith86]

ok thanks, one more doubt regarding the same.
The ++ operator is overloaded as shown in the above post. I tried this. And it worked

int j = obj++;

But later i thought of overloading + operator instead of ++ to do the same job. Now the below line gave compiler error.

int j = obj2+;

But this worked after I added another integer after +

int j = obj2+100;

I verified 100 is not going into the overloaded operator function. The function was simply returning a int which could have been assigned to j. It didn't happen. But when another constant is there on the right side it worked. Why is it so?

[Narue]

operator+ is still a binary operator, whether you use the other operand or not. You could overload the unary operator+, but it's a prefix operator.