Question

I was going through some book and I decided to write my own implementation of post-increment operator for user defined type. Here is the code.

#include <iostream>

using namespace std;

class X
{
    int a;

    public:
    X(int x=1):a(x){}

    X operator++(int)
    {
        X oldobj = *this;
        (*this).a++;
        return oldobj;
    }

int get(){return a;}

};

int main()
{
    X obj,obj2;
    obj++ = obj2;
    cout<< obj.get() << endl;
    return 0;
}   

I would expect the output to be 1 since obj2’s value will be copied after the increment is done. But the output was 2.

Thoughts?

P.S. I know this code will not win any medals and its fallacies. It is just for my understanding. Incidentally, ++obj = obj2 returns 1;

Is the behavior undefined?

Answer 1

As your syntax tells you, the postfix operator returns a copy of the old value, so that’s what gets incremented, not your object.

Basically,

obj++ = obj2;

Will do this:

X tempObj = obj;
obj ++;
tempObj = obj2;

You’re assigning obj2 to a temporary variable.