Why is the copy constructor being called here?

Question

#include <iostream>
using namespace std;
class A
{
    int x;
public:

    A(int a)
    {   
        x = a;
        cout << "CTOR CALLED";
    }
    A(A &t)
    {
        cout << "COPY CTOR CALLED";
    }
    void display()
    {
        cout << "Random stuff";
    }
    A operator = (A &d)
    {   
        d.x = x;
        cout << "Assignment operator called";
        return *this;
    }
};

int main()
{
    A a(3), b(4);
    a = b;
    return 0;
}

The output of this code is :

CTOR CALLED
CTOR CALLED
Assignment operator called
COPY CTOR CALLED

When I used the watch in visual studio it showed that the value of x in a had been changed even before the overloaded assignment operator was called.

So why is the copy constructor even being called here?

Answer 1

Because you return by value from the assignment operator. It should return a reference :

A& operator = (A &d) { ... }

Answer 2

As @SomeProgrammerDude already said, it's because you return by value, not by reference like you usually do with an assignment operator. I would like to add that your assignment operator is currently the wrong way around:

A operator = (A &d)
{   
    d.x = x;
    cout << "Assignment operator called";
    return *this;
}

Usually you pass d by const reference because we want to alter the members of this . You are changing the members of d . Which would functionally turn your a = b; into b = a; since a = b; is actually changing members of b now !

Making d a const& prevents these mistakes.

You should change it to:

A& operator = (A const &d)
{   
    x = d.x;
    cout << "Assignment operator called";
    return *this;
}