Why does the address of the assigned object not change in C++?

Question

In this C++ example, a class C has a default constructor, a copy constructor and an assignment operator:

struct C {
    C();
    C(const C& c);
    C& operator=(const C& c);
};

The implementation is as follows with some output for tracking the objects. I added some example addresses in comments on the lines as reference to the main program below.

#include "C.h"
#include <iostream>
using namespace std;

C::C() {
    cout << "Standard constructor." << endl;
}

C::C(const C& c) {
    cout << "Copy constructor." << endl;
}

C& C::operator=(const C& c) {
    cout << "Address of reference argument: &c = " << &c << endl;      // F9B4
    cout << "Address of this: &*this =           " << &*this << endl;  // F8D4
    return *this;
}

I'll be using a factory method that returns anonymous objects:

C foo() {
    return C();
}

Now in the following program, I create a named object c (see (1)) and create an anonymous object (2) which I assign to c (3). I expected that c then has the address of the anonymous object after the assignment.

int main()
{
    C c;  // (1)
    cout << "Address of initial c from main: &c = " << &c << endl;                  // F8D4
    c=foo();  // (2) and (3)
    cout << "Address of initial c from main after assignment: &c = " << &c << endl; // Expected F9B4 but is F8D4
}

Answer 1

Objects in C++ never change addresses.

An assignment operator is no different from any other method, normally it just goes over all member variables and assigns new values to them, copied from the corresponding members of the other object.