Question

#include <iostream>
using namespace std;

int main ()
{
    int    x = 0;
    int    y = 1;

    int& z = x;
    z = x;
    z = y;

    cout << "\nx: " << x;
    cout << "\ny: " << y;

    cout << "\nz: " << z;
    return 0;
}

**

EDIT:

**
This code returns 1 for all 3 cases. Shouldn’t this be an error instead?

8.5.3 section of C++ standard says:

A reference cannot be changed to refer to another object after
initialization. Note that initialization of a 2 reference is treated
very differently from assignment to it. Argument passing (5.2.2) and
function value return (6.6.3) are initializations.

Answer 1

No, in your code you aren’t changing what z references, instead you’re changing the contents of z (and in turn what it references, x).

You can see this with the following code:

x = 5;
cout << x;
cout << z;

Both x and z will have the value 5, since z remains a reference to x.