c++ does pass address of an reference as a pointer gives undefined behavior?

I have following code,

#include <iostream>
using namespace std;

void swap(int *x, int *y)
{
    int z = *x;
    *x = *y;
    *y = z;
}

// Driver Code
int main()
{
    int a = 45, b = 35;
    auto& a_ref = a;
    cout << "a = " << a_ref << " b = " << b << "\n";

    swap(&a_ref, &b);

    cout << "After Swap with pass by pointer\n";
    cout << "a = " << a << " b = " << b << "\n";
}

As can be seen, I first get a reference of a then pass it as a pointer, I am wondering will this give undefined sometimes?

>Solution :

You can’t take the address of a reference. When a variable of reference type appears in an expression, it always evaluates to the object (or function) to which it refers. So the expression &a_ref is, by all means, equivalent to &a.

[expr.type]/1:

If an expression initially has the type “reference to T”, […]. The expression designates the object or function denoted by the reference