How to get a pointer from a reference?

Yes, applying the address-of operator to the reference is the same as taking the address of the original object.

#include <iostream>

struct foo {};

void bar( const foo& obj )
{
  std::cout << &obj << std::endl;
}

int main()
{
  foo obj;
  std::cout << &obj << std::endl;
  bar( obj );

  return 0;
}

Result:

0x22ff1f
0x22ff1f

Any operator applied to a reference will actually apply to the object it refers to (§5/5 [expr]); the reference can be thought of as another name for the same object. Taking the address of a reference will therefore give you the address of the object that it refers to.

It as actually unspecified whether or not a reference requires storage (§8.3.2/4 [dcl.ref]) and so it wouldn't make sense to take the address of the reference itself.

As an example:

int x = 5;
int& y = x;
int* xp = &x;
int* yp = &y;

In the above example, xp and yp are equal - that is, the expression xp == yp evaluates to true because they both point to the same object.

The general solution is to use std::addressof, as in:

#include <type_traits>

void foo(T & x)
{
    T * p = std::addressof(x);
}

This works no matter whether T overloads operator& or not.