Question

From this question, and consequently, from the Standard (ISO C++-03):

It is unspecified whether or not a reference requires storage (3.7).

In some answers in that thread, it’s said that references have, internally, the same structure of a pointer, thus, having the same size of it (32/64 bits).

What I’m struggling to grasp is: how would a reference come not to require storage?

Any sample code exemplifying this would be greatly appreciated.

Edit:
From @JohannesSchaub-litb comment, is there anything like, if I’m not using a const &, or if I’m using a const & with default value, it requires allocation? It seems to me, somehow, that there should be no allocations for references at all — except, of course, when there are explicit allocations involved, like:

A& new_reference(*(new A())); // Only A() instance would be allocated,
                              // not the new_reference itself

Is there any case like this?

Answer 1

Take something simple:

int foo() {
  int  x = 5;
  int& r = x;
  r = 10;
  return x;
}

The implementation may use a pointer to x behind the scenes to implement that reference, but there’s no reason it has to. It could just as well translate the code to the equivalent form of:

int foo() {
  int x = 10
  return x;
}

Then no pointers are needed whatsoever. The compiler can just bake it right into the executable that r is the same as x, without storing and dereferencing a pointer that points at x.

The point is, whether the reference requires any storage is an implementation detail that you shouldn’t need to care about.