Why is it called operator overloading?

Question

If the following class, Foo, is defined. It is said it overloads the unary ampersand (&) operator:

class Foo {
public:
   Foo* operator&() { return nullptr; }
};

I think in this case, (reglardless of the fact that you can get the address of such an object by means of std::addressof() and other idiomatic tricks) there is no way to access/choose the original unary ampersand operator that returns the address of the object called on, am I wrong?

By overloading however, I understand that there is a set of functions of which one will be selected at compile-time based on some criteria. But this thinking doesn't seem to match the scenario above.

Why is it then called overloading and not something else like redefining or replacing?

Answer 1

Consider the following code:

int x;
Foo y;
&x; // built-in functionality
&y; // y.operator&();

We have two variables of different types. We apply the same & operator to both of them. For x it uses the built-in address-of operator whereas for y it calls your user-defined function.

That's exactly what you're describing as overloading: There are multiple functions (well, one of them is the built-in functionality, not really a "function") and they're selected based on the type of the operand.