Why is it called operator overloading?

Question

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

class Fo         


        

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.