How does dereferencing of a function pointer happen?

Question

Why and how does dereferencing a function pointer just \"do nothing\"?

This is what I am talking about:

#include

void hello() { print         


        

Answer 1

It's not quite the right question. For C, at least, the right question is

What happens to a function value in an rvalue context?

(An rvalue context is anywhere a name or other reference appears where it should be used as a value, rather than a location—basically anywhere except on the left-hand side of an assignment. The name itself comes from the right-hand side of an assignment.)

OK, so what happens to a function value in an rvalue context? It is immediately and implicitly converted to a pointer to the original function value. If you dereference that pointer with *, you get the same function value back again, which is immediately and implicitly converted into a pointer. And you can do this as many times as you like.

Two similar experiments you can try:

• What happens if you dereference a function pointer in an lvalue context—the left-hand side of an assignment. (The answer will be about what you expect, if you keep in mind that functions are immutable.)

• An array value is also converted to a pointer in an lvalue context, but it is converted to a pointer to the element type, not to a pointer to the array. Dereferencing it will therefore give you an element, not an array, and the madness you show doesn't occur.

Hope this helps.

P.S. As to why a function value is implicitly converted to a pointer, the answer is that for those of use who use function pointers, it's a great convenience not to have to use &'s everywhere. There's a dual convenience as well: a function pointer in call position is automatically converted to a function value, so you don't have to write * to call through a function pointer.

P.P.S. Unlike C functions, C++ functions can be overloaded, and I'm not qualified to comment on how the semantics works in C++.