Question

Why can you call C/C++ functions like this:

(&myAwesomefunction)(arg1,arg2,arg3...);

Surely taking the address of a function gives you a pointer/address. And as I understand it pointers are not callable as functions. What’s really happening?

Does it mean that I can call any address like a function?

// Is this legal?
int x = 69;
(&x)(3,78,69.456,'c','o','n','d','o','m');

What are the implications of THAT?

Answer 1

On the contrary, pointers to functions are the only things callable as functions. The definition of the function call operator starts out:

The expression that denotes the called function shall have type
pointer to function returning void or returning an object type other
than an array type.

(This means that your second hypothetical is not legal, because &x is does not have a type that is a pointer to a function of any sort).

So, you might ask, why can you also call a function like this (where myAwesomefunction is an identifier declared as a function)?

myAwesomefunction(arg1,arg2,arg3...);

The answer is that a identifier like myAwesomefunction is a primary expression that evaluates to a function designator. The definition of a function designator says:

A function designator is an expression that has function type. Except
when it is the operand of the sizeof operator or the unary &
operator, a function designator with type ‘‘function returning type’’
is converted to an expression that has type ‘‘pointer to function
returning type’’.

So formally, when you write myAwesomefunction(arg1,arg2,arg3...);, myAwesomefunction evaluates to a function designator, which then is converted to a pointer to the function, which is then used to call the function.

This conversion of a function designator to a pointer to the function is similar to the conversion of an expression with array type to a pointer to the array’s first element.