函数名称和函数指针

Question

Did the name of the function represent an pointer to this function like array?

If I declare an function as follow void fu (void); and an array of pointer to function like that void(*ptr_fn[8])(void); so can I do that ptr_fn[0] = fu;

Answer 1

No, and the name of an array does not represent a pointer, either.

The name represents the function, but there's an implicit conversion when you use the function name in an expression.

See 4.3 Function-to-pointer conversion:

An lvalue of function type T can be converted to a prvalue of type “pointer to T.” The result is a pointer to the function.

Answer 2

Yes. It is also equivalent to:

ptr_fn[0] = &fu ;

ie the & is optional.

Equally when calling the function through the pointer

ptr_fn[0]() ;

is equivalent to:

(*ptr_fn[0])() ;

The use of the & and * operators serves to emphasise perhaps that you are dealing with a function-pointer, and so aids maintenance and comprehension by humans, but has no effect on the compiler's code generation.