ctypedeffunction-declaration

Defining function pointer in C?

  1. Recently I came across a surprising way of defining a function pointer in C:
typedef void (func_type)(void);
func_type *func_ptr;

Is this a correct way of defining a function pointer?

  1. If we define func_obj as below, what is the type of this object and what this object can be used for?
#include <stdio.h>

typedef void (func_type)(void);
func_type *func_ptr;

func_type func_obj;

int main()
{
  printf("Size of func_obj: %zu\n", sizeof func_obj);
  printf("Size of func_ptr: %zu\n", sizeof func_ptr);
  return 0;
}

The code prints:

Size of func_obj: 1
Size of func_ptr: 8
Solution

  • These declarations

    typedef void (func_type)(void);
func_type *func_ptr;

    are similar to declarations like for example

    typedef int T;
T *int_ptr;

    So there is nothing wrong with the typedef that intrduces an alias for a function type.

    Pay attention to that this typedef declaration

    typedef void (func_type)(void);

    is equivalent to

    typedef void func_type(void);

    This declaration

    func_type func_obj;

    declares a function with the name func_obj of the type void( void ).

    Such a declaration allows for example to list functions in one line without repeating their parameter lists.

    This statement

    printf("Size of func_obj: %zu\n", sizeof func_obj);

    is invalid according to the C Standard because you may not apply the sizeof operator to a functon.

    From the C Standard (6.5.3.4 The sizeof and alignof operators)

    1 The sizeof operator shall not be applied to an expression that has function type or an incomplete type, to the parenthesized name of such a type, or to an expression that designates a bit-field member. The alignof operator shall not be applied to a function type or an incomplete type.

    Some compilers however can have their own extensions that contradict the C Standard.