typedef a function pointer type

Question

I want to declare a pointer type which point to a function, so I try:

typedef void (*print)(void); works perfect

void (*print)(void);

Answer 1

1. No, you don't have to use a typedef to create an object of type 'pointer to function':

   void somefunc(void (*pointer)(void))
   {
       (*pointer)();
       pointer();
   }
   

   However, there is no way to create a name for a type other than by using a typedef. I suppose you could indulge in macro hackery to generate a 'pointer to function', but after the macro is expanded, you'd have a 'pointer to function' written out:

   #define PTR_FUNC(func) void (*func)(void)
   void somefunc(PTR_FUNC(pointer)) { ... }
   

2. The (void) notation as the type name is wrong. You don''t write: (int) x; and expect to declare a variable x -— it is a type cast. Same with the notation you're using in your typedef.

3. You can't write typedef void (*)(void) Print; because it is not an allowed C syntax. You also can't write typedef [32] int name; either — it isn't valid C syntax.

Answer 2

The correct way is:

typedef void (*print_function_ptr)(void)

and its usage for variable/parameter declaration is:

print_function_ptr p;

1. You don't need a typedef to declare a variable. You can directly write void (*p)(void) to declare a variable p pointing to a function taking void and returning void. However to declare a type alias / name for a pointer to function, typedef is the tool.

2. It does not mean anything it is not a valid C syntax.

3. Because it is not how C works. Typedefs in C mimics how variables are declared or defined.