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typedef a function pointer type

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

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

void (*print)(void); p is a ponter variable , not a type.

typedef (void) (*print)(void); error expected identifier or ‘(’ before ‘void’

typedef void (*)(void) Print;

error: expected ‘=’, ‘,’, ‘;’, ‘asm’ or ‘_ attribute _’ before ‘Print’ .

My question is:

  1. Do I have to use typedef to declare a function pointer type ?

  2. Why typedef (void) (*print)(void); is wrong ? what () means here?

  3. Why I can't write in this way:typedef void (*)(void) Print ?

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Lidong Guo Avatar asked Dec 12 '22 11:12

Lidong Guo


2 Answers

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.

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Juraj Blaho Avatar answered Dec 17 '22 05:12

Juraj Blaho


  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.

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Jonathan Leffler Avatar answered Dec 17 '22 03:12

Jonathan Leffler