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); 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:

Do I have to use typedef to declare a function pointer type ?
Why typedef (void) (*print)(void); is wrong ? what () means here?
Why I can't write in this way:typedef void (*)(void) Print ?

All your questions can pretty much be answered with "because the language designers decided it that way" — Per Johansson
– Per Johansson, Commented Sep 5, 2013 at 14:26
In an answer to another question, I tried to explain how to declare a function pointer. See stackoverflow.com/a/6905987/396551 — Sjoerd
– Sjoerd, Commented Sep 5, 2013 at 14:46

Juraj Blaho · Accepted Answer · 2013-09-05 14:31:30Z

10

The correct way is:

typedef void (*print_function_ptr)(void)

and its usage for variable/parameter declaration is:

print_function_ptr p;

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.
It does not mean anything it is not a valid C syntax.
Because it is not how C works. Typedefs in C mimics how variables are declared or defined.

edited Sep 5, 2013 at 14:31

answered Sep 5, 2013 at 14:25

Juraj Blaho

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4 Comments

Lidong Guo Over a year ago

I hope declare a function pointer type , not a variable, is there a way to do this without typedef?

Jonathan Leffler Over a year ago

@LidongGuo: what do you mean by 'declare a function pointer type'? You can declare an object that is of type 'pointer to function' without using a typedef. There is no way other than using typedef to create a type name that you can use subsequently.

Lidong Guo Over a year ago

I hope to declare a type like a struct name, then I can use it declare lots of function pointer'

blubberdiblub Over a year ago

@LidongGuo then this answer shows you how to do that. After the typedef, you can just do this to have 5 function pointers with the same signature: print_function_ptr p1, p2, p3, p4, p5;

Jonathan Leffler · Accepted Answer · 2013-09-05 14:34:15Z

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)) { ... }
```
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.
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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