What does a typedef with parenthesis like “typedef int (f)(void)” mean? Is it a function prototype?

Question

typedef int (fc_name) (void);

Here fc_name is any valid C symbol.

How different is this from a function pointer typedef ?

Answer 1

It's a typedef to a function type. The intent is to use it for function pointers, but in this case the syntax to use it would be:

int bar(void);

fc_name* foo = bar; /* Note the * */

Update: As mentioned in the comments to Jonathan Leffler's answer , the typedef can be used to declare functions. One use could be for declaring a set of callback functions:

typedef int (callback)(int, void*);

callback onFoo;
callback onBar;
callback onBaz;
callback onQux;

Answer 2

The first parentheses are superfluous - it is equivalent to:

typedef int fc_name(void);

I don't think this does anything useful, though I can't get GCC to complain about it on its own.

This means that fc_name is an alias for a function type that takes no arguments and returns an int . It isn't directly all that useful, though you can declare, for example, the rand() function using:

fc_name rand;

You cannot use the typedef in a function definition.

A pointer to function typedef would read:

typedef int (*fc_name)(void);

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This code shows that the typedefs without the asterisk are not function pointers (addressing a now-deleted alternative answer):

static int function(void)
{
    return 0;
}

typedef int   fc_name1 (void);
typedef int  (fc_name2)(void);
typedef int (*fc_name3)(void);

fc_name1 x = function;
fc_name2 y = function;
fc_name3 z = function;

When compiled, 'gcc' says:

gcc -Wextra -Wall -pedantic -c -O x.c
x.c:10:1: error: function ‘x’ is initialized like a variable
x.c:11:1: error: function ‘y’ is initialized like a variable

---

And this code demonstrates that you can indeed use fc_name *var = funcname; as suggested by jamesdlin :

static int function(void)
{
    return 0;
}

typedef int   fc_name1 (void);
typedef int  (fc_name2)(void);
typedef int (*fc_name3)(void);

fc_name1  x_0 = function;
fc_name1 *x_1 = function;
fc_name2  y_0 = function;    // Damn Bessel functions - and no <math.h>
fc_name2 *y_1 = function;    // Damn Bessel functions - and no <math.h>
fc_name3  z   = function;

Using y0, y1 generates GCC warnings:

x.c:12:11: warning: conflicting types for built-in function ‘y0’
x.c:13:11: warning: built-in function ‘y1’ declared as non-function

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And, building on the comment from schot :

static int function(void)
{
    return 0;
}

typedef int   fc_name1 (void);
typedef int  (fc_name2)(void);
typedef int (*fc_name3)(void);

fc_name1  x_0 = function;   // Error
fc_name1 *x_1 = function;   // x_1 is a pointer to function
fc_name1  x_2;              // Declare int x_2(void);
fc_name1 *x_3 = x_2;        // Declare x_3 initialized with x_2

fc_name2  y_0 = function;   // Damn Bessel functions - and no <math.h>
fc_name2 *y_1 = function;   // Damn Bessel functions - and no <math.h>
fc_name1  y_2;              // Declare int y_2(void);
fc_name1 *y_3 = x_2;        // Declare y_3 initialized with y_2

fc_name3  z   = function;

Interesting - the dark corners of C are murky indeed.

Answer 3

Interesting! A typedef declaration is a declaration with typedef as the storage class.

typedef int   fc_name1 (void);   
// this defines a function type called fc_name1 
// which takes no parameter and returns int

later on, you could define a function like following,

fc_name1 myFunc;
// this is equivalent to the next line
// int myFunc(void);

You should be able figure this out from the c/c++ standard!

Answer 4

I've never before seen this done to a typedef name , but parentheses around the name of a function are useful to prevent its being expanded as a function-like macro of the same name. For instance, the isxxx functions in ctype.h are defined as both functions and macros. This is so you can take a pointer to isalpha . But how does the C library define the out-of-line isalpha ? Probably like this:

#include <ctype.h>

int
(isalpha)(int c)
{
    return isalpha(c);
}

The use of isalpha in the function body is expanded as a macro, the use in the function header isn't.

Answer 5

  1 #include <stdio.h>
  2 
  3 
  4 typedef int (fc_name)(void);
  5 
  6 
  7 
  8 int test_func1 ()
  9 {
 10     printf("\n test_func1 called\n");
 11 
 12     return 0;
 13 }
 14 
 15 int test_func2 (void)
 16 {
 17     printf("\n test_func2 called\n");
 18     return 0;
 19 }
 20 
 21 int handler_func(fc_name *fptr)
 22 {
 23     //Call the actual function
 24     fptr();
 25 }
 26 
 27 int main(void)
 28 {
 29     fc_name  *f1, *f2;
 30 
 31     f1 = test_func1;
 32     f2 = test_func2;
 33 
 34     handler_func(f1);
 35     handler_func(f2);
 36 
 37     printf("\n test complete\n");
 38 
 39     return 0;
 40 }

OUTPUT:-

 test_func1 called

 test_func2 called

 test complete

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So the typedef which I questioned about, (Line # 4 here) represents a function type and is not the same as function pointer typedef. This kind of typedef does not have much significance. These are used as a style standard or simply to create obfuscation intentionally ;-)

Answer 6

The correct form is:

typedef int (*myfunc)(void);

You can define a function like follows:

int helloword(void) {
    printf("hello, world\n");
}

and then define a variable point to this function:

myfunc hw_func;
hw_func = helloworld;

and call the function by the function pointer:

int ret = (*hw_func)();

The reason we need function pointer is that C language doesn't have predefined function pointer and use void * pointer to call a function is illegal in C language.