typedef int (fc_name) (void);
here fc_name
is any valid C symbol.
how different is this from a function pointer typedef
?
typedef int (fc_name) (void);
here fc_name
is any valid C symbol.
how different is this from a function pointer typedef
?
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.
A pointer to function typedef would read:
typedef int (*fc_name)(void);
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
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.
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 * */
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
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 is not much significance. These are used as a style standard or simply to create obfuscation intentionally ;-)