Function pointer as an argument
Is it possible to pass a function pointer as an argument to a function in C?
If so, how would I declare and define a function which takes a function pointer as an argument?
Definitely.
void f(void (*a)()) {
a();
}
void test() {
printf("hello world\n");
}
int main() {
f(&test);
return 0;
}
Let say you have function
int func(int a, float b);
So pointer to it will be
int (*func_pointer)(int, float);
So than you could use it like this
func_pointer = func;
(*func_pointer)(1, 1.0);
/*below also works*/
func_pointer(1, 1.0);
To avoid specifying full pointer type every time you need it you coud typedef
it
typedef int (*FUNC_PTR)(int, float);
and than use like any other type
void executor(FUNC_PTR func)
{
func(1, 1.0);
}
int silly_func(int a, float b)
{
//do some stuff
}
main()
{
FUNC_PTR ptr;
ptr = silly_func;
executor(ptr);
/* this should also wotk */
executor(silly_func)
}
I suggest looking at the world-famous C faqs.
This is a good example :
int sum(int a, int b)
{
return a + b;
}
int mul(int a, int b)
{
return a * b;
}
int div(int a, int b)
{
return a / b;
}
int mathOp(int (*OpType)(int, int), int a, int b)
{
return OpType(a, b);
}
int main()
{
printf("%i,%i", mathOp(sum, 10, 12), mathOp(div, 10, 2));
return 0;
}
The output is : '22, 5'
As said by other answers, you can do it as in
void qsort(void *base, size_t nmemb, size_t size,
int (*compar)(const void *, const void *));
However, there is one special case for declaring an argument of function pointer type: if an argument has the function type, it will be converted to a pointer to the function type, just like arrays are converted to pointers in parameter lists, so the former can also be written as
void qsort(void *base, size_t nmemb, size_t size,
int compar(const void *, const void *));
Naturally this applies to only parameters, as outside a parameter list int compar(const void *, const void *);
would declare a function.