Function pointer as an argument

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.

Definitely.

void f(void (*a)()) {
    a();
}

void test() {
    printf("hello world\n");
}

int main() {
     f(&test);
     return 0;
}