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'