Function pointer as an argument

Question

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?

Answer 1

Definitely.

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

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

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

Answer 2

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.

Answer 3

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'

Answer 4

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.

Answer 5

Check qsort()

void qsort(void *base, size_t nmemb, size_t size,
           int (*compar)(const void *, const void *));

The last argument to the function is a function pointer. When you call qsort() in a program of yours, the execution "goes into the library" and "steps back into your own code" through the use of that pointer.