How do I give arguments to a function that accepts function pointer

Question

I am trying to do something like this:

void print (int number)
{
    printf("Argument \"%i\" has been given", number);
}

void foo (void (*ptr)(int arg))
{
    ptr(arg);
}

int main (void)
{
    foo(print(10));

    return 0;
}

Unlikely this will work, because print(10) should return void, not an actual function address.. but at least I hope my question is understandable as it is very hard for me to explain it with plain words.

---

How do I transfer arguments in a function pointer like that?

Answer 1

You cannot "transfer arguments in a function pointer like that". What you can do instead is pass an additional argument that match the argument expected by your function pointer.

This would looks like this:

void print (int number)
{
    printf("Argument \"%i\" has been given", number);
}

void foo (void (*ptr)(int), int arg) // the new argument is here.
{
    ptr(arg);
}

int main (void)
{
    foo(print, 10);

    return 0;
}

Alternative

What we are doing here is packing the argument and the function pointer together in a struct, and passing this struct to foo() . This is a bit similar to c++'s std::bind but simplified and less powerful.

typedef struct bind_s
{
  void (*ptr)(int);
  int arg;
} bind_t;

void print (int number)
{
  printf("Argument \"%i\" has been given", number);
}

void foo (bind_t call)
{
  call.ptr(call.arg);
}

int main (void)
{
  bind_t call = {print, 10};
  foo(call);

  return 0;
}