How to pass a function as a parameter?

Question

How is it possible to give a function (B) a function (A) as a parameter?

So that I can use function A in the function B.

Like the Variable B in the following example:

foo(int B) { ... }

function pointer is the thing you're looking for, but be warned: It's a bit messy and in my opinion not suitable for beginners. — stefan
– stefan, Commented Sep 13, 2012 at 10:54

unwind · Accepted Answer · 2012-09-13 11:03:11Z

4

By using function pointers. Look at the qsort() standard library function, for instance.

Example:

#include <stdlib.h>

int op_add(int a, int b)
{
  return a + b;
}

int operate(int a, int b, int (*op)(int, int))
{
  return op(a, b);
}

int main(void)
{
  printf("12 + 4 is %d\n", operate(12, 4, op_add));

  return EXIT_SUCCESS;
}

Will print 12 + 4 is 16.

The operation is given as a pointer to a function, which is called from within the operate() function.

edited Sep 13, 2012 at 11:03

answered Sep 13, 2012 at 10:54

unwind

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2 Comments

stefan Over a year ago

I would suggest not using "operator" as a variable name to avoid confusion with actual operators in C++. But of course it is valid C.

unwind Over a year ago

@stefan I normally don't care much about keeping my C valid C++, but sure, why not. I renamed the argument, thanks.

felix021 · Accepted Answer · 2012-09-13 10:58:52Z

2

write function pointer type in another function's parameter list looks a little weird, especially when the pointer is complicated, so typedef is recommended.

EXAMPLE

#include <stdio.h>

typedef int func(int a, int b);

int add(int a, int b) { return a + b; }

int operate(int a, int b, func op)
{
    return op(a, b);
}

int main()
{
    printf("%d\n", operate(3, 4, add));
    return 0;
}

answered Sep 13, 2012 at 10:58

felix021

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2 Comments

Paul R Over a year ago

+1: interesting - today I learned something - I never knew that you could use that syntax - I've only ever used explicit function pointers, e.g. typedef int (*func)(int a, int b);

felix021 Over a year ago

yeah. that is designed to simply the usage of function pointer.

Some programmer dude · Accepted Answer · 2012-09-13 10:56:01Z

2

Lets say you have a function you want to call:

void foo() { ... }

And you want to call it from bar:

void bar(void (*fun)())
{
    /* Call the function */
    fun();
}

Then bar can be called like this:

bar(foo);

answered Sep 13, 2012 at 10:56

Some programmer dude

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