What are function pointers used for, and how would I use them?

Question

I understand I can use pointers for functions.

Can someone explain why one would use them, and how? Short example code would be very helpful to me.

Answer 1

Let's do a map-like function for C.

void apply(int *arr, size_t len, int (*func)(int))
{
    for(size_t i = 0; i < len; i++)
        arr[i] = func(arr[i]);
}

That way, we can transform a function that works on integers to work on arrays of integers. We could also do a similar version:

void apply_enumerated(int *arr, size_t len, int (*func)(size_t, int))
{
    for(size_t i = 0; i < len; i++)
        arr[i] = func(i, arr[i]);
}

This does the same thing, but allows our function to know which element it's on. We could use this, for example:

int cube(int i) { return i * i * i }

void print_array(int *array, size_t len, char *sep)
{
    if(sep == NULL) sep = ", ";
    printf("%d", *array);
    for(size_t i = 1; i < len; i++) printf("%s%d", sep, array[i])
}

#define ARRAY_SIZE(a) (sizeof(a)/sizeof((a)[0]))

int main(void)
{
    int array[5] = { 1, 2, 3, 4, 5 };
    print_array(array, ARRAY_SIZE(array), NULL);
    apply(array, ARRAY_SIZE(array), cube);
    print_array(array, ARRAY_SIZE(array), NULL);
    return 0;
}

That code will print:

1, 2, 3, 4, 5
1, 8, 27, 64, 125

For our enumeration example:

int mult(size_t i, int j) { return i * j }

// print_array and ARRAY_SIZE as before

int main(void)
{
    int array[5] = { 1, 2, 3, 4, 5 };
    print_array(array, ARRAY_SIZE(array), NULL);
    apply_enumerated(array, ARRAY_SIZE(array), mult);
    print_array(array, ARRAY_SIZE(array), NULL);
    return 0;
}

This prints:

1, 2, 3, 4, 5
0, 2, 6, 12, 20

As a more real world example, a string library could have a function that applies a function that operates on single characters to all the characters in the string. An example of such functions are the standard-library toupper() and tolower() functions in ctype.h - we could use this string_apply() function to make a string_toupper() function easily.