How to declare an array that is received like an parameter in a function that also receive a pointer to a function in C and that function is using the values from the array? The function that use the values from the array doesn't receive a parameter. It's something like:

int find_array(void)

and the function that receive this function like parameter also receive the array. I think that declaration should look something like:

int function ( int (*pmin)(void), int* array )

How the function find_array knows that it should work with the array received like parameter? How to declare this array?

CodePudding user response：

int function should receive just 2 parameters, a pointer to function find_array and the array. I should also to find a way to declare the size of the array

It had to be extended to three parameters in order to give the size of the array:

#include <stdio.h>

// simple function working on an array
int find_minimum(int *array, int n) {
    int minimum = array[0];
    for (int i = 1; i < n; i  ) {
        if (array[i] < minimum) {
            minimum = array[i];
        }
    }
    return minimum;
}

// function receiving an array and a function to process it
int function(int (*pmin)(int *array, int n), int *array, int n) {
    return pmin(array, n);
}

int main() {
    int array[] = {1, 2, -3, 4, 5};
    int min = function(find_minimum, array, sizeof array / sizeof(int));
    printf("minimum: %d\n", min);
return 0;
}

Looks good:

$ gcc -Wall -Werror array.c
$ ./a.out                  
minimum: -3
$ 

Regarding the parameter for the function to be give as parameter you might want to use ellipsis for a variable amount of parameters, see https://en.wikipedia.org/wiki/Ellipsis_(computer_programming).