C pass void pointer array to function

Question

I want to pass a void pointer array to a function which fills it with the result it works fine with integer but I have troubles returning char*

int function(void *arr[]);
int main(void) {
    void *arr[3];
    function(arr);
    printf("%d\n", *(int *)&arr[0]);
    printf("%d\n", *(int *)&arr[1]);
    printf("%s\n", (char *)&arr[2]);
}
int function(void *arr[]){
    arr[0] = (void*)(int)4;
    arr[1] = (void*)(int)6;
    char* test = "abc";
    arr[2] = (void*)test;
    return 0;
}

for the String I don't get the right value

Answer 1

You may pass an array of voids, and assign that array elements to a dynamically allocated memory region that stores the pointed-to value.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int function(void *arr[]){    
    arr[0] = malloc(sizeof(int));
    if (arr[0] == NULL) goto ERR_MALLOC_0;
    *(int*)arr[0] = 4;

    arr[1] = malloc(sizeof(int));
    if (arr[1] == NULL) goto ERR_MALLOC_1;
    *(int*)arr[1] = 6;

    const char *const test = "abc";
    arr[2] = malloc(strlen(test));
    if (arr[2] == NULL) goto ERR_MALLOC_2;
    strcpy(arr[2], test);

    return 0;

    // remember about error checking
    free(arr[2]);
    ERR_MALLOC_2:
    free(arr[1]);
    ERR_MALLOC_1:
    free(arr[0]);
    ERR_MALLOC_0:
    return -1;
}

int main(void) {
    void *arr[3];
    int err = function(arr);
    if (err == -1) abort();

    // & is the address of element, not it's element
    printf("%d\n", *(int*)arr[0]);
    printf("%d\n", *(int*)arr[1]);
    printf("%s\n", (char*)arr[2]);

    // remember to free memory
    for (int i = 0; i < 3; ++i) {
        // funny how we do not need to know the effective type
        free(arr[i]);
    }
}

But such function is just plainly confusing and will result in many, many bugs and problems. Instead just actually use variables of proper type:

#include <stdlib.h>
#include <string.h>

int function(int *val1, int *val2, char **string) {
    *val1 = 3;
    *val2 = 6;
    const char *test = "abc";
    *string = malloc(strlen(test));
    if (*string == NULL) return -1;
    strcpy(*string, test);
    return 0;
}

int main(void) {
    int val1; 
    int val2;
    char *string;
    int err = function(&val1, &val2, &string);
    if (err == -1) abort();

    printf("%d\n", val1);
    printf("%d\n", val2);
    printf("%s\n", string);

    free(string);
}

If you are really striving for implementing some virtual representation and operations on different data types, using plain "array of void pointers" will get you nowhere, because such array doesn't know what is the underlying type of the value that is being stored - you, as a programmer, have to know what is inside that array (ie. that arr[0] is an int and arr[2] is a char*), so as you know it you might as well just use variables of proper types from the start.