Trouble understanding void arrays and using them in function calls

i'm attempting a question which asks me to create a function called randSwap, that takes in 2 arrays of void pointers, and with a 50% chance, it will swap 2 values of the same index between the 2 arrays. Below is my code, but whenever i go to compile, it gives me several errors discussing:

warning: initialization makes pointer from integer without a cast.

void *array1[4] = {3,4,2,5};

and several others relating to array1 and 2.

When i run the program, the values of array1 becomes corrupted producing a large single integer, and array 2 prints its original values both times.

I am very unfamiliar with using void types, as I think this is where the problem lies.

#include <stdio.h>

int randSwap(void *array1[], void *array2[], int length)
{
    static int numofswaps = 0;
    int value;
    int toswap;
    void *temp;


    value = rand() % 2;
    if(value == 2)
    {
        toswap = rand() % length;
        temp = array1[toswap];
        array1[toswap] = array2[toswap];
        array2[toswap] = temp;

        numofswaps++; 
    }

    return numofswaps;
}

int main(void)
{
    int i;
    void *array1[4] = {3,4,2,5};
    void *array2[4] = {6,3,7,4};
    int length = 4;
    int numofswaps;
    srand(time(NULL));

    printf("Array1\n");
    for(i=0; i<length; i++);
    {

        printf("%d\t", (int*)array1[i]);
    }

    printf("\nArray2\n");
    for(i=0; i<length; i++)
    {
        printf("%d\t", (int*)array2[i]);
    }

    numofswaps = randSwap(array1, array2, length);
    numofswaps = randSwap(array1, array2, length);

    printf("\nArray1\n");
    for(i=0; i<length; i++);
    {

        printf("%d\t", (int*)array1[i]);
    }

    printf("\nArray2\n");
    for(i=0; i<length; i++)
    {
        printf("%d\t", (int*)array2[i]);
    }
    printf("\n");
    return 0;
 }