Why my code doesn't work...

Question

//I don't understand why my compiler (Code::Blocks) gives me a warning when I execute the following code:
#include <stdio.h>

void *square(const void *num);
int main()
{
int a = 5;
int result;
result = square(&a);
printf("%d squared is %d", a, result);
return 0;
}
void *square(const void *num)
{
int r;
r = (*(int*)num)*(*(int*)num);
return r;
}
//Can anybody explain why? When I change from return r; to return (int*)r; in the 'square' function and from square(&a) to (int)square(&a) in the 'main' function, it works. Also, should I declare the variable 'r' in the 'square' function to be static because it is being passed out of the function (local variable)? Thanks.

Answer 1

When you said I implicitly converting from pointer to int, and vice versa. Do you mean the variable 'r' in the 'square' function? If so, I need to typecast the variable 'r' to be return (int*)r so the function will know I'm returning an integer value? Also, when we type casting the void pointer why do we put an asterisk inside the bracket? I'm really confusing...