在C语言中，为什么我们需要将char *强制转换为void？

Question

I read a unit test, that checks for invalid free or double-free:

int main() {
    char *a = (char*) my_malloc(200);
    char *b = (char*) my_malloc(50);
    char *c = (char*) my_malloc(200);
    char *p = (char*) my_malloc(3000);

    (void) a, (void) c;

    memcpy(p, b - 200, 450);
    my_free(p + 200);
    printstatistics();
}

Why do we need to cast char* to void and what happens in memory when we do this cast?

Answer 1

(void) a, (void) c;

is a common way to get rid of compiler warnings about unused variables. Since those two variables are initialised only and not used later, most compilers would issue warnings about it. Since this is apparently some kind of a test of memory allocation they are not used on purpose, so someone decided to silence warnings.