Casting a pointer - What is the difference at runtime?

Question

Consider the following small example code:

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

int main()
{
    int *i;
    char *c1, *c2;

    i = malloc(4);

    *i = 65535;

    c1 = i;
    c2 = (char *)i;

    printf("%p %p %p\n", i, c1, c2);
    printf("%d %d", *c1, *c2);

    free(i);

    return 0;
}

In the example, I allocate memory to store an integer, which is pointed by i. Then, I store the value 65535 (1111 1111 1111 1111) in *i. The next thing I do is make two char* pointers also point to the integer. I do it two times, but in two different ways: c1 = i; and c2 = (char *)i;. Finally, I print all pointers and all values in the screen. The three pointers are pointing to the same address, and the two values *c1 and *c2 are correct (-1).

However, the compiler generates a warning in this line: c1 = i;. The warning is generated because I did not use the (char *) cast to do the assignment.

What I would like to ask is why the compiler generates this warning, since I do not see any difference in using c1 = i; or c2 = (char *)i;. In both cases, the result is the same address with the same size in bytes. And this is valid for all casts, even if it is a (int *) cast, (float *) cast, (short *) cast, etc. All of them generate the same value, but the compiler will only accept it without a warning if the cast being used is of the pointer's type.

I really would like to know why the compiler asks for that cast, even if the result is always the same.

Answer 1

When you use:

c2 = i;

the compiler warns you about the assignment of type int* to char*. It could potentially be an inadvertent error. The compiler is warning you with the hope that, if it is indeed an inadvertent error, you have the chance to fix it.

When you use:

c2 = (char *)i;

you are telling the compiler that you, as the programmer, know what you are doing.