pointer-to-const conversion in C

Question

The following code compiles without warning on GCC but gives a warning in Visual Studio 2005.

const void * x = 0;
char * const * p = x;

x points to a constant object of unknown type, and p points to a constant pointer to char. Why should the assignment to p result in a warning?

Again, this is C, not C++. Thanks.

Answer 1

It happens because when you make a pointer of one type point to another type, sometimes it is done unintentionally (bug), so the compiler warns you about it.

So, in order to tell the compiler that you actually intent to do it, you have to do explicit casting, like this:

        const void * x = 0;
        char * const * p = (char * const * )x;

---

P.S. At the first place I wrote "most of the times is done unintentionally", but AndreyT made me reconsider it when he rightfully said that void * exists specifically for that purpose.

Answer 2

What if x would point to, say, a struct Thing as opposed to a char? In that case, you'd be doing something with unspecified behavior. GCC tends to let you do this because it assumes that you're smart enough not to shoot yourself in the foot, but there is good reason for the warning.

Answer 3

warning C4090: 'initializing' : different 'const' qualifiers

You cannot cast const void to char *const or even char *. By dereferencing p, you can now modify *(char *)(*x). This is a little known subtlety about pointers in C.

A working type for p would be:

char const **p = x;

And yes, I put const on the right like a man.

Answer 4

You have to read the following from right to left.
char * const * p = x;

For Example:
P points to a const pointer of type char.

Answer 5

The C code is valid and a conforming compiler shouldn't warn as const ness is correctly preserved and conversion of void * to any pointer type (function pointers aside) is implicit.

A C++ compiler is supposed to warn about the implicit conversion, but a warning about discarding the const qualifier is wrong and should be considered a compiler bug.

Answer 6

const void * x = 0;
char * const * p = x;

At first I assumed you meant to take the address of x and wrote the code for that. Then I decided that x points to a pointer to char []. Both anyway, and it's still quite confusing:

#include <stdio.h>
int main()
{
        char s [] = "Hello World!";
        const void * x = s;
        const char * const * const p = (const char * const *)&x;
        printf("%s\n", *p);
/* won't compile
        *p++;      // increments x to point to 'e'
        (**p)++;   // increments 'H' to 'I'
*/
        const char * y = s;
        x = &y;
        const char * const * const q = x;
        printf("%s\n", *q);
/* won't compile
        *q++;
        (**q)++;
*/
        return 0;
}

The extra const before the char in the declaration of p prevents (**p)++ from compiling. However, the added const before the declaration of q (in GCC) shadows warning: dereferencing ‘void *’ pointer with error: increment of read-only location ‘**q’ for (**q)++. Hope that helps, it has helped me a little :-)