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 p

Answer 1

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 
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 :-)