void * arithmetic

Question

#include
int main(int argc,char *argv[])
{
   int i=10;
   void *k;
   k=&i;

   k++;
   printf(\"%p\\n%p\\n\",&i,k);
   return 0;
}
         


        

Answer 1

It is a GCC extension.

In GNU C, addition and subtraction operations are supported on pointers to void and on pointers to functions. This is done by treating the size of a void or of a function as 1.

If you add the -pedantic flag it will produce the warning:

warning: wrong type argument to increment

If you want to abide to the standard, cast the pointer to a char*:

k = 1 + (char*)k;

---

The standard specifies one cannot perform addition (k+1) on void*, because:

1. Pointer arithmetic is done by treating k as the pointer to the first element (#0) of an array of void (C99 §6.5.6/7), and k+1 will return element #1 in this "array" (§6.5.6/8).

2. For this to make sense, we need to consider an array of void. The relevant info for void is (§6.2.5/19)

   The void type comprises an empty set of values; it is an incomplete type that cannot be completed.

3. However, the definition of array requires the element type cannot be incomplete (§6.2.5/20, footnote 36)

   Since object types do not include incomplete types, an array of incomplete type cannot be constructed.

Hence k+1 cannot be a valid expression.