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

Arithmetic on void* is a GCC extension. When I compile your code with clang -Wpointer-arith the output is :

test.c:9:4: warning: use of GNU void* extension [-Wpointer-arith]
k++;
~^

The usual behavior of a pointer increment is to add the size of the pointee type to the pointer value. For instance :

int *p;
char *p2;
p++; /* adds sizeof(int) to p */
p2 += 2; /* adds 2 * sizeof(char) to p2 */

Since void has no size, you shouldn't be able to perform pointer arithmetic on void* pointers, but GNU C allows it.