Pointer arthmetic confusion

Question

I see the below code snippet always return 1 instead of 4 Not able to really make out what is wrong

#include <stdio.h>
int main(void) {
   int a[4] = {1,2,3,4};
   int *p = a;

   p++;
   printf("%ld\n",(long int)(p-a));
   return 0;
}

If you assign a which points to the first element to the pointer p then increment p points to 2 and if you again subtract a from it you're back at 1 ?! — Rizier123, Mar 23 '15 at 10:38
Try this instead: printf("%ld\n",(long int)((char *)p-(char *)a)); — sansuiso, Mar 23 '15 at 10:42
You calculating how many blocks apart the pointer and the array are, not how many bytes. — Mariano Macchi, Mar 23 '15 at 11:33

Giorgi Moniava · Accepted Answer · 2015-03-23T10:47:33.780

This is the basics of pointer arithmetic. When you have:

int a[4] = {0};
int *p = a;

when you do p++ - compiler automatically increases p with four bytes (in case size of integer is four). Same happens with subtraction if you subtract 1 from p compiler will automatically subtract four bytes. But to more precisely answer your question it seems - operator when applied to pointer types divides result on size of element type to which pointer points to.

Pointer arthmetic confusion

1 Answers1