C Pointers and integers — Pointer-behavior in a certain case

Question

#include <stdio.h>

int main() {

    int arr [6] = {22,3,30,1};
    int * p = arr ;
    p++;
    int ** p2 = &p;
    int x = 50 &(** p2 );
    printf("\n\n%d\n\n", x);
}

Can someone explain what happens in the second last row? printf prints 2.

What do *you* think it does? That really belongs in your question as well. — WhozCraig, May 10 '17 at 22:28
Possible duplicate of [What does \*\* do in C language](http://stackoverflow.com/questions/35021521/what-does-do-in-c-language) — Manav Dubey, May 10 '17 at 22:28
Can you see that 50 & 3 = 2? `p++;` increments `p` so that it points to the second element of `arr[]`. — Weather Vane, May 10 '17 at 22:31
Okay, the ampersand is in that case a binary compare operator and has nothing to do with the pointer. thanks. — nan0, May 10 '17 at 22:41

score 2 · Accepted Answer · answered May 10 '17 at 22:47

x is assigned the value of 50 bitwise-and'd with the integer pointed to by the pointer pointed to by p2.

Or in other terms, it is bitwise-and-ing 50 and 3. 50 is in binary 00110010 where 3 is 00000011. The only bit that they both have a 1 in is the second-least-significant. Therefore, the result is 00000010, or 2.

C Pointers and integers — Pointer-behavior in a certain case

1 Answers1