Question

What does this line (x = n & 5;) mean in the code below? As far as I know ampersand is used for pointers. I was not expecting this code to compiled, but it compiled and ran fine. The results I got is

0,1,0,1,4,5,4,5,0,1,

#include <stdio.h>

int main(void){
    int x, n;
    for (n = 0; n < 10; n++){
        x = n & 5;
        printf("%d,", x);
    }
    printf("\n");
    return 0;
}

Answer 1

In this case it’s bitwise AND.

x = n & 5;

will AND 5 (which is 0b101) with whatever is in n.

AND works on the bits that represent the values. The result will have a 1 if both values have a 1 in that position, 0 otherwise.

Since we’re ANDing with 5, and there are only 4 values you can make with the two bits in 0b101, the only possible values for x are 1, 4, 5 and 0. Here’s an illustration:

n           &    5     = x
1 (0b0001)  &  0b0101  = 1  (0b0001)
2 (0b0010)  &  0b0101  = 0  (0b0000)
3 (0b0011)  &  0b0101  = 1  (0b0001)
4 (0b0100)  &  0b0101  = 4  (0b0100)
5 (0b0101)  &  0b0101  = 5  (0b0101)
6 (0b0110)  &  0b0101  = 4  (0b0100)
7 (0b0111)  &  0b0101  = 4  (0b0100)
8 (0b1000)  &  0b0101  = 0  (0b0000)
9 (0b1001)  &  0b0101  = 1  (0b0001)