Bitwise and doesn't work on unsigned long

Question

I have this code where I use "bitwise and" (&) to try and change x:

#include <stdio.h>

int main (void){

    unsigned long x = 0;
    printf("x = %lu\n", x);

    x &= 1;
    printf("x = %lu\n", x);

    return 0;

}

I compile and execute the code in order to get output:

x = 0
x = 0

But I expected:

x = 0
x = 1

Why?

Does this answer your question? [What does &= mean?](https://stackoverflow.com/questions/20948621/what-does-mean) — OldProgrammer, Mar 08 '22 at 14:09

Vlad from Moscow · Accepted Answer · 2022-03-08T14:20:03.357

3

It seems you mean the bitwise inclusive OR operator that can be used to set bits

x |= 1;
printf("x = %lu\n", x);

Or you could use another bitwise operator: the bitwise exclusive OR operator that can be used to toggle bits

x ^= 1;
printf("x = %lu\n", x);

Otherwise 0 & 1 will give 0 because initially x was initialized by 0

unsigned long x = 0;

Here is a demonstration program

#include <stdio.h>

int main( void )
{
    unsigned long x = 0;
    printf( "x = %lu\n", x );

    x |= 1;
    printf( "x = %lu\n", x );

    x = 0;
    x ^= 1;
    printf( "x = %lu\n", x );
}

The program output is

x = 0
x = 1
x = 1

edited Mar 08 '22 at 14:20

answered Mar 08 '22 at 14:08

Vlad from Moscow

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This answer seems to imply that `x |= 1` and `x ^= 1` are interchangeable. It would be better to stress that `|=` *sets* a bit, while `^=` *toggles* the bit. – chepner Mar 08 '22 at 14:19

Bitwise and doesn't work on unsigned long

1 Answers1