Bitwise Operators - C

Integers are stored in binary. Bitwise operators work on each bit separately, which makes them fast and useful for flags, masks, low-level hardware code and many GATE questions.

Operator	Name	Rule for each bit	Example (a = 12 = 1100₂, b = 10 = 1010₂)
`a & b`	AND	1 if both bits are 1	1000₂ = 8
`a \| b`	OR	1 if either bit is 1	1110₂ = 14
`a ^ b`	XOR	1 if the bits differ	0110₂ = 6
`~a`	NOT	flips every bit	−13 (two's complement)
`a << n`	left shift	moves bits left, fills with 0	12 << 2 = 48 (× 4)
`a >> n`	right shift	moves bits right	12 >> 2 = 3 (÷ 4)

Open in playground

#include <stdio.h>

void print_bits(unsigned int x) {
    for (int i = 7; i >= 0; i--) {
        putchar((x >> i) & 1 ? '1' : '0');
    }
}

int main(void) {
    int a = 12, b = 10;
    printf("a & b = %d\n", a & b);
    printf("a | b = %d\n", a | b);
    printf("a ^ b = %d\n", a ^ b);
    printf("~a    = %d\n", ~a);
    printf("a << 2 = %d, a >> 2 = %d\n", a << 2, a >> 2);
    printf("12 in binary: ");
    print_bits(12);
    printf("\n");
    return 0;
}

Output

a & b = 8
a | b = 14
a ^ b = 6
~a    = -13
a << 2 = 48, a >> 2 = 3
12 in binary: 00001100

~a is −13 because C stores negative numbers in two's complement: ~x = −x − 1 for every integer x.

Common bit tricks

Task	Expression
Is bit k set?	`(x >> k) & 1`
Set bit k	`x \| (1 << k)`
Clear bit k	`x & ~(1 << k)`
Toggle bit k	`x ^ (1 << k)`
Is x odd?	`x & 1`
Is x a power of two? (x > 0)	`(x & (x - 1)) == 0`
Turn off the lowest set bit	`x & (x - 1)`

Open in playground

#include <stdio.h>

int count_ones(unsigned int x) {
    int count = 0;
    while (x) {
        x &= x - 1;      /* removes the lowest 1 bit each time */
        count++;
    }
    return count;
}

int main(void) {
    int x = 0;
    x |= 1 << 3;                      /* set bit 3 */
    x |= 1 << 0;                      /* set bit 0 */
    printf("x = %d\n", x);
    printf("bit 3 set? %d\n", (x >> 3) & 1);
    x &= ~(1 << 3);                   /* clear bit 3 */
    printf("after clearing bit 3: %d\n", x);
    printf("ones in 255: %d, ones in 1000: %d\n", count_ones(255), count_ones(1000));
    printf("64 is a power of two? %d\n", (64 & 63) == 0);

    int p = 5, q = 9;                 /* swap with XOR, no temporary */
    p ^= q;
    q ^= p;
    p ^= q;
    printf("p = %d, q = %d\n", p, q);
    return 0;
}

Output

x = 9
bit 3 set? 1
after clearing bit 3: 1
ones in 255: 8, ones in 1000: 6
64 is a power of two? 1
p = 9, q = 5

Warning: Bitwise operators have lower precedence than comparison operators. x & 1 == 0 means x & (1 == 0), which is always 0. Write (x & 1) == 0.

GATE-style practice questions

GATE-styleNAT · numerical answer

What does the following program print?

#include <stdio.h>
int main(void) {
    int a = 13, b = 6;
    printf("%d\n", (a & b) | (a ^ b));
    return 0;
}

Your answer

Show answer and explanation

Answer: 15

13 = 1101₂ and 6 = 0110₂. a & b = 0100₂ = 4 and a ^ b = 1011₂ = 11. 4 | 11 = 1111₂ = 15. (In fact (a & b) | (a ^ b) always equals a | b.)

GATE-styleNAT · numerical answer

What does the following program print?

#include <stdio.h>
int main(void) {
    unsigned int x = 0xF0;
    x = (x >> 4) | (x << 4);
    printf("%u\n", x & 0xFF);
    return 0;
}

Your answer

Show answer and explanation

Answer: 15

0xF0 >> 4 = 0x0F and 0xF0 << 4 = 0xF00. ORing gives 0xF0F, and keeping only the low 8 bits with & 0xFF leaves 0x0F = 15.

GATE-styleMCQ · one correct answer

What does the following program print?

#include <stdio.h>
int main(void) {
    int x = 5;
    if (x & 1 == 0)
        printf("even\n");
    else
        printf("odd\n");
    return 0;
}

even
odd
Compilation error
Nothing is printed

Show answer and explanation

Answer: B

== binds tighter than &, so the condition is x & (1 == 0) = 5 & 0 = 0, which is false, and the else branch runs. Here the answer happens to be right for 5, but for x = 4 the program would also print "odd". Always bracket bitwise tests: (x & 1) == 0.

GATE-styleMSQ · one or more correct answers

For a positive integer x, which expressions are non-zero exactly when x is a power of two? (Select all that apply.)

(x & (x - 1)) == 0
(x & -x) == x
(x | (x - 1)) == x
(x ^ 1) == 0

Show answer and explanation

Answer: A, B

A power of two has a single 1 bit. x & (x − 1) clears the lowest 1 bit, so the result is 0 exactly when there was only one. x & −x isolates the lowest 1 bit, which equals x exactly when it is the only one. The other two fail, for example for x = 4.