A curated list of awesome bitwise operations and tricks

Maintainer - Keon Kim Please feel free to pull requests

Set n^th bit

x | (1<<n)

Unset n^th bit

x & ~(1<<n)

Toggle n^th bit

x ^ (1<<n)

Round up to the next power of two

unsigned int v; //only works if v is 32 bit
v--;
v |= v >> 1;
v |= v >> 2;
v |= v >> 4;
v |= v >> 8;
v |= v >> 16;
v++;

Round down / floor a number

n >> 0

5.7812 >> 0 // 5

Get the maximum integer

int maxInt = ~(1 << 31);
int maxInt = (1 << 31) - 1;
int maxInt = (1 << -1) - 1;
int maxInt = -1u >> 1;

Get the minimum integer

int minInt = 1 << 31;
int minInt = 1 << -1;

Get the maximum long

long maxLong = ((long)1 << 127) - 1;

Multiply by 2

n << 1; // n*2

Divide by 2

n >> 1; // n/2

Multiply by the m^th power of 2

n << m;

Divide by the m^th power of 2

n >> m;

Check Equality

_{This is 35% faster in Javascript}

(a^b) == 0; // a == b
!(a^b) // use in an if

Check if a number is odd

(n & 1) == 1;

Exchange (swap) two values

//version 1
a ^= b;
b ^= a;
a ^= b;

//version 2
a = a ^ b ^ (b = a)

Get the absolute value

//version 1
x < 0 ? -x : x;

//version 2
(x ^ (x >> 31)) - (x >> 31);

Get the max of two values

b & ((a-b) >> 31) | a & (~(a-b) >> 31);

Get the min of two values

a & ((a-b) >> 31) | b & (~(a-b) >> 31);

Check whether both numbers have the same sign

(x ^ y) >= 0;

Flip the sign

i = ~i + 1; // or
i = (i ^ -1) + 1; // i = -i

Calculate 2ⁿ

1 << n;

Whether a number is power of 2

n > 0 && (n & (n - 1)) == 0;

Modulo 2ⁿ against m

m & ((1 << n) - 1);

Get the average

(x + y) >> 1;
((x ^ y) >> 1) + (x & y);

Get the m^th bit of n (from low to high)

(n >> (m-1)) & 1;

Set the m^th bit of n to 0 (from low to high)

n & ~(1 << (m-1));

Check if n^th bit is set

if (x & (1<<n)) {
  n-th bit is set
} else {
  n-th bit is not set
}

Isolate (extract) the right-most 1 bit

x & (-x)

Isolate (extract) the right-most 0 bit

~x & (x+1)

Set the right-most 0 bit to 1

x | (x+1)

Set the right-most 1 bit to 0

x & (x-1)

n + 1

-~n

n - 1

~-n

Get the negative value of a number

~n + 1;
(n ^ -1) + 1;

if (x == a) x = b; if (x == b) x = a;

x = a ^ b ^ x;

Swap Adjacent bits

((n & 10101010) >> 1) | ((n & 01010101) << 1)

Different rightmost bit of numbers m & n

(n^m)&-(n^m) // returns 2^x where x is the position of the different bit (0 based)

Common rightmost bit of numbers m & n

~(n^m)&(n^m)+1 // returns 2^x where x is the position of the common bit (0 based)

These are techniques inspired by the fast inverse square root method. Most of these are original.

Turn a float into a bit-array (unsigned uint32_t)

#include <stdint.h>
typedef union {float flt; uint32_t bits} lens_t;
uint32_t f2i(float x) {
  return ((lens_t) {.flt = x}).bits;
}

_{Caveat: Type pruning via unions is undefined in C++; use std::memcpy instead.}

Turn a bit-array back into a float

float i2f(uint32_t x) {
  return ((lens_t) {.bits = x}).flt;
}

Approximate the bit-array of a positive float using frexp

frexp gives the 2ⁿ decomposition of a number, so that man, exp = frexp(x) means that man * 2^exp = x and 0.5 <= man < 1.

man, exp = frexp(x);
return (uint32_t)((2 * man + exp + 125) * 0x800000);

_{Caveat: This will have at most 2^-16 relative error, since man + 125 clobbers the last 8 bits, saving the first 16 bits of your mantissa.}

Fast Inverse Square Root

return i2f(0x5f3759df - f2i(x) / 2);

_{Caveat: We're using the i2f and the f2i functions from above instead.}

See this Wikipedia article for reference.

Fast n^th Root of positive numbers via Infinite Series

float root(float x, int n) {
#DEFINE MAN_MASK 0x7fffff
#DEFINE EXP_MASK 0x7f800000
#DEFINE EXP_BIAS 0x3f800000
  uint32_t bits = f2i(x);
  uint32_t man = bits & MAN_MASK;
  uint32_t exp = (bits & EXP_MASK) - EXP_BIAS;
  return i2f((man + man / n) | ((EXP_BIAS + exp / n) & EXP_MASK));
}

See this blog post regarding the derivation.

Fast Arbitrary Power

return i2f((1 - exp) * (0x3f800000 - 0x5c416) + f2i(x) * exp)

_{Caveat: The 0x5c416 bias is given to center the method. If you plug in exp = -0.5, this gives the 0x5f3759df magic constant of the fast inverse root method.}

See these set of slides for a derivation of this method.

Fast Geometric Mean

The geometric mean of a set of n numbers is the n^th root of their product.

#include <stddef.h>
float geometric_mean(float* list, size_t length) {
  // Effectively, find the average of map(f2i, list)
  uint32_t accumulator = 0;
  for (size_t i = 0; i < length; i++) {
    accumulator += f2i(list[i]);
  }
  return i2f(accumulator / n);
}

See here for its derivation.

Fast Natural Logarithm

#DEFINE EPSILON 1.1920928955078125e-07
#DEFINE LOG2 0.6931471805599453
return (f2i(x) - (0x3f800000 - 0x66774)) * EPSILON * LOG2

_{Caveat: The bias term of 0x66774 is meant to center the method. We multiply by ln(2) at the end because the rest of the method computes the log2(x) function.}

See here for its derivation.

Fast Natural Exp

return i2f(0x3f800000 + (uint32_t)(x * (0x800000 + 0x38aa22)))

_{Caveat: The bias term of 0x38aa22 here corresponds to a multiplicative scaling of the base. In particular, it corresponds to z such that 2^z = e}

See here for its derivation.

Convert letter to lowercase:

OR by space => (x | ' ')
Result is always lowercase even if letter is already lowercase
eg. ('a' | ' ') => 'a' ; ('A' | ' ') => 'a'

Convert letter to uppercase:

AND by underline => (x & '_')
Result is always uppercase even if letter is already uppercase
eg. ('a' & '_') => 'A' ; ('A' & '_') => 'A'

Invert letter's case:

XOR by space => (x ^ ' ')
eg. ('a' ^ ' ') => 'A' ; ('A' ^ ' ') => 'a'

Letter's position in alphabet:

AND by chr(31)/binary('11111')/(hex('1F') => (x & "\x1F")
Result is in 1..26 range, letter case is not important
eg. ('a' & "\x1F") => 1 ; ('B' & "\x1F") => 2

Get letter's position in alphabet (for Uppercase letters only):

AND by ? => (x & '?') or XOR by @ => (x ^ '@')
eg. ('C' & '?') => 3 ; ('Z' ^ '@') => 26

Get letter's position in alphabet (for lowercase letters only):

XOR by backtick/chr(96)/binary('1100000')/hex('60') => (x ^ '`')
eg. ('d' ^ '`') => 4 ; ('x' ^ '`') => 24

Fast color conversion from R5G5B5 to R8G8B8 pixel format using shifts

R8 = (R5 << 3) | (R5 >> 2)
G8 = (R5 << 3) | (R5 >> 2)
B8 = (R5 << 3) | (R5 >> 2)

Note: using anything other than the English letters will produce garbage results

• Bit Twiddling Hacks
• Floating Point Hacks
• Hacker's Delight
• The Bit Twiddler