Summary

Name Operator Description
AND & Both bits are 1, return 1. Otherwise, return 0.
OR \ One of both bit is 1, return 1. Otherwise, return 0.
XOR ^ Two bits are different return 1. Otherwise, return 0.
NOT ~ Flip bit, 0 becomes 1, 1 becomes 0
Shift left << Shifts all the bits to the left
Right left >> Shifts all the bits to the right

Application

Integer

Change bit

//Set nth bit
x |= (1 << n);

//Set the right-most 0 bit to 1
x |= (x+1);

//Unset nth bit
x &= ~(1 << n);

//Set the right-most 1 bit to 0
x &= (x-1);

//Toggle nth bit
x ^= (1 << n);

//Get the mth bit of n
(x >> n) & 1;

//Swap Adjacent bits
((x & 10101010) >> 1) | ((x & 01010101) << 1);

Multiplication / Division x by $2^n$

x << n //multiplication
x >> n //division

Round up to the next power of two

x--;
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
x++;

Round down to the next power of two

x--;
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
x++;
x = x >> 1; //the same with the code above but added this line

Floor x

x >> 0;
6.723 >> 0; //6

Flip the sign

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

Absolute value of x

int abs = (x ^ (x >> 31)) - (x >> 31);

The greatest power of 2 that divides x

x & (-x);

Get the minimum / maximum integer (and long)

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

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

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

The min / max of two values

//First way
int min = a & ((a-b) >> 31) | b & (~(a-b) >> 31);
int max = b & ((a-b) >> 31) | a & (~(a-b) >> 31);

//Second way
int min = (b ^ (a ^ b) & -(a < b));
int max = (a ^ (a ^ b) & -(a < b));

Average

int avg = (x + y) >> 1;
int avg = ((x ^ y) >> 1) + (x & y);

Swap 2 variables

//First way
a ^= b;
b ^= a;
a ^= b;

//Second way
a = a ^ b ^ (b = a)

Quick conditional assignment

In some cases. you may require some conditional assignment such as

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

You can use bitwise XOR operator for these type of assignment.

x ^= a ^ b;

Check if…

  • nth bit is set

x & (1 << n)

  • x is an odd number

(x & 1) == 1;

  • x is a power of 2

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

  • a = b

(35% faster in Javascript)

(a ^ b) == 0;

  • both number have the same sign

(a ^ b) >= 0;

String

Convert letter to lowercase or uppercase

c & '_'

If the converting letter is uppercase, nothing change. Eg:

char c;
c = 'a' & '_';  //c = 'A'
c = 'A' & '_';  //c = 'A'

Convert letter to an uppercase or lowercase

c | ' '

If the converting letter is a lowercase, nothing change. Eg:

char c;
c = 'A' | ' ';  //c = 'a'
c = 'a' | ' ';  //c = 'a'

Invert letter’s case

c ^ ' '

For example:

char c;
c = 'A' ^ ' '; //c = 'a'
c = 'a' ^ ' '; //c = 'A'

Letter’s position in alphabet

//For lowercase only
c ^ '`'

//For uppercase only
c ^ '@'

//Letter case is not important
c & "\x1F"

Miscellaneous

Use XOR for basic encryption and decryption

#include<iostream>
#define key 5

int main(){
    std::string s;
    std::cin >> s;
    
    //Mã hóa
    for(char &i : s){
        i ^= key;
    }
    std::cout << s + "\n";
    
    //Giải mã
    for(char &i : s){
        i ^= key;
    }
    std::cout << s + "\n";
}

Input:  Hello
Output: M`iij
        Hello

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)

I’m gonna lose my mind after finished this post, thanks for reading.