# Bitwise operations cheat sheet

*Recommendations and additions to this cheat sheet are welcome.*

This cheat sheet is mostly suitable for most common programming languages, but the target usage is C/C++ on x86 platform.

Bitmap `i`

is unsigned 32 bit integers. For 64 bit operands, the suffix `L`

should be added to integer literals, e.g. `1`

should be `1L`

.

All calculation demonstration is done with 8 bits integers for readability.

## Truth table

AND0 0 | 0

0 1 | 0

1 0 | 0

1 1 | 1OR0 0 | 0

0 1 | 1

1 0 | 1

1 1 | 1XOR0 0 | 0

0 1 | 1

1 0 | 1

1 1 | 0

## Operators

`AND &`

OR |

NOT ~

XOR ^

Left shift <<

Right shift >>

## Get a bit

`(i >> n) & 1`

## Set a bit to 1

`i | (1 << n)`

## Set a bit to 0

`i & ~(1 << n)`

## Store a bit

The bit to be stored is `v`

which is either `0`

or `1`

.

`(i & ~(1 << n)) | (v << n)`

## Toggle a bit

`i ^ (1 << n)`

## Get least significant bit

`i & -i`

Note: this gives you really the lowest bit but not the index of the lowest bit.

`-i`

is equivalent to `~i + 1`

`~i 10100111`

~i+1 10101000

i 01011000

i&-i 00001000

## Get most significant bit

`unsigned int get_msb(unsigned int i){`

i |= i >> 1;

i |= i >> 2;

i |= i >> 4;

i |= i >> 8;

i |= i >> 16;

return (i + 1) >> 1;

}

How it works:

`i 01000010`

i|=i>>1 01100011

i|=i>>2 01111011

i|=i>>4 01111111

...

i|=i>>16 01111111

i+1 10000000

(i+1)>>1 01000000

## Get index of most significant bit

`inline unsigned int get_bit_index(const unsigned int i){`

unsigned int r;

asm ( "bsr %1, %0\n"

: "=r"(r)

: "r" (i)

);

return r;

}

Yah, just one instruction. This instruction supports 16/32/64 bit integers, source and destination type must have the same size.

## Change endianess

Convert from big-endian to little-endian or vice-versa.

Well you shouldn’t need to handcraft this function but anyway FYR:

`((i>>24) & 0xFF) | // Move byte 3 to byte 0`

((i<<8) & 0xFF0000) | // Move byte 1 to byte 2

((i>>8) & 0xFF00) | // Move byte 2 to byte 1

((i<<24) & 0xFF000000) // Move byte 0 to byte 3

## Bit reversal

Lookup table is actually faster:

`static const unsigned char BitReverseTable256[] = `

{

0x00, 0x80, 0x40, 0xC0, 0x20, 0xA0, 0x60, 0xE0, 0x10, 0x90, 0x50, 0xD0, 0x30, 0xB0, 0x70, 0xF0,

0x08, 0x88, 0x48, 0xC8, 0x28, 0xA8, 0x68, 0xE8, 0x18, 0x98, 0x58, 0xD8, 0x38, 0xB8, 0x78, 0xF8,

0x04, 0x84, 0x44, 0xC4, 0x24, 0xA4, 0x64, 0xE4, 0x14, 0x94, 0x54, 0xD4, 0x34, 0xB4, 0x74, 0xF4,

0x0C, 0x8C, 0x4C, 0xCC, 0x2C, 0xAC, 0x6C, 0xEC, 0x1C, 0x9C, 0x5C, 0xDC, 0x3C, 0xBC, 0x7C, 0xFC,

0x02, 0x82, 0x42, 0xC2, 0x22, 0xA2, 0x62, 0xE2, 0x12, 0x92, 0x52, 0xD2, 0x32, 0xB2, 0x72, 0xF2,

0x0A, 0x8A, 0x4A, 0xCA, 0x2A, 0xAA, 0x6A, 0xEA, 0x1A, 0x9A, 0x5A, 0xDA, 0x3A, 0xBA, 0x7A, 0xFA,

0x06, 0x86, 0x46, 0xC6, 0x26, 0xA6, 0x66, 0xE6, 0x16, 0x96, 0x56, 0xD6, 0x36, 0xB6, 0x76, 0xF6,

0x0E, 0x8E, 0x4E, 0xCE, 0x2E, 0xAE, 0x6E, 0xEE, 0x1E, 0x9E, 0x5E, 0xDE, 0x3E, 0xBE, 0x7E, 0xFE,

0x01, 0x81, 0x41, 0xC1, 0x21, 0xA1, 0x61, 0xE1, 0x11, 0x91, 0x51, 0xD1, 0x31, 0xB1, 0x71, 0xF1,

0x09, 0x89, 0x49, 0xC9, 0x29, 0xA9, 0x69, 0xE9, 0x19, 0x99, 0x59, 0xD9, 0x39, 0xB9, 0x79, 0xF9,

0x05, 0x85, 0x45, 0xC5, 0x25, 0xA5, 0x65, 0xE5, 0x15, 0x95, 0x55, 0xD5, 0x35, 0xB5, 0x75, 0xF5,

0x0D, 0x8D, 0x4D, 0xCD, 0x2D, 0xAD, 0x6D, 0xED, 0x1D, 0x9D, 0x5D, 0xDD, 0x3D, 0xBD, 0x7D, 0xFD,

0x03, 0x83, 0x43, 0xC3, 0x23, 0xA3, 0x63, 0xE3, 0x13, 0x93, 0x53, 0xD3, 0x33, 0xB3, 0x73, 0xF3,

0x0B, 0x8B, 0x4B, 0xCB, 0x2B, 0xAB, 0x6B, 0xEB, 0x1B, 0x9B, 0x5B, 0xDB, 0x3B, 0xBB, 0x7B, 0xFB,

0x07, 0x87, 0x47, 0xC7, 0x27, 0xA7, 0x67, 0xE7, 0x17, 0x97, 0x57, 0xD7, 0x37, 0xB7, 0x77, 0xF7,

0x0F, 0x8F, 0x4F, 0xCF, 0x2F, 0xAF, 0x6F, 0xEF, 0x1F, 0x9F, 0x5F, 0xDF, 0x3F, 0xBF, 0x7F, 0xFF

};

unsigned int bit_reversal(const unsigned int i){

return (BitReverseTable256[i & 0xFF] << 24) |

(BitReverseTable256[(i >> 8) & 0xFF] << 16) |

(BitReverseTable256[(i >> 16) & 0xFF] << 8) |

(BitReverseTable256[(i >> 24) & 0xFF]);

}

## Count the number of bits

Hamming weight is faster than lookup table according to *Google’s “Director of Engineering” Hiring Test*. The actual performance test is here.

`const uint64_t m1 = 0x55555555; // 0101...`

const uint64_t m2 = 0x33333333; // 00110011...

const uint64_t m4 = 0x0F0F0F0F; // 0000111100001111...

const uint64_t m8 = 0x00FF00FF; // 8 zeros, 8 ones...

const uint64_t m16 = 0x0000FFFF; // 16 zeros, 16 ones...

const uint64_t m32 = 0x00000000ffffffff; // 32 zeros, 32 ones

const uint64_t h01 = 0x0101010101010101; //the sum of 256 to the power of 0,1,2,3...

//This is a naive implementation, shown for comparison,

//and to help in understanding the better functions.

//This algorithm uses 24 arithmetic operations (shift, add, and).

int popcount64a(uint64_t x)

{

x = (x & m1 ) + ((x >> 1) & m1 ); //put count of each 2 bits into those 2 bits

x = (x & m2 ) + ((x >> 2) & m2 ); //put count of each 4 bits into those 4 bits

x = (x & m4 ) + ((x >> 4) & m4 ); //put count of each 8 bits into those 8 bits

x = (x & m8 ) + ((x >> 8) & m8 ); //put count of each 16 bits into those 16 bits

x = (x & m16) + ((x >> 16) & m16); //put count of each 32 bits into those 32 bits

x = (x & m32) + ((x >> 32) & m32); //put count of each 64 bits into those 64 bits

return x;

}

//This uses fewer arithmetic operations than any other known

//implementation on machines with slow multiplication.

//This algorithm uses 17 arithmetic operations.

int popcount64b(uint64_t x)

{

x -= (x >> 1) & m1; //put count of each 2 bits into those 2 bits

x = (x & m2) + ((x >> 2) & m2); //put count of each 4 bits into those 4 bits

x = (x + (x >> 4)) & m4; //put count of each 8 bits into those 8 bits

x += x >> 8; //put count of each 16 bits into their lowest 8 bits

x += x >> 16; //put count of each 32 bits into their lowest 8 bits

x += x >> 32; //put count of each 64 bits into their lowest 8 bits

return x & 0x7f;

}

//This uses fewer arithmetic operations than any other known

//implementation on machines with fast multiplication.

//This algorithm uses 12 arithmetic operations, one of which is a multiply.

int popcount64c(uint64_t x)

{

x -= (x >> 1) & m1; //put count of each 2 bits into those 2 bits

x = (x & m2) + ((x >> 2) & m2); //put count of each 4 bits into those 4 bits

x = (x + (x >> 4)) & m4; //put count of each 8 bits into those 8 bits

return (x * h01) >> 56; //returns left 8 bits of x + (x<<8) + (x<<16) + (x<<24) + ...

}