awesome-bits 位运算黑魔法

awesome-bits 位运算黑魔法

awesome-bits 位运算黑魔法

2018年09月26日 发布，来源：github.com

awesome-bits A curated list of awesome bitwise operations and tricks

Maintainer - Keon Kim Please feel free to pull requests

Integers

Set nth bit

x | (1<<n)

Unset nth bit

x & ~(1<<n)

Toggle nth 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++;

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 mth power of 2

n << m;

Divide by the mth 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 mth bit of n (from low to high)

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

Set the mth bit of n to 0 (from low to high)

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

Check if nth 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;

((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)

Floats

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 * 2exp = 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 nth Root of positive numbers via Infinite Series

float root(float x, int n) {
#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 nth 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 2z = e

See here for its derivation.

Strings

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

Miscellaneous

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 