What USEFUL bitwise operator code tricks should a developer know about? [closed]

Question

I must say I have never had cause to use bitwise operators, but I am sure there are some operations that I have performed that would have been more efficiently done with them. How have "shifting" and "OR-ing" helped you solve a problem more efficiently?

Answer 1

Using bitwise operations on strings (characters)

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' ^ '`') => 25

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

Answer 2

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• Bitwise operations on integers(int)

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Get the maximum integer

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

Get the minimum integer

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

Get the maximum long

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

Multiplied by 2

n << 1; // n*2 

Divided by 2

n >> 1; // n/2 

Multiplied by the m-th power of 2

n << m; // (3<<5) ==>3 * 2^5 ==> 96 

Divided by the m-th power of 2

n >> m; // (20>>2) ==>(20/( 2^2) ==> 5 

Check odd number

(n & 1) == 1; 

Exchange two values

a ^= b; b ^= a; a ^= b; 

Get absolute value

(n ^ (n >> 31)) - (n >> 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 have the same sign

(x ^ y) >= 0; 

Calculate 2^n

2 << (n-1); 

Whether is factorial of 2

n > 0 ? (n & (n - 1)) == 0 : false; 

Modulo 2^n against m

m & (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)); 

n + 1

-~n 

n - 1

~-n 

Get the contrast number

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

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

x = a ^ b ^ x;