• 找到最低位的 1

n & -n;
/*
  6 ^ -6 = 110 ^ 010 = 010
*/

• 消掉最低位的 1

n & (n - 1);

• 檢查是不是$2^n$

const isPowerOfTwo = x !== 0 && n & (n - 1 === 0);

XOR

• Swap two varibles

a = 4;
b = 6;
a = a ^ b;
b = a ^ b;
a = a ^ b;
console.log(a, b); // 6 4

Reference

• Bit manipulation
• Bitwise operation
• Bit Twiddling Hacks