当n为2^k时，如何证明m%n等价于m&(n-1)？

Question

Notice that '%' is a remainder operator and '&' is a bitwise AND operator, and k is an integer bigger than 0.

Example:

33%16=1 equivalent to 33&(16-1)=1

I found this equivalent in JDK1.8 ThreadLocalMap. I know it is correct but don't how to prove it is correct. I would be appreciated if you could give some help.

Answer 1

m % n is the remainder of dividing m by n and is therefore a number between 0 and n-1, or between 0 and 2^k - 1.

2^k in binary is one followed by k zeroes. 2^k -1 is k contiguous ones.

m & n is m & (2^k - 1) is a number in which, in binary, the 1 bits must occur in the rightmost k bits. It is therefore in the range 0 to 2^k - 1.

QED.