按位和代替模运算符 [英] Bitwise and in place of modulus operator
问题描述
我们知道，例如，2 的幂的模可以这样表示:
We know that for example modulo of power of two can be expressed like this:
x % 2 inpower n == x & (2 inpower n - 1).
示例:
x % 2 == x & 1
x % 4 == x & 3
x % 8 == x & 7
两个数的一般非幂呢?
让我们说:
x % 7==?
推荐答案
首先，这样说其实不太准确
First of all, it's actually not accurate to say that
x % 2 == x & 1
简单的反例:x = -1.在许多语言中，包括 Java，-1 % 2 == -1.也就是说，% 不一定是模数的传统数学定义.例如，Java 称其为余数运算符".
Simple counterexample: x = -1. In many languages, including Java, -1 % 2 == -1. That is, % is not necessarily the traditional mathematical definition of modulo. Java calls it the "remainder operator", for example.
关于按位优化，在按位算术中只能轻松"完成2的模幂.一般而言，只有以 b 为底的数的模幂才能轻松"完成以 b 为底的数字表示.
With regards to bitwise optimization, only modulo powers of two can "easily" be done in bitwise arithmetics. Generally speaking, only modulo powers of base b can "easily" be done with base b representation of numbers.
例如，在基数 10 中，对于非负 N，N mod 10^k 只是取最低有效的 k 位.
In base 10, for example, for non-negative N, N mod 10^k is just taking the least significant k digits.
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