检查偶数或奇数`1`位在若干 [英] Checking even or odd `1` bits in a number

问题描述

我要检查是否有数量全部偶数或奇数位设置且仅他们。对于例如：

I want to check if number has all even or odd bits set to one and only them. For eg.:

数 42 是正确的，因为在二进制code 101010 它拥有一切，只有偶数位套 1 。
数 21 也是正确的， 10101 。

Number 42 is correct, because in a binary code 101010 it has all and only even bits sets to 1. Number 21 is also correct, 10101.

数 69 为如。 1000101 是不正确的，因为只有三奇位集来 1 。

Number 69 for eg. 1000101 is not correct, because there are only three odd bits sets to 1.

我已经用不同的操作与试过^＆安培;，＆GT;＆GT中，＆lt;＆LT; ，我仍然不知道我怎么可以这样用这些运营商。是的，我需要在 C 。

I've tried using different operations with ^, &, >>, << and i still have no idea how can i do this using these operators. Yes, i need to do this using logical operators in C.

推荐答案

这些数字有该属性（X ^（X GT;＆GT; 1））+ 1 是2的功率和是是2的幂，如果 Y'放大器; （Y - 1）== 0

Those numbers have that property that (x ^ (x >> 1)) + 1 is a power of 2. And y is a power of 2 if y & (y - 1) == 0

所以，一个测试可能是（（X ^（X GT;＆GT; 1））+ 1）及（X ^（X GT;＆GT; 1））== 0 这将适用于任何规模的数

So one test could be ((x ^ (x >> 1)) + 1) & (x ^ (x >> 1)) == 0 which would work for numbers of any size.

这篇关于检查偶数或奇数`1`位在若干的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！