如何确定一个数字中的所有设置位是否也都设置在另一个数字中? [英] How to determine if all set bits in one number is also set in another number?

问题描述

给出2个长度相同的二进制字符串 a 和 b (例如 111000 和 110000 )，是否可以使用按位操作检查 b 中的所有设置位是否也在 a 中设置?在上面的示例中， b = 110000 在位置1和2(从左到右)中有两个设置位，这些位也都在 a = 111000 中设置./p>

我可以按字符比较两个字符串，但这太慢了.我知道我可以使用 Integer.parseInt(a，2); 之类的东西将字符串转换为数字，但是我不知道用于完成任务的按位运算.

解决方案

屏蔽位，然后检查它们是否仍然全部设置.

  int a = 0b111000;int b = 0b110000;如果((a& b)== b){...} 

Given 2 binary strings a and b of the same length (e.g. 111000 and 110000), is it possible to check if all set bits in b is also set in a using bitwise operations? In the above example, b = 110000 has two set bits in position 1 and 2 (from left to right), and those bits are also set in a = 111000.

I can compare two strings character-wise but that would be too slow. I know I can turn the strings into number with something like Integer.parseInt(a, 2); but I don't know the bitwise operations to use to achieve the task.

解决方案

Mask the bits and then check if they're still all set.

int a = 0b111000;
int b = 0b110000;

if ((a & b) == b) {
    ...
}

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