获取低阶N位 [英] Getting Lower-Order N Bits

问题描述

有没有任何方法可以获取任何整数的低阶 n 位(其中 n 可以是1到32之间的任何数字)而无需预先计算32位掩码，每个订单一个，并使用& 运算符?我也不希望使用具有两个幂的％，而只是按位运算.

Is there any way you can get any integer's lower-order n bits (where n can be any number between 1 and 32) without previously pre-computing 32 bitmasks, one for each order, and using the & operator? I also don't want to be using % with powers of two, just bitwise operations.

例如，假设用户输入一个整数 Num 和另一个整数 ShiftCount ，值的范围是1到32我想将操作 Num>>中丢失的位存储在第三个变量中.ShiftCount .

Say, for example, that a user enters an integer Num and another integer ShiftCount with a value ranging from 1 to 32. I want to store in a third variable the bits that are lost in the operation Num >> ShiftCount.

推荐答案

类似于 Num&((1<< ShiftCount)-1)?

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