在C中的参数给定范围内的屏蔽位 [英] Masking bits within a range given in parameter in C

问题描述

我是C语言编程的新手，不能确定是否已有很好的解释，如果可以的话，对不起.我正在尝试将位设置在给我的范围内.函数签名看起来像:

I am new to C programming and not sure that there is already a good explanation for how to do this, if so I am sorry. I am trying to set the bits within a range given to me. the function signature looks like:

unsigned int setBits(int low, int high, unsigned int source) {

source是要操作的数字，low是该范围内的最低位，而high是该范围内的最高位.我了解在尝试具体获取最后4位或前4位或其任意组合时的位移很好，但不了解如何从将在参数中更改的范围中获取位.任何帮助将不胜感激.

source being the number to be operated on, low being the lowest bit in the range, and high being the highest bit in the range. I understand bit-shifting just fine when trying to get specifically the last 4 bits or first 4 or any combination thereof, but do not understand how to get the bits from a range that will be changed in the parameter. Any help would be greatly appreciated.

推荐答案

2种方法:将source中的位从low设置为high的迭代方法:

2 approaches: Iterative method to set bit in source from low to high:

unsigned int setBitsI(int low, int high, unsigned int source) {
  while (low <= high) {
    source |= 1u << low;
    low++;
  }
  return source;
}

非迭代方法:

unsigned int setBitsNI(int low, int high, unsigned int source) {
  unsigned setmask = 1u << (high - low);
  setmask <<= 1;
  setmask--;
  setmask <<= low;
  return source | setmask;
}

当high为"bit-width-1"且low为0，1u << bit_width为UB时，避免使用1u << (1u + high - low)十分重要.

Important to avoid 1u << (1u + high - low) for when high is "bit-width-1" and low is 0, 1u << bit_width is UB.

low或high的值应超出位范围，否则会出现问题.

Should low or high have an value outside the bit range, problems occur.

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