如何忽略变量[闭合]中的位

始终使用5位的解决方案可以是

new_value = ((data & 0x002) >> 1) |
            ((data & 0x008) >> 2) |
            ((data & 0x020) >> 3) |
            ((data & 0x080) >> 4) |
            ((data & 0x200) >> 5);

但这里有另一个解决方案,它不是使用固定数量的位(即在您的情况下为5位),而是使用一个函数,允许您指定要保留的位的数量。

#include <stdio.h>
#include <stdlib.h>

unsigned keepOddBits(const unsigned data, const unsigned number_of_bits_to_keep)
{
  unsigned new_value = 0;
  unsigned mask = 0x2;
  int i;
  for (i=0; i < number_of_bits_to_keep; ++i)
  {
    if (mask & data)
    {
      new_value = new_value | ((mask & data) >> (i + 1));
    }
    mask = mask << 2;
  }
  return new_value;
}

int main()
{
  printf("data 0x%x becomes 0x%x\n", 0x3ff, keepOddBits(0x3ff, 5));
  printf("data 0x%x becomes 0x%x\n", 0x2aa, keepOddBits(0x2aa, 5));
  printf("data 0x%x becomes 0x%x\n", 0x155, keepOddBits(0x155, 5));
  return 0;
}

将 输出 :

data 0x3ff becomes 0x1f
data 0x2aa becomes 0x1f
data 0x155 becomes 0x0

改变 main

int main()
{
  printf("data 0x%x becomes 0x%x\n", 0x3ff, keepOddBits(0x3ff, 3));
  printf("data 0x%x becomes 0x%x\n", 0x2aa, keepOddBits(0x2aa, 3));
  printf("data 0x%x becomes 0x%x\n", 0x155, keepOddBits(0x155, 3));
  return 0;
}

将 :

data 0x3ff becomes 0x7
data 0x2aa becomes 0x7
data 0x155 becomes 0x0

手动创建位,如下所示:

//Option 1
uint8_t result = 0;
result |= (inputValue >> 1) & 1;
result |= ((inputValue >> 3) & 1) << 1;
result |= ((inputValue >> 5) & 1) << 2;
result |= ((inputValue >> 7) & 1) << 3;
result |= ((inputValue >> 8) & 1) << 4;
result |= ((inputValue >> 9) & 1) << 5;

union
{
    struct
    {
        unsigned b0 : 1;
        unsigned b1 : 1;
        unsigned b2 : 1;
        unsigned b3 : 1;
        unsigned b4 : 1;
        unsigned b5 : 1;
        unsigned b6 : 1;

        unsigned bx : 3;
    }
    uint16_t raw;
}raw;

    raw.raw = value;

    uint8_t without_skipped = raw.b1 | (raw.b3 << 1) | (raw.b5 << 2) | (raw.bx << 3) ;

这是给小恩的