c - C 中恰好一个字节的位掩码

Question

我的目标是拯救 long四字节 like this :

unsigned char bytes[4];
unsigned long n = 123;

bytes[0] = (n >> 24) & 0xFF;
bytes[1] = (n >> 16) & 0xFF;
bytes[2] = (n >> 8) & 0xFF;
bytes[3] = n & 0xFF;

但我希望代码是可移植的，所以我使用 CHAR_BIT来自<limits.h> :

unsigned char bytes[4];
unsigned long n = 123;

bytes[0] = (n >> (CHAR_BIT * 3)) & 0xFF;
bytes[1] = (n >> (CHAR_BIT * 2)) & 0xFF;
bytes[2] = (n >> CHAR_BIT) & 0xFF;
bytes[3] = n & 0xFF;

问题在于位掩码 0xFF只占8位，并不一定相当于1个字节。有没有办法让上层代码完全移植到所有平台？

Answer 1

怎么样:

unsigned long mask = 1;
mask<<=CHAR_BIT;
mask-=1;

然后使用它作为掩码而不是0xFF？

测试程序:

#include <stdio.h>

int main() {
    #define MY_CHAR_BIT_8 8
    #define MY_CHAR_BIT_9 9
    #define MY_CHAR_BIT_10 10
    #define MY_CHAR_BIT_11 11
    #define MY_CHAR_BIT_12 12
    {
        unsigned long mask = 1;
        mask<<=MY_CHAR_BIT_8;
        mask-= 1;
        printf("%lx\n", mask);
    }
    {
        unsigned long mask = 1;
        mask<<=MY_CHAR_BIT_9;
        mask-= 1;
        printf("%lx\n", mask);
    }
    {
        unsigned long mask = 1;
        mask<<=MY_CHAR_BIT_10;
        mask-= 1;
        printf("%lx\n", mask);
    }
    {
        unsigned long mask = 1;
        mask<<=MY_CHAR_BIT_11;
        mask-= 1;
        printf("%lx\n", mask);
    }
    {
        unsigned long mask = 1;
        mask<<=MY_CHAR_BIT_12;
        mask-= 1;
        printf("%lx\n", mask);
    }
}

Output :

ff
1ff
3ff
7ff
fff