Bitmask for exactly one byte in C

Question

My goal is to save a long in four bytes 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;

But I want the code to be portable, so I use CHAR_BIT from <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;

The problem is that the bitmask 0xFF only accounts for eight bits, which is not necessarily equivalent to one byte. Is there a way to make the upper code completely portable for all platforms?

Answer 1

How about something like:

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

and then using this as the mask instead of 0xFF?

Test program:

#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