有效地设置数字 N 中的 X 位,从 MSB 的位置 P 开始

Effectively set X-bits in a number N, starting from position P from MSB

下面的代码在数字N中设置X位从LSB开始位置P(N是32位数字)

//gcc 5.4.0
#include  <stdio.h>
unsigned int set_bits_pos(int N, int X, int P)
{
    unsigned int mask = ((1UL << X)-1) << P; 
    return (N|mask);
}

int main(void)
{
    unsigned int bits = 3, pos=5, num=0x0F;       
    printf("0x%X\n", set_bits_pos(num,bits,pos));
    return 0;
}

输出:

0xEF

如何将上述函数转换为设置数字N中的X位,从MSB开始的位置P?最好的方法。谢谢

编辑:根据建议从 MSB 和 Arch Independent 设置 32 位和 64 位 N

//gcc 5.4.0
#include  <stdio.h>
size_t set_bits_pos(size_t N, int X, int P)
{
    size_t mask = ((1UL << X)-1) << ((8*sizeof(N)) - P - X); 
    return (N|mask);
}

int main(void)
{
    size_t num=0x0F;
    int bits = 3, pos=5;       
    printf("0x%X\n", set_bits_pos(num,bits,pos));
    return 0;
}

输出:

0x0700000F

我建议使用精确宽度的整数类型:

#include <inttypes.h>

uint32_t  set_low32(const uint32_t     value,
                    const unsigned int bits,
                    const unsigned int skip)
{
    uint32_t  mask = (~(uint32_t)0) >> (32 - bits);
    return value | (mask << skip);
}

uint64_t  set_low64(const uint64_t     value,
                    const unsigned int bits,
                    const unsigned int skip)
{
    uint64_t  mask = (~(uint64_t)0) >> (64 - bits);
    return value | (mask << skip);
}

uint32_t  set_high32(const uint32_t     value,
                     const unsigned int bits,
                     const unsigned int skip)
{
    uint32_t  mask = (~(uint32_t)0) << (32 - bits);
    return value | (mask >> skip);
}

uint64_t  set_high64(const uint64_t     value,
                     const unsigned int bits,
                     const unsigned int skip)
{
    uint64_t  mask = (~(uint64_t)0) << (64 - bits);
    return value | (mask >> skip);
}

bits 是要设置的位数(有问题的 X),skip 是要跳过的 low/high 位数(P 有问题)。