求数字中1的个数

最近还是在学习《redis设计与实现》及源码，学习到位操作这里，这里谈及了用汉明权重swar算法获取二进制中1的个数，因此复习特此相关算法。直接上代码如下：

/*一个一个数看最末位是否是1*/
int ones_func1(int num)
{
    int count = 0;
    int tmp_num = num;
    while(tmp_num > 0)
    {
        count += tmp_num & 0x1;
        tmp_num = tmp_num >> 1;
    }
    return count;
}

/*用特性n&(n-1)消除最末位1*/
int ones_func2(int num)
{
    int count = 0;
    int tmp_num = num;
    while(tmp_num > 0)
    {
        count++;
        tmp_num = tmp_num & (tmp_num - 1);
    }
    return  count;
}

/*swar 算法计算汉明重量*/
int ones_func3(int num)
{
    int count = num;
    count = (count & 0x55555555) + ((count >> 1) & 0x55555555);
    count = (count & 0x33333333) + ((count >> 2) & 0x33333333);
    count = (count & 0x0F0F0F0F) + ((count >> 4) & 0x0F0F0F0F);
    count = (count * (0x01010101) >> 24);
    return count;
}

static const unsigned char bitsinbyte[256] =
                                            {
                                             0,1,1,2,1,2,2,3,
                                             1,2,2,3,2,3,3,4,
                                             1,2,2,3,2,3,3,4,
                                             2,3,3,4,3,4,4,5,
                                             1,2,2,3,2,3,3,4,
                                             2,3,3,4,3,4,4,5,
                                             2,3,3,4,3,4,4,5,
                                             3,4,4,5,4,5,5,6,
                                             1,2,2,3,2,3,3,4,
                                             2,3,3,4,3,4,4,5,
                                             2,3,3,4,3,4,4,5,
                                             3,4,4,5,4,5,5,6,
                                             2,3,3,4,3,4,4,5,
                                             3,4,4,5,4,5,5,6,
                                             3,4,4,5,4,5,5,6,
                                             4,5,5,6,5,6,6,7,
                                             1,2,2,3,2,3,3,4,
                                             2,3,3,4,3,4,4,5,
                                             2,3,3,4,3,4,4,5,
                                             3,4,4,5,4,5,5,6,
                                             2,3,3,4,3,4,4,5,
                                             3,4,4,5,4,5,5,6,
                                             3,4,4,5,4,5,5,6,
                                             4,5,5,6,5,6,6,7,
                                             2,3,3,4,3,4,4,5,
                                             3,4,4,5,4,5,5,6,
                                             3,4,4,5,4,5,5,6,
                                             4,5,5,6,5,6,6,7,
                                             3,4,4,5,4,5,5,6,
                                             4,5,5,6,5,6,6,7,
                                             4,5,5,6,5,6,6,7,
                                             5,6,6,7,6,7,7,8
                                            };
/*查表法*/
int ones_func4(int num)
{
    int tmp_num = num;
    int count = 0;
    if (tmp_num  < 256 )
        count = bitsinbyte[tmp_num];
    else
    {
        while (tmp_num >= 256)
        {
            count += bitsinbyte[tmp_num & 0xff];
            tmp_num = tmp_num >> 8;
        }
        count += bitsinbyte[tmp_num];
    }
    return count;
}

位运算用途多，比如通过n&(n-1)是否等于0判断是否一个数为2的次幂，通过亦或判断两个数的汉明距离等。

最后

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