bitcount函数统计其整数参数的值为1的二进制位的个数

/* bitcount:  count 1 bits in x */
int bitcount(unsigned x)
{
    int b;
 
    for (b = 0; x != 0; x >>= 1)
        if (x & 01)
            b++;
    return b;
}

然后升级版本：

根据：表达式 x & =(x - 1) 可以删除x中最右边值为1的一个二进制位

解释：

Answer: If x is odd, then (x-1) has the same bit representation as x except that the rightmost 1-bit is now a 0. In this case, (x & (x-1)) == (x-1). If x is even, then the representation of (x-1) has the rightmost zeros of x becoming ones and the rightmost one becoming a zero. Anding the two clears the rightmost 1-bit in x and all the rightmost 1-bits from (x-1).

所以升级后：

/* bitcount:  count 1 bits in x */
int bitcount(unsigned x)
{
    int b;
 
    for (b = 0; x != 0; x &= (x-1))
        b++;
    return b;
} 

加快了执行速度