/* 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;
}然後升級版本：
根據：表達式 x & =(x - 1) 可以刪除x中最右邊值爲1的一個二進制位
解釋：
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).所以升級後：
/* bitcount: count 1 bits in x */
int bitcount(unsigned x)
{
int b;
for (b = 0; x != 0; x &= (x-1))
b++;
return b;
}