这是一个小分析(“可读形式”版本):
usnigned int nx = ~x; // I suppose it's unsigned
int a = nx & (nx >> 1);
// a will be 0 if there are no 2 consecutive "1" bits.
// or it will contain "1" in position N1 if nx had "1" in positions N1 and N1 + 1
if (a == 0) return 0; // we don't have set bits for the following algorithm
int b = a ^ (a & (a - 1));
// a - 1 : will reset the least 1 bit and will set all zero bits (say, NZ) that were on smaller positions
// a & (a - 1) : will leave zeroes in all (NZ + 1) LSB bits (because they're only bits that has changed
// a ^ (a & (a - 1)) : will cancel the high part, leaving only the smallest bit that was set in a
// so, for a = 0b0100100 we'll obtain a power of two: b = 0000100
return b | (b << 1);
// knowing that b is a power of 2, the result is b + b*2 => b*3
似乎该算法正在寻找变量中的前 2 个(从 LSB 开始)连续0
位。x
如果没有,则结果为 0。如果找到它们,例如在 position 上PZ
,则结果将包含两个设置位:PZ
和PZ+1
。