用Hamming weight统计二进制中1的个数的实现原理

[修电视的攻城狮]

在网上找到一种据说是性能最好的统计二进制中1的个数方法，用的是Hamming weight，实现如下：

/* ===========================================================================
* Problem: 
*   The fastest way to count how many 1s in a 32-bits integer.
*
* Algorithm:
*   The problem equals to calculate the Hamming weight of a 32-bits integer,
*   or the Hamming distance between a 32-bits integer and 0. In binary cases,
*   it is also called the population count, or popcount.[1]
*
*   The best solution known are based on adding counts in a tree pattern
*   (divide and conquer). Due to space limit, here is an example for a
*   8-bits binary number A=01101100:[1]
* | Expression            | Binary   | Decimal | Comment                    |
* | A                     | 01101100 |         | the original number        |
* | B = A & 01010101      | 01000100 | 1,0,1,0 | every other bit from A     |
* | C = (A>>1) & 01010101 | 00010100 | 0,1,1,0 | remaining bits from A      |
* | D = B + C             | 01011000 | 1,1,2,0 | # of 1s in each 2-bit of A | 
* | E = D & 00110011      | 00010000 | 1,0     | every other count from D   |
* | F = (D>>2) & 00110011 | 00010010 | 1,2     | remaining counts from D    |
* | G = E + F             | 00100010 | 2,2     | # of 1s in each 4-bit of A |
* | H = G & 00001111      | 00000010 | 2       | every other count from G   |
* | I = (G>>4) & 00001111 | 00000010 | 2       | remaining counts from G    |
* | J = H + I             | 00000100 | 4       | No. of 1s in A             |
* Hence A have 4 1s.
*
* [1] http://en.wikipedia.org/wiki/Hamming_weight
*
* ===========================================================================
*/
请问下，这个方法的原理是什么？为什么经过上面求A、B、C、D、E、F、G、H、I、J就得出了结果？

[天下第一好大人]

就是归并来计算求和，先是1bit和1bit的和存入2bit里，然后2bit和2bit的和存入4bit里，然后4bit和4bit的和存入8bit里。而在每个阶段，求和以后结果的范围都不会超越那几个bit的最大值。

移位的作用就是为了相临的bits相加时对齐用的。

[showjim]

Intel SSE4有专用指令POPCNT
这个算法的改良版第一次归并用的是减法，比直接两两归并少一两条指令