数字中设置的位数 [英] Number of bits set in a number

问题描述

以下神奇的公式给出了设置的数字位数(汉明权重).

The following the magical formula which gives the number of bits set in a number (Hamming weight).

/*Code to Calculate count of set bits in a number*/
int c;
int v = 7;
v = v - ((v >> 1) & 0x55555555);                    // reuse input as temporary
v = (v & 0x33333333) + ((v >> 2) & 0x33333333);     // temp
c = ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24; // count
printf(" Number of Bits is %d",c);
/*-----------------------------------*/

来自: http://graphics.stanford.edu/~seander/bithacks.html

任何人都可以向我解释其背后的原理吗?

Can anyone please explain me the rationale behind this?

推荐答案

这确实是非常聪明的代码，而且显然比简单的幼稚循环更难理解.

It's really quite clever code, and is obviously a lot more difficult to understand than a simple naive loop.

对于第一行，让我们只取一个四位的数量，并将其称为abcd.该代码基本上是这样做的:

For the first line, let's just take a four-bit quantity, and call it abcd. The code basically does this:

abcd - ((abcd >> 1) & 0101) = abcd - (0abc & 0101) = abcd - 0a0c

因此，在每个两位的组中，它减去高位的值.这对我们有什么好处?

So, in each group of two bits, it subtracts the value of the high bit. What does that net us?

11 - 1 -> 10 (two bits set)
10 - 1 -> 01 (one bit set)
01 - 0 -> 01 (one bit set)
00 - 0 -> 00 (zero bits set)

因此，第一行将每个连续的两位组设置为原始值中包含的位数-它计算以两位为一组的设置位数.将所得的四位数称为ABCD.

So, that first line sets each consecutive group of two bits to the number of bits contained in the original value -- it counts the bits set in groups of two. Call the resulting four-bit quantity ABCD.

下一行:

(ABCD & 0011) + ((ABCD>>2) & 0011) = 00CD + (AB & 0011) = 00CD + 00AB

因此，它采用两位的组并将对相加在一起.现在，每个四位组包含在输入的相应四位中设置的位数.

So, it takes the groups of two bits and adds pairs together. Now, each four-bit group contains the number of bits set in the corresponding four bits of the input.

在下一行中，v + (v >> 4) & 0xF0F0F0F(被解析为(v + (v >> 4)) & 0xf0f0f0f)执行相同的操作，将成对的四位组加在一起，以便每个八位组(字节)都包含对位的计数.相应的输入字节.现在我们有一个像0x0e0f0g0h这样的数字.

In the next line, v + (v >> 4) & 0xF0F0F0F (which is parsed as (v + (v >> 4)) & 0xf0f0f0f) does the same, adding pairs of four-bit groups together so that each eight-bit group (byte) contains the bit-set count of the corresponding input byte. We now have a number like 0x0e0f0g0h.

请注意，将任意位置的字节乘以0x01010101会将该字节复制到最高有效字节(以及将一些副本保留在较低字节中).例如，0x00000g00 * 0x01010101 = 0x0g0g0g00.因此，通过乘以0x0e0f0g0h，我们将把e+f+g+h保留在最上面的字节中.最后的>>24会提取该字节并留下答案.

Note that multiplying a byte in any position by 0x01010101 will copy that byte up to the most-significant byte (as well as leaving some copies in lower bytes). For example, 0x00000g00 * 0x01010101 = 0x0g0g0g00. So, by multiplying 0x0e0f0g0h, we will leave e+f+g+h in the topmost byte; the >>24 at the end extracts that byte and leaves you with the answer.

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