这个C程序背后的逻辑是什么？

Question

Here is a small piece of program (14 lines of program) which counts the number of bits set in a number .

Input-Output --> 0-->0(0000000), 5-->2(0000101), 7-->3(0000111)

int CountBits (unsigned int x)
{
  static unsigned int mask[] = { 0x55555555,
      0x33333333,
      0x0F0F0F0F,
      0x00FF00FF,
      0x0000FFFF
      } ;

      int i ;
      int shift ; /* Number of positions to shift to right*/
      for (i =0, shift =1; i < 5; i ++, shift *= 2)
              x = (x & mask[i ])+ ( ( x >> shift) & mask[i]);
      return x;
}

Can someone explain the algorithm used here/why this works?

Answer 1

This post , by Ian Ashdown, explains it in more detail:

Freed's numbers are members of a sequence, where the Nth number of the sequence is itself an infinite sequence from right to left of 2* N 1's followed by 2 *N 0's, followed 2**N 1's, and so on. The initial numbers are:

...0101010101010101
...0011001100110011
...0000111100001111
...0000000011111111
...

For a word size of 16 bits then, we have four "B-constants":

B[1] = 0101010101010101
B[2] = 0011001100110011
B[3] = 0000111100001111
B[4] = 0000000011111111

So that's what those numbers in mask[] are, eg. 0x55555555 is the hexadecimal representation of the bit pattern 1010101010101010101010101010101 .

The algorithm itself does this:

1. Interpret adjacent bits as numbers (0 or 1) and add them. The results are numbers that can be represented with two bits (ie. 0 to 3).
2. Interpret adjacent pairs of bits as numbers (0 to 3) and add them. The results can be represented with four bits (ie. 0 to 15).
3. Interpret adjacent groups-of-4 bits as numbers (0 to 15) and add them. The results can be represented with eight bits (ie. 0 to 255).

...and so on, until you have a result that is as wide as however many bits you need.

I suggest that you try it on paper, by hand, with a few numbers using the binary masks above. Then you might get a feel for the algorithm being expressed by that code.