C 中的位计数类似于 bit twiddling hack

Question

I need to make a routine that counts bits in a word that does not involve loops (only bit operations), and does not use large constants.

int x = 0xFFFFFFFF;
x += (~((x >> 1) & 0x55555555)+1);
x = (((x >> 2) & 0x33333333) + (x & 0x33333333));
x = (((x >> 4) + x) & 0x0F0F0F0F);
x += (x >> 8);
x += (x >> 16);
return(x & 0x0000003F);

This I found on bit twiddling hacks, but the largest constant I can use is 0xFF... Not sure how to do this otherwise.

Thanks folks.

Answer 1

You can for example use a constant array COUNTS[16] which is the number of set bits in the binary representation of numbers from 0 to 15. Then:

static inline int byte_count (int x) {
  static const int COUNTS[16] = { 0, 1, 1, 2, 1, /* fill in the rest manually */ };
  return COUNTS[x & 15] + COUNTS[x >> 4];
}

int count(int x) {
  return byte_count(x >> 24) + byte_count((x >> 16) & 255) + byte_count((x >> 8) & 255) + byte_count(x & 255);
}

No loops and no constants larger than 255.

Answer 2

Using your algorithm:

int x = 0xFF;
x |= (x << 8);  // x = 0xFFFF
x |= (x << 16); // x = 0xFFFFFFFF

and then the rest of the code - provided it works.

Recursive solution:

int foo ( int x )
{
    if ( x == 0 )
        return 0;
    return (x & 1) + foo ( x/2 );
}

Answer 3

your question is already answered here

int NumberOfSetBits(int i)
{
    i = i - ((i >> 1) & 0x55555555);
    i = (i & 0x33333333) + ((i >> 2) & 0x33333333);
    return (((i + (i >> 4)) & 0x0F0F0F0F) * 0x01010101) >> 24;
}