32位无符号整数中1的个数

      4 int bitcount(unsigned int n)
      5 {
      6     n = n - ((n>>1)&033333333333) - ((n>>2)&011111111111);
      7     n = (n + (n>>3))&030707070707;

      8     n = n % 0x3f;
      9     return n;
      10 }

   17 int bitcount(unsigned int n)
     18 {
     19     n = n - ((n>>1)&0x77777777) - ((n>>2)&0x33333333)-((n>>3)&0x11111111);
     20     n = (n + (n>>4))&0xf0f0f0f;

     21     n = n + ((n >> 8) & 0xf) + ((n >> 16) & 0xf) + ((n >> 24) & 0xf);
     22     n = n & 0xff;
     23     return n;                                                                                             
     24 }

     4 int bitcount(unsigned int n)
      5 {
      6     unsigned int tmp;
      7     tmp = n & 0x33333333;
      8     n = n - tmp;
      9     n = (n >> 2) & 0x33333333;
     10     tmp = tmp - ((tmp >> 1) & 0x33333333);
     11     n = n - ((n >> 1) & 0x33333333);
     12     tmp = tmp + n;
     13     n = tmp + (tmp >> 16);
     14     tmp = (n & 0xf) + ((n >> 4) & 0xf) +
     15           ((n >> 8) & 0xf) + ((n >> 12) & 0xf);
     16     tmp = tmp & 0xff;
     17     return tmp;
     18 }

    int bitcount(unsigned int n)

{

  unsigned x;

      x = (n >> 1) & 0x77777777;

      n = n - x;

      x = (x >> 1) & 0x77777777;

      n = n - x;

      x = (x >> 1) & 0x77777777;

      n = n -x;

      x = (n + (n >> 16));

      n =  (n & 0xf) + ((n >> 4) & 0xf) + ((n >> 8) & 0xf) + ((n >> 12) & 0xf);

       x = n & 0xff;

      return x;

}

原文地址:https://www.cnblogs.com/openix/p/2687774.html