更快bobhash, 比time33快 (memcached也使用)

http://burtleburtle.net/bob/hash/doobs.html

Bob优化它的第二版本hash, 速度提高了3倍,http://burtleburtle.net/bob/c/lookup3.c

下面我提取的一个变长key, 小端版本(intel机器)

#include <stdint.h>     /* defines uint32_t etc */
#include <sys/param.h>  /* attempt to define endianness */
#ifdef linux
# include <endian.h>    /* attempt to define endianness */
#endif

/*
* My best guess at if you are big-endian or little-endian.  This may
* need adjustment.
*/
#if (defined(__BYTE_ORDER) && defined(__LITTLE_ENDIAN) && \
     __BYTE_ORDER == __LITTLE_ENDIAN) || \
    (defined(i386) || defined(__i386__) || defined(__i486__) || \
     defined(__i586__) || defined(__i686__) || defined(vax) || defined(MIPSEL))
# define HASH_LITTLE_ENDIAN 1
# define HASH_BIG_ENDIAN 0
#elif (defined(__BYTE_ORDER) && defined(__BIG_ENDIAN) && \
       __BYTE_ORDER == __BIG_ENDIAN) || \
      (defined(sparc) || defined(POWERPC) || defined(mc68000) || defined(sel))
# define HASH_LITTLE_ENDIAN 0
# define HASH_BIG_ENDIAN 1
#else
# define HASH_LITTLE_ENDIAN 0
# define HASH_BIG_ENDIAN 0
#endif

 

#define hashsize(n) ((uint32_t)1<<(n))
#define hashmask(n) (hashsize(n)-1)
#define rot(x,k) (((x)<<(k)) | ((x)>>(32-(k))))


#define mix(a,b,c) \
{ \
  a -= c;  a ^= rot(c, 4);  c += b; \
  b -= a;  b ^= rot(a, 6);  a += c; \
  c -= b;  c ^= rot(b, 8);  b += a; \
  a -= c;  a ^= rot(c,16);  c += b; \
  b -= a;  b ^= rot(a,19);  a += c; \
  c -= b;  c ^= rot(b, 4);  b += a; \
}

#define final(a,b,c) \
{ \
  c ^= b; c -= rot(b,14); \
  a ^= c; a -= rot(c,11); \
  b ^= a; b -= rot(a,25); \
  c ^= b; c -= rot(b,16); \
  a ^= c; a -= rot(c,4);  \
  b ^= a; b -= rot(a,14); \
  c ^= b; c -= rot(b,24); \
}


uint32_t bob_hash( const void *key, size_t length, uint32_t initval)
{
  uint32_t a,b,c;                                          /* internal state */
  union { const void *ptr; size_t i; } u;     /* needed for Mac Powerbook G4 */

  /* Set up the internal state */
  a = b = c = 0xdeadbeef + ((uint32_t)length) + initval;

  u.ptr = key;
  if (HASH_LITTLE_ENDIAN && ((u.i & 0x3) == 0)) {
    const uint32_t *k = (const uint32_t *)key;         /* read 32-bit chunks */
    const uint8_t  *k8;

    /*------ all but last block: aligned reads and affect 32 bits of (a,b,c) */
    while (length > 12)
    {
      a += k[0];
      b += k[1];
      c += k[2];
      mix(a,b,c);
      length -= 12;
      k += 3;
    }

    /*----------------------------- handle the last (probably partial) block */
    /*
     * "k[2]&0xffffff" actually reads beyond the end of the string, but
     * then masks off the part it's not allowed to read.  Because the
     * string is aligned, the masked-off tail is in the same word as the
     * rest of the string.  Every machine with memory protection I've seen
     * does it on word boundaries, so is OK with this.  But VALGRIND will
     * still catch it and complain.  The masking trick does make the hash
     * noticably faster for short strings (like English words).
     */
#ifndef VALGRIND

    switch(length)
    {
    case 12: c+=k[2]; b+=k[1]; a+=k[0]; break;
    case 11: c+=k[2]&0xffffff; b+=k[1]; a+=k[0]; break;
    case 10: c+=k[2]&0xffff; b+=k[1]; a+=k[0]; break;
    case 9 : c+=k[2]&0xff; b+=k[1]; a+=k[0]; break;
    case 8 : b+=k[1]; a+=k[0]; break;
    case 7 : b+=k[1]&0xffffff; a+=k[0]; break;
    case 6 : b+=k[1]&0xffff; a+=k[0]; break;
    case 5 : b+=k[1]&0xff; a+=k[0]; break;
    case 4 : a+=k[0]; break;
    case 3 : a+=k[0]&0xffffff; break;
    case 2 : a+=k[0]&0xffff; break;
    case 1 : a+=k[0]&0xff; break;
    case 0 : return c;              /* zero length strings require no mixing */
    }

#else /* make valgrind happy */

    k8 = (const uint8_t *)k;
    switch(length)
    {
    case 12: c+=k[2]; b+=k[1]; a+=k[0]; break;
    case 11: c+=((uint32_t)k8[10])<<16;  /* fall through */
    case 10: c+=((uint32_t)k8[9])<<8;    /* fall through */
    case 9 : c+=k8[8];                   /* fall through */
    case 8 : b+=k[1]; a+=k[0]; break;
    case 7 : b+=((uint32_t)k8[6])<<16;   /* fall through */
    case 6 : b+=((uint32_t)k8[5])<<8;    /* fall through */
    case 5 : b+=k8[4];                   /* fall through */
    case 4 : a+=k[0]; break;
    case 3 : a+=((uint32_t)k8[2])<<16;   /* fall through */
    case 2 : a+=((uint32_t)k8[1])<<8;    /* fall through */
    case 1 : a+=k8[0]; break;
    case 0 : return c;
    }

#endif /* !valgrind */

  } else if (HASH_LITTLE_ENDIAN && ((u.i & 0x1) == 0)) {
    const uint16_t *k = (const uint16_t *)key;         /* read 16-bit chunks */
    const uint8_t  *k8;

    /*--------------- all but last block: aligned reads and different mixing */
    while (length > 12)
    {
      a += k[0] + (((uint32_t)k[1])<<16);
      b += k[2] + (((uint32_t)k[3])<<16);
      c += k[4] + (((uint32_t)k[5])<<16);
      mix(a,b,c);
      length -= 12;
      k += 6;
    }

    /*----------------------------- handle the last (probably partial) block */
    k8 = (const uint8_t *)k;
    switch(length)
    {
    case 12: c+=k[4]+(((uint32_t)k[5])<<16);
             b+=k[2]+(((uint32_t)k[3])<<16);
             a+=k[0]+(((uint32_t)k[1])<<16);
             break;
    case 11: c+=((uint32_t)k8[10])<<16;     /* fall through */
    case 10: c+=k[4];
             b+=k[2]+(((uint32_t)k[3])<<16);
             a+=k[0]+(((uint32_t)k[1])<<16);
             break;
    case 9 : c+=k8[8];                      /* fall through */
    case 8 : b+=k[2]+(((uint32_t)k[3])<<16);
             a+=k[0]+(((uint32_t)k[1])<<16);
             break;
    case 7 : b+=((uint32_t)k8[6])<<16;      /* fall through */
    case 6 : b+=k[2];
             a+=k[0]+(((uint32_t)k[1])<<16);
             break;
    case 5 : b+=k8[4];                      /* fall through */
    case 4 : a+=k[0]+(((uint32_t)k[1])<<16);
             break;
    case 3 : a+=((uint32_t)k8[2])<<16;      /* fall through */
    case 2 : a+=k[0];
             break;
    case 1 : a+=k8[0];
             break;
    case 0 : return c;                     /* zero length requires no mixing */
    }

  } else {                        /* need to read the key one byte at a time */
    const uint8_t *k = (const uint8_t *)key;

    /*--------------- all but the last block: affect some 32 bits of (a,b,c) */
    while (length > 12)
    {
      a += k[0];
      a += ((uint32_t)k[1])<<8;
      a += ((uint32_t)k[2])<<16;
      a += ((uint32_t)k[3])<<24;
      b += k[4];
      b += ((uint32_t)k[5])<<8;
      b += ((uint32_t)k[6])<<16;
      b += ((uint32_t)k[7])<<24;
      c += k[8];
      c += ((uint32_t)k[9])<<8;
      c += ((uint32_t)k[10])<<16;
      c += ((uint32_t)k[11])<<24;
      mix(a,b,c);
      length -= 12;
      k += 12;
    }

    /*-------------------------------- last block: affect all 32 bits of (c) */
    switch(length)                   /* all the case statements fall through */
    {
    case 12: c+=((uint32_t)k[11])<<24;
    case 11: c+=((uint32_t)k[10])<<16;
    case 10: c+=((uint32_t)k[9])<<8;
    case 9 : c+=k[8];
    case 8 : b+=((uint32_t)k[7])<<24;
    case 7 : b+=((uint32_t)k[6])<<16;
    case 6 : b+=((uint32_t)k[5])<<8;
    case 5 : b+=k[4];
    case 4 : a+=((uint32_t)k[3])<<24;
    case 3 : a+=((uint32_t)k[2])<<16;
    case 2 : a+=((uint32_t)k[1])<<8;
    case 1 : a+=k[0];
             break;
    case 0 : return c;
    }
  }

  final(a,b,c);
  return c;
}

测试

1000w个53字节长的key, 结果

real    0m0.790s
user    0m0.788s
sys     0m0.000s

time33是:

real    0m1.041s
user    0m1.028s
sys     0m0.004s

 


posted @ 2010-12-22 17:22  napoleon_liu  阅读(2086)  评论(1编辑  收藏  举报