Jenkins哈希的Javascript实现？

Question

有没有我可以使用的詹金斯哈希的JavaScript实现 - 而不是自己实现它？

我知道有一个python实现我可以用来写我自己的js。但我不是JavaScript专家，因此我宁愿有人执行。

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更新：（我曾尝试转换http://www.burtleburtle.net/bob/c/lookup3.c） 更新2：此代码为您提供了相同的哈希值作为hashlittle2在lookup3.c

var Jenkins = { 
rot: function(x,k) { 
    return (x<<k) | (x>>>(32-k)); 
}, 

mix: function(a,b,c) { 
    a = (a - c) | 0; a ^= Jenkins.rot(c, 4); c = (c + b) | 0; 
    b = (b - a) | 0; b ^= Jenkins.rot(a, 6); a = (a + c) | 0; 
    c = (c - b) | 0; c ^= Jenkins.rot(b, 8); b = (b + a) | 0; 
    a = (a - c) | 0; a ^= Jenkins.rot(c,16); c = (c + b) | 0; 
    b = (b - a) | 0; b ^= Jenkins.rot(a,19); a = (a + c) | 0; 
    c = (c - b) | 0; c ^= Jenkins.rot(b, 4); b = (b + a) | 0; 
    return {a : a, b : b, c : c}; 
}, 

final: function(a,b,c) { 
    c ^= b; c -= Jenkins.rot(b,14) | 0; 
    a ^= c; a -= Jenkins.rot(c,11) | 0; 
    b ^= a; b -= Jenkins.rot(a,25) | 0; 
    c ^= b; c -= Jenkins.rot(b,16) | 0; 
    a ^= c; a -= Jenkins.rot(c,4) | 0; 
    b ^= a; b -= Jenkins.rot(a,14) | 0; 
    c ^= b; c -= Jenkins.rot(b,24) | 0; 
    return {a : a, b : b, c : c}; 
}, 

hashlittle2: function(k, initval, initval2) { 
    var length = k.length; 
    a = b = c = 0xdeadbeef + length + initval; 
    c += initval2; 

    offset = 0; 
    while (length > 12) { 
     a += (k.charCodeAt(offset+0) + (k.charCodeAt(offset+1)<<8) + (k.charCodeAt(offset+2)<<16) + (k.charCodeAt(offset+3)<<24)); a = a>>>0; 
     b += (k.charCodeAt(offset+4) + (k.charCodeAt(offset+5)<<8) + (k.charCodeAt(offset+6)<<16) + (k.charCodeAt(offset+7)<<24)); b = b>>>0; 
     c += (k.charCodeAt(offset+8) + (k.charCodeAt(offset+9)<<8) + (k.charCodeAt(offset+10)<<16) + (k.charCodeAt(offset+11)<<24)); c = c>>>0; 
     o = Jenkins.mix(a,b,c); 
     a = o.a; b = o.b; c = o.c; 
     length -= 12; 
     offset += 12; 
    } 

    switch(length) { 
     case 12: c += (k.charCodeAt(offset+8) + (k.charCodeAt(offset+9)<<8) + (k.charCodeAt(offset+10)<<16) + (k.charCodeAt(offset+11)<<24)); b += (k.charCodeAt(offset+4) + (k.charCodeAt(offset+5)<<8) + (k.charCodeAt(offset+6)<<16) + (k.charCodeAt(offset+7)<<24)); a += (k.charCodeAt(offset+0) + (k.charCodeAt(offset+1)<<8) + (k.charCodeAt(offset+2)<<16) + (k.charCodeAt(offset+3)<<24)); break; 
     case 11: c += (k.charCodeAt(offset+8) + (k.charCodeAt(offset+9)<<8) + (k.charCodeAt(offset+10)<<16)); b += (k.charCodeAt(offset+4) + (k.charCodeAt(offset+5)<<8) + (k.charCodeAt(offset+6)<<16) + (k.charCodeAt(offset+7)<<24)); a += (k.charCodeAt(offset+0) + (k.charCodeAt(offset+1)<<8) + (k.charCodeAt(offset+2)<<16) + (k.charCodeAt(offset+3)<<24)); break; 
     case 10: c += (k.charCodeAt(offset+8) + (k.charCodeAt(offset+9)<<8)); b += (k.charCodeAt(offset+4) + (k.charCodeAt(offset+5)<<8) + (k.charCodeAt(offset+6)<<16) + (k.charCodeAt(offset+7)<<24)); a += (k.charCodeAt(offset+0) + (k.charCodeAt(offset+1)<<8) + (k.charCodeAt(offset+2)<<16) + (k.charCodeAt(offset+3)<<24)); break; 
     case 9: c += (k.charCodeAt(offset+8)); b += (k.charCodeAt(offset+4) + (k.charCodeAt(offset+5)<<8) + (k.charCodeAt(offset+6)<<16) + (k.charCodeAt(offset+7)<<24)); a += (k.charCodeAt(offset+0) + (k.charCodeAt(offset+1)<<8) + (k.charCodeAt(offset+2)<<16) + (k.charCodeAt(offset+3)<<24)); break; 
     case 8: b += (k.charCodeAt(offset+4) + (k.charCodeAt(offset+5)<<8) + (k.charCodeAt(offset+6)<<16) + (k.charCodeAt(offset+7)<<24)); a += (k.charCodeAt(offset+0) + (k.charCodeAt(offset+1)<<8) + (k.charCodeAt(offset+2)<<16) + (k.charCodeAt(offset+3)<<24)); break; 
     case 7: b += (k.charCodeAt(offset+4) + (k.charCodeAt(offset+5)<<8) + (k.charCodeAt(offset+6)<<16)); a += (k.charCodeAt(offset+0) + (k.charCodeAt(offset+1)<<8) + (k.charCodeAt(offset+2)<<16) + (k.charCodeAt(offset+3)<<24)); break; 
     case 6: b += ((k.charCodeAt(offset+5)<<8) + k.charCodeAt(offset+4)); a += (k.charCodeAt(offset+0) + (k.charCodeAt(offset+1)<<8) + (k.charCodeAt(offset+2)<<16) + (k.charCodeAt(offset+3)<<24)); break; 
     case 5: b += (k.charCodeAt(offset+4)); a += (k.charCodeAt(offset+0) + (k.charCodeAt(offset+1)<<8) + (k.charCodeAt(offset+2)<<16) + (k.charCodeAt(offset+3)<<24)); break; 
     case 4: a += (k.charCodeAt(offset+0) + (k.charCodeAt(offset+1)<<8) + (k.charCodeAt(offset+2)<<16) + (k.charCodeAt(offset+3)<<24)); break; 
     case 3: a += (k.charCodeAt(offset+0) + (k.charCodeAt(offset+1)<<8) + (k.charCodeAt(offset+2)<<16)); break; 
     case 2: a += (k.charCodeAt(offset+0) + (k.charCodeAt(offset+1)<<8)); break; 
     case 1: a += (k.charCodeAt(offset+0)); break; 
     case 0: return {b : b, c : c}; 
    } 

    o = Jenkins.final(a,b,c); 
    a = o.a; b = o.b; c = o.c; 

    return {b : b>>>0, c : c>>>0}; 
}  

}

Answer 1

这里有一个JavaScript端口，我做了lookup3.c个人项目，在JavaScript中实现了Bloom filter。我不能确定它是否会对C代码产生相同的结果。

不直接转换成JavaScript的主要事情之一是指针算术在指向键输入的指针上执行。在下面的代码中查找offset这个词，看看我是如何处理这个问题的。

如果您希望输出如同整数未经签名，则可以使用returnValue >>> 0。

var BloomFilter = { 
    // Convert a JavaScript string into an array of 32-bit words. 
    // This preserves the UTF-16 encoding, padding with the null character if necessary. 
    stringToWords: function(s) { 
     var b = []; 
     if(s.length & 1) { 
      s += "\u0000"; 
     } 
     for (var i = 0; i < s.length; i += 2) { 
      b.push((s.charCodeAt(i) << 16) | (s.charCodeAt(i + 1))); 
     } 
     return b; 
    }, 

    // Hash an array of multiple 32-bit words to a single word. 
    // Adapted from "lookup3.c, by Bob Jenkins, May 2006, Public Domain." 
    // as retrieved 2010-07-03 from http://burtleburtle.net/bob/c/lookup3.c 

    hashWord: function(k, initval) { 
     // definition of bitwise rotate function 
     function rot(x, k) { 
      return (x << k) | (x >>> (32 - k)); 
     } 

     // initialization 
     var a, b, c, length = k.length, offset = 0; 
     a = b = c = (0xdeadbeef + (length << 2) + initval) | 0; 

     // handle most of the key 
     while(length > 3) { 
      a = (a + k[offset]) | 0; 
      b = (b + k[offset + 1]) | 0; 
      c = (c + k[offset + 2]) | 0; 

      // mixing function 
      a = (a - c) | 0; a ^= rot(c, 4); c = (c + b) | 0; 
      b = (b - a) | 0; b ^= rot(a, 6); a = (a + c) | 0; 
      c = (c - b) | 0; c ^= rot(b, 8); b = (b + a) | 0; 
      a = (a - c) | 0; a ^= rot(c,16); c = (c + b) | 0; 
      b = (b - a) | 0; b ^= rot(a,19); a = (a + c) | 0; 
      c = (c - b) | 0; c ^= rot(b, 4); b = (b + a) | 0; 

      length -= 3; 
      offset += 3; 
     } 

     // handle the final words if left over; fall-through is intended 
     switch(length) { 
      case 3: c = (c + k[offset + 2]) | 0; 
      case 2: b = (b + k[offset + 1]) | 0; 
      case 1: a = (a + k[offset]) | 0; 

      // final mixing 
      c ^= b; c = (c - rot(b,14)) | 0; 
      a ^= c; a = (a - rot(c,11)) | 0; 
      b ^= a; b = (b - rot(a,25)) | 0; 
      c ^= b; c = (c - rot(b,16)) | 0; 
      a ^= c; a = (a - rot(c, 4)) | 0; 
      b ^= a; b = (b - rot(a,14)) | 0; 
      c ^= b; c = (c - rot(b,24)) | 0; 

      case 0: break; // nothing left to do 
     } 

     // return the result 
     return c; 
    }, 

    // Hash a string by converting to UTF-16 before using the lookup3 algorithm. 
    hashString: function(s) { 
     return BloomFilter.hashWord(BloomFilter.stringToWords(s), 0); 
    } 
} 

Answer 2

只需几修改，C code on Wikipedia将在JavaScript中完美工作。只是摆脱数据类型，并使用key.length，而不是必须通过的长度。