我可以使用Jenkins Hash的javascript实现 - 而不是自己实现吗？
我知道有一个python实现我可以用来编写自己的js。但我不是Javascript专家，因此我更喜欢有人在实施。
更新:(我确实尝试转换http://www.burtleburtle.net/bob/c/lookup3.c） 更新2：此代码在lookup3.c
中为您提供与hashlittle2相同的哈希值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};
}
}
答案 0 :(得分：3)
这是我为lookup3.c做的JavaScript端口，用于在JavaScript中实现Bloom filter的个人项目。我不能肯定地说它是否会产生与C代码相同的结果。
直接转换为JavaScript的主要内容之一是对指向键输入的指针执行的指针算术。在下面的代码中查找单词offset，看看我是如何处理的。
如果您希望输出就像整数是无符号的那样，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);
}
}
答案 1 :(得分：2)
只需进行一些修改，C code on Wikipedia将在JavaScript中完美运行。只需删除数据类型并使用key.length而不必传递长度。
答案 2 :(得分：1)
这是我对lookup3（2006）所做的优化端口：
function lookup3(k, init = 0, init2 = 0) {
var len = k.length, o = 0,
a = 0xdeadbeef + len + init | 0,
b = 0xdeadbeef + len + init | 0,
c = 0xdeadbeef + len + init + init2 | 0;
while (len > 12) {
a += k[o] | k[o+1] << 8 | k[o+2] << 16 | k[o+3] << 24;
b += k[o+4] | k[o+5] << 8 | k[o+6] << 16 | k[o+7] << 24;
c += k[o+8] | k[o+9] << 8 | k[o+10] << 16 | k[o+11] << 24;
a -= c; a ^= c<<4 | c>>>28; c = c+b | 0;
b -= a; b ^= a<<6 | a>>>26; a = a+c | 0;
c -= b; c ^= b<<8 | b>>>24; b = b+a | 0;
a -= c; a ^= c<<16 | c>>>16; c = c+b | 0;
b -= a; b ^= a<<19 | a>>>13; a = a+c | 0;
c -= b; c ^= b<<4 | b>>>28; b = b+a | 0;
len -= 12, o += 12;
}
if(len > 0) { // final mix only if len > 0
switch (len) { // incorporate trailing bytes before fmix
case 12: c += k[o+11] << 24;
case 11: c += k[o+10] << 16;
case 10: c += k[o+9] << 8;
case 9: c += k[o+8];
case 8: b += k[o+7] << 24;
case 7: b += k[o+6] << 16;
case 6: b += k[o+5] << 8;
case 5: b += k[o+4];
case 4: a += k[o+3] << 24;
case 3: a += k[o+2] << 16;
case 2: a += k[o+1] << 8;
case 1: a += k[o];
}
c ^= b; c -= b<<14 | b>>>18;
a ^= c; a -= c<<11 | c>>>21;
b ^= a; b -= a<<25 | a>>>7;
c ^= b; c -= b<<16 | b>>>16;
a ^= c; a -= c<<4 | c>>>28;
b ^= a; b -= a<<14 | a>>>18;
c ^= b; c -= b<<24 | b>>>8;
}
// use c as 32-bit hash; add b for 64-bit hash. a is not mixed well.
return [b >>> 0, c >>> 0];
}
这里是lookup2（1996），它非常相似（但速度较慢）：
function lookup2(k, init = 0) {
var len = k.length, o = 0,
a = 0x9e3779b9 | 0,
b = 0x9e3779b9 | 0,
c = init | 0;
while (len >= 12) {
a += k[o] | k[o+1] << 8 | k[o+2] << 16 | k[o+3] << 24;
b += k[o+4] | k[o+5] << 8 | k[o+6] << 16 | k[o+7] << 24;
c += k[o+8] | k[o+9] << 8 | k[o+10] << 16 | k[o+11] << 24;
a -= b; a -= c; a ^= c >>> 13;
b -= c; b -= a; b ^= a << 8;
c -= a; c -= b; c ^= b >>> 13;
a -= b; a -= c; a ^= c >>> 12;
b -= c; b -= a; b ^= a << 16;
c -= a; c -= b; c ^= b >>> 5;
a -= b; a -= c; a ^= c >>> 3;
b -= c; b -= a; b ^= a << 10;
c -= a; c -= b; c ^= b >>> 15;
len -= 12, o += 12;
}
c += k.length;
switch(len) {
case 11: c += k[o+10] << 24;
case 10: c += k[o+9] << 16;
case 9: c += k[o+8] << 8;
// the first byte of c is reserved for the length
case 8: b += k[o+7] << 24;
case 7: b += k[o+6] << 16;
case 6: b += k[o+5] << 8;
case 5: b += k[o+4];
case 4: a += k[o+3] << 24;
case 3: a += k[o+2] << 16;
case 2: a += k[o+1] << 8;
case 1: a += k[o];
}
a -= b; a -= c; a ^= c >>> 13;
b -= c; b -= a; b ^= a << 8;
c -= a; c -= b; c ^= b >>> 13;
a -= b; a -= c; a ^= c >>> 12;
b -= c; b -= a; b ^= a << 16;
c -= a; c -= b; c ^= b >>> 5;
a -= b; a -= c; a ^= c >>> 3;
b -= c; b -= a; b ^= a << 10;
c -= a; c -= b; c ^= b >>> 15;
// c is intended as 32-bit hash only
return c >>> 0;
}