我想在一個小項目中使用SuperFastHash,我似乎無法理解爲什麼它給同一個字符串不同的哈希。如果指針和字符串都相同,它只輸出相同的散列。有任何想法嗎?代碼證明如下。SuperFastHash返回不同的值爲相同的字符串
// SuperFastHash, taken from http://www.azillionmonkeys.com/qed/hash.html
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#undef get16bits
#if (defined(__GNUC__) && defined(__i386__)) || defined(__WATCOMC__) \
|| defined(_MSC_VER) || defined (__BORLANDC__) || defined (__TURBOC__)
#define get16bits(d) (*((const uint16_t *) (d)))
#endif
#if !defined (get16bits)
#define get16bits(d) ((((uint32_t)(((const uint8_t *)(d))[1])) << 8)\
+(uint32_t)(((const uint8_t *)(d))[0]))
#endif
uint32_t SuperFastHash (const char * data, int len);
int main(void)
{
char* str = "a\0a";
printf("%s\n", &str[0]); // a
printf("%s\n", &str[2]); // a
printf("%i\n", SuperFastHash(&str[0], 25)); // -1120168156
printf("%i\n", SuperFastHash(&str[2], 25)); // -280310739
}
uint32_t SuperFastHash (const char * data, int len) {
uint32_t hash = len, tmp;
int rem;
if (len <= 0 || data == NULL) return 0;
rem = len & 3;
len >>= 2;
/* Main loop */
for (;len > 0; len--) {
hash += get16bits (data);
tmp = (get16bits (data+2) << 11)^hash;
hash = (hash << 16)^tmp;
data += 2*sizeof (uint16_t);
hash += hash >> 11;
}
/* Handle end cases */
switch (rem) {
case 3: hash += get16bits (data);
hash ^= hash << 16;
hash ^= ((signed char)data[sizeof (uint16_t)]) << 18;
hash += hash >> 11;
break;
case 2: hash += get16bits (data);
hash ^= hash << 11;
hash += hash >> 17;
break;
case 1: hash += (signed char)*data;
hash ^= hash << 10;
hash += hash >> 1;
}
/* Force "avalanching" of final 127 bits */
hash ^= hash << 3;
hash += hash >> 5;
hash ^= hash << 4;
hash += hash >> 17;
hash ^= hash << 25;
hash += hash >> 6;
return hash;
}
我覺得很蠢。謝謝! – user1637451
四隻眼睛看到兩個以上。別客氣! – Wolf