而其计算字符串hash值的方法如下，将源码摘出来以供查备：
复制代码 代码如下:
static inline ulong zend_inline_hash_func(const char *arkey, uint nkeylength)
{
register ulong hash = 5381; //此处初始值的设置有什么玄机么？
/* variant with the hash unrolled eight times */
for (; nkeylength >= 8; nkeylength -= 8) { //这种step=8的方式是为何？
hash = ((hash << 5) + hash) + *arkey++;
hash = ((hash << 5) + hash) + *arkey++;
hash = ((hash << 5) + hash) + *arkey++;
hash = ((hash << 5) + hash) + *arkey++; //比直接*33要快
hash = ((hash << 5) + hash) + *arkey++;
hash = ((hash << 5) + hash) + *arkey++;
hash = ((hash << 5) + hash) + *arkey++;
hash = ((hash << 5) + hash) + *arkey++;
}
switch (nkeylength) {
case 7: hash = ((hash << 5) + hash) + *arkey++; /* fallthrough... */ //此处是将剩余的字符hash
case 6: hash = ((hash << 5) + hash) + *arkey++; /* fallthrough... */
case 5: hash = ((hash << 5) + hash) + *arkey++; /* fallthrough... */
case 4: hash = ((hash << 5) + hash) + *arkey++; /* fallthrough... */
case 3: hash = ((hash << 5) + hash) + *arkey++; /* fallthrough... */
case 2: hash = ((hash << 5) + hash) + *arkey++; /* fallthrough... */
case 1: hash = ((hash << 5) + hash) + *arkey++; break;
case 0: break;
empty_switch_default_case()
}
return hash;//返回hash值
}
ps：对于以下函数，仍有两点不明：
hash = 5381设置的理由？
这种step=8的循环方式是为了效率么？
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