PHP中用hash实现的数组
PHP中使用最多的非Array莫属了，那Array是如何实现的？在PHP内部Array通过一个hashtable来实现，其中使用链接法解决hash冲突的问题，这样最坏情况下，查找Array元素的复杂度为O(N)，最好则为1.
而其计算字符串hash值的方法如下，将源码摘出来以供查备：
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？
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？
