日期:2014-05-20 浏览次数:20893 次
/**
* Applies a supplemental hash function to a given hashCode, which
* defends against poor quality hash functions. This is critical
* because HashMap uses power-of-two length hash tables, that
* otherwise encounter collisions for hashCodes that do not differ
* in lower bits. Note: Null keys always map to hash 0, thus index 0.
*/
static int hash(int h) {
// This function ensures that hashCodes that differ only by
// constant multiples at each bit position have a bounded
// number of collisions (approximately 8 at default load factor).
h ^= (h >>> 20) ^ (h >>> 12);
return h ^ (h >>> 7) ^ (h >>> 4);
}
for (Entry<K,V> e = table[indexFor(hash, table.length)];
e != null;
e = e.next) {
Object k;
if (e.hash == hash && ((k = e.key) == key || key.equals(k)))
return e.value;
}
------解决方案--------------------
哦,我是想说这个函数的意义就是用来确定要取的entry在entry数组中的位置的,至于它为什么这么那我就不知道了,望后面的大牛来指点……
------解决方案--------------------
就是为了纠正hashcode函数的缺陷,因为hashmap的capacity的值是2的指数个,如果两个对象的hashCode值的低位相同,很有可能导致hashCode/capacity的值相同,就会出现冲突。
0101 0000 0000 1111 = 20495
0111 0000 0000 1111 = 28687
假如hashmap的capacity是16,那么20495%16 = 15,28687%16=15,就冲突了
------解决方案--------------------
hashmap数据的存储就是按照hashcode来的,相同hashcode的数据是放在链的同一位置
------解决方案--------------------
hashmap 应该是先用对象的hash值去找对象,找到的值是个数组即具有相同的hash值的,然后再用equal方法去比较二个是否相等,相等的则返回