原文讲解的HashMap JDK1.6
http://www.jianshu.com/p/8b372f3a195d/
该文没有很细的说明hash值是怎么生成的
/**
* Computes key.hashCode() and spreads (XORs) higher bits of hash
* to lower. Because the table uses power-of-two masking, sets of
* hashes that vary only in bits above the current mask will
* always collide. (Among known examples are sets of Float keys
* holding consecutive whole numbers in small tables.) So we
* apply a transform that spreads the impact of higher bits
* downward. There is a tradeoff between speed, utility, and
* quality of bit-spreading. Because many common sets of hashes
* are already reasonably distributed (so don't benefit from
* spreading), and because we use trees to handle large sets of
* collisions in bins, we just XOR some shifted bits in the
* cheapest possible way to reduce systematic lossage, as well as
* to incorporate impact of the highest bits that would otherwise
* never be used in index calculations because of table bounds.
*/
static final int hash(Object key) {
int h;
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
该代码列表是截取自jdk1.8 HashMap中 hash的生成方式:在分析之前应该清楚Object 的native hash方法,没有深入到native 源码。关于源码的链接:
其中
首先通过传入的key获取hashcode 记为var1
然后在var1基础上无符号右移16位,int类型四个字节,共32位,也就是去高16位,记为var2
最后var1 ^ var2 做亦或操作 得到最终hash值返回
import lombok.AllArgsConstructor;
import lombok.Data;
import lombok.ToString;
import java.util.HashMap;
/**
* Created by yanghuanqing@wdai.com on 30/06/2017.
*/
public class HashMapTest {
public static void main(String[] args) {
int i = "yang".hashCode();
System.out.println(i);
System.out.println(i+">>>16 ="+( "yang".hashCode()>>>16));
int off = i^(i>>>16);
System.out.println(off);
}
}
//没有lombok 可以自己敲set get construct
@Data
@AllArgsConstructor
@ToString
class Student{
private String name;
private int age;
}
3701441 // yang 的内置hash值
3701441>>>16 =56 //位置之后的值
3701497 //取异或操作的值
此外有string的哈希的生成算法
Returns a hash code for this string. The hash code for a String object is computed as
s[0]31^(n-1) + s[1]31^(n-2) + … + s[n-1]
String.clss
public int hashCode() {
int h = hash;
if (h == 0 && value.length > 0) {
char val[] = value;
for (int i = 0; i < value.length; i++) {
h = 31 * h + val[i];
}
hash = h;
}
return h;
}
补充HashMap put操作
//第一次参数是 key的Object.hashCode方法生成的hash值
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
boolean evict) {
Node<K,V>[] tab; Node<K,V> p; int n, i;
if ((tab = table) == null || (n = tab.length) == 0)
//如果第一次插入数据,那么会通过resize()函数范围一个默认的大小为16,该
//函数也是扩容过程中一个重要的函数
n = (tab = resize()).length;
//这一步就是在寻找桶的过程,就是上图总数组,根据容量取如果容量是16 对hash值取低16位,那么下标范围就在容量大小范围内了。
if ((p = tab[i = (n - 1) & hash]) == null)
//如果对应数据内没有对应值,就加入新节点。
tab[i] = newNode(hash, key, value, null);
else {
Node<K,V> e; K k;
//哈希值一样,key地址一样,key不为空,key的数据一样
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
else {
for (int binCount = 0; ; ++binCount) {
if ((e = p.next) == null) {
p.next = newNode(hash, key, value, null);
//如果连表的超过8采用二叉树存储结构
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
treeifyBin(tab, hash);
break;
}
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
break;
p = e;
}
}
//在key的哈希碰撞的情况下,旧值会被替换掉
if (e != null) { // existing mapping for key
V oldValue = e.value;
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
//该参数记录什么???数据的大小??还是整个hashMap中总的元素个数??
++modCount;
//应该是size来记录数组的大小
if (++size > threshold)
//超出数组就要重新分配大的数组
resize();
//如果扩容了,需要清理一些原来的数据
afterNodeInsertion(evict);
return null;
}
下面为map取数的过程
final Node<K,V> getNode(int hash, Object key) {
Node<K,V>[] tab; Node<K,V> first, e; int n; K k;
if ((tab = table) != null && (n = tab.length) > 0 &&
(first = tab[(n - 1) & hash]) != null) {
if (first.hash == hash && // always check first node
((k = first.key) == key || (key != null && key.equals(k))))
return first;
if ((e = first.next) != null) {
if (first instanceof TreeNode)
return ((TreeNode<K,V>)first).getTreeNode(hash, key);
do {
//定位到对应的数组,然后第一次没有命中,就直接一次往下遍历
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
return e;
} while ((e = e.next) != null);
}
}
return null;
}