[TOC]
1 HashMap为什么线程不安全？
对于大部分公司来说：java8已经能够满足业务需求，同时以稳定性为主，不轻易升级版本。
/**
* Associates the specified value with the specified key in this map.
* If the map previously contained a mapping for the key, the old
* value is replaced.
*
* @param key key with which the specified value is to be associated
* @param value value to be associated with the specified key
* @return the previous value associated with {@code key}, or
* {@code null} if there was no mapping for {@code key}.
* (A {@code null} return can also indicate that the map
* previously associated {@code null} with {@code key}.)
*/
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)
n = (tab = resize()).length;
if ((p = tab[i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
else {
Node<K,V> e; K k;
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);
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;
}
}
if (e != null) { // existing mapping for key
V oldValue = e.value;
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
++modCount;
if (++size > threshold)
resize();
afterNodeInsertion(evict);
return null;
}
底层实现：数组加链表
首先通过hashcode()函数计算出hash值，然后取余得到下标，当hash冲突的时候1.7采用的是头插法，头插法速度快一些，不需要遍历到尾节点，这时候get的时候就不能取得头插法的元素，因为你的头结点是周瑜，所以这里就必须每次插入之后移动链表
分析源码：首先看构造方法，无参，指定数组容量，总的属性16*0.75=12；
public static int highestOneBit(int i) {
return i & (MIN_VALUE >>> numberOfLeadingZeros(i));
}
public static int numberOfLeadingZeros(int i) {
// HD, Count leading 0's
if (i <= 0)
return i == 0 ? 32 : 0;
int n = 31;
if (i >= 1 << 16) { n -= 16; i >>>= 16; }
if (i >= 1 << 8) { n -= 8; i >>>= 8; }
if (i >= 1 << 4) { n -= 4; i >>>= 4; }
if (i >= 1 << 2) { n -= 2; i >>>= 2; }
return n - (i >>> 1);
}8421 先翻译成二进制数，有一个位数为一，其他都不为一 0000 0101 代表10 put和get环环相扣的，必须是2的N次幂，才能通过-1加上与运算得到小于length的下标，配套使用 不管高四位怎么变，最终的结果的都不会变，但是int类型很多不一样，所以就会有冲突，影响get的元素，遍历比较慢 右移且进行了异或运算的原因，提高返回来的hashcode的散列性，
扩容的目的就是提高搜索效率
长链表变成短链表
头插法1，2，3扩容之后变成3，2，1
单线程的会怎么样？扩容resize()两个线程都会newTable公用
调用Put的时候，多线程扩容的时候会导致循环链表，因为可以会阻塞
记得先看构造函数，看各种方法，一层层往里面挖
记得看注释说明
熟悉数据结构与算法看源码会简单很多
代码量膨胀了一倍来解决这些问题
数组+链表/红黑树
扩容时插入顺序的改进
函数方法 forEach compute系列
Map新的API merge replace
java 7 中经典的哈希表实现
存在的问题：非常容易碰到死锁，潜在的安全隐患
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.
final Node<K,V>[] resize() {
Node<K,V>[] oldTab = table;
int oldCap = (oldTab == null) ? 0 : oldTab.length;
int oldThr = threshold;
int newCap, newThr = 0;
if (oldCap > 0) {
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;
}
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
oldCap >= DEFAULT_INITIAL_CAPACITY)
newThr = oldThr << 1; // double threshold
}
else if (oldThr > 0) // initial capacity was placed in threshold
newCap = oldThr;
else { // zero initial threshold signifies using defaults
newCap = DEFAULT_INITIAL_CAPACITY;
newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
}
if (newThr == 0) {
float ft = (float)newCap * loadFactor;
newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
(int)ft : Integer.MAX_VALUE);
}
threshold = newThr;
@SuppressWarnings({"rawtypes","unchecked"})
Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
table = newTab;
if (oldTab != null) {
for (int j = 0; j < oldCap; ++j) {
Node<K,V> e;
if ((e = oldTab[j]) != null) {
oldTab[j] = null;
if (e.next == null)
newTab[e.hash & (newCap - 1)] = e;
else if (e instanceof TreeNode)
((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
else { // preserve order
Node<K,V> loHead = null, loTail = null;
Node<K,V> hiHead = null, hiTail = null;
Node<K,V> next;
do {
next = e.next;
if ((e.hash & oldCap) == 0) {
if (loTail == null)
loHead = e;
else
loTail.next = e;
loTail = e;
}
else {
if (hiTail == null)
hiHead = e;
else
hiTail.next = e;
hiTail = e;
}
} while ((e = next) != null);
if (loTail != null) {
loTail.next = null;
newTab[j] = loHead;
}
if (hiTail != null) {
hiTail.next = null;
newTab[j + oldCap] = hiHead;
}
}
}
}
}
return newTab;
}
扩充HashMap的时候，不需要计算hash,只需要看原来的hash值新增的bit是1还是0，是0的话索引不变，是1的话变成“原索引+旧容量”，正是这样巧妙地rehash方式，即省去了重新计算hash值的时间，而且同时，由于新增的1bit是0还是1可以认为是随机的，均匀地分散冲突的节点到新的桶中
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