1、首先看下HashMap的put方法。

 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)// 1.1 初始化数组长度n = (tab = resize()).length;// 1.2 查询数组对应下标是否有值，下标计算tab[i = (n - 1) & hash] // hash表示通过key.hashcode和key.hashcode的高16位进行异或操作，为什么要这样做？,n-1表示15，二进制表示01111/*** 解释：比如key.hashcode = 10010110101010,那么如何直接拿 key.hashcode & （n-1） 此处n-1 = 15*   10010110101010*            01111*-----------------------*            01010* 此时是否发现key.hashcode的高16位压根就没有进行计算，那么只用低位计算是不是冲突的概率较大？* 这就是为什么将key.hashcode的高位和低位进行^操作后在与15进行&操作*/if ((p = tab[i = (n - 1) & hash]) == null)// 1.3 将对应的值放入数组对应下标tab[i] = newNode(hash, key, value, null);else {// hash 发生碰撞Node<K,V> e; K k;// 并且发现key已存在，那么将新值和旧值互换if (p.hash == hash &&((k = p.key) == key || (key != null && key.equals(k))))e = p;else if (p instanceof TreeNode)// 红黑二叉树e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);else {// 链表for (int binCount = 0; ; ++binCount) {// 判断链表的next是否有值if ((e = p.next) == null) {p.next = newNode(hash, key, value, null);// 当链表的数量大于等于8时，链表会自动转换成红黑二叉树,因为链表过长会导致链表的查询效率过低if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1sttreeifyBin(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 keyV oldValue = e.value;if (!onlyIfAbsent || oldValue == null)e.value = value;afterNodeAccess(e);return oldValue;}}++modCount;// 当map中的数据大于16*0.75=12 map的初始化大小位16,负载因子0.75if (++size > threshold)resize();afterNodeInsertion(evict);return null;}

2、 通过解读源码发现hashmap不时线程安全
       那么此时hashtable登场 通过源码发现hashtable的put方法上加了synchronized关键字，HashTable容器使用synchronized来保证线程安全，但在线程竞争激烈的情况下HashTable的效率非常低下。因为当一个线程访问HashTable的同步方法时，其他线程访问HashTable的同步方法时，可能会进入阻塞或轮询状态。如线程1使用put进行添加元素，线程2不但不能使用put方法添加元素，并且也不能使用get方法来获取元素，所以竞争越激烈效率越低，此时会发现多线程操作时效率明显过低对吧！那么有办法解决吗？

        public synchronized V put(K key, V value) {// Make sure the value is not nullif (value == null) {throw new NullPointerException();}// Makes sure the key is not already in the hashtable.Entry<?,?> tab[] = table;int hash = key.hashCode();int index = (hash & 0x7FFFFFFF) % tab.length;@SuppressWarnings("unchecked")Entry<K,V> entry = (Entry<K,V>)tab[index];for(; entry != null ; entry = entry.next) {if ((entry.hash == hash) && entry.key.equals(key)) {V old = entry.value;entry.value = value;return old;}}addEntry(hash, key, value, index);return null;}

3、 当前有办法了，ConcurrentHashMap闪亮登场

 /** Implementation for put and putIfAbsent */final V putVal(K key, V value, boolean onlyIfAbsent) {if (key == null || value == null) throw new NullPointerException();int hash = spread(key.hashCode());int binCount = 0;for (Node<K,V>[] tab = table;;) {Node<K,V> f; int n, i, fh;if (tab == null || (n = tab.length) == 0)// 初始化数组,刚才我们提到多线程,难道每个线程都去初始化？下面单独拉出来看下tab = initTable();else if ((f = tabAt(tab, i = (n - 1) & hash)) == null) {// 利用cas无锁机制，cas原子操作，在指定位置设定值if (casTabAt(tab, i, null,new Node<K,V>(hash, key, value, null)))break;                   // no lock when adding to empty bin}else if ((fh = f.hash) == MOVED)// 此处map扩容后，数据转移tab = helpTransfer(tab, f);else {// 锁住当前node节点，这样其他线程继续操作时，因为node对象不是同一个了，// 所以不会影响其他线程的操作，这样效率就提高了。V oldVal = null;synchronized (f) {if (tabAt(tab, i) == f) {if (fh >= 0) {binCount = 1;for (Node<K,V> e = f;; ++binCount) {K ek;if (e.hash == hash &&((ek = e.key) == key ||(ek != null && key.equals(ek)))) {oldVal = e.val;if (!onlyIfAbsent)e.val = value;break;}Node<K,V> pred = e;if ((e = e.next) == null) {pred.next = new Node<K,V>(hash, key,value, null);break;}}}else if (f instanceof TreeBin) {Node<K,V> p;binCount = 2;if ((p = ((TreeBin<K,V>)f).putTreeVal(hash, key,value)) != null) {oldVal = p.val;if (!onlyIfAbsent)p.val = value;}}}}if (binCount != 0) {if (binCount >= TREEIFY_THRESHOLD)treeifyBin(tab, i);if (oldVal != null)return oldVal;break;}}}addCount(1L, binCount);return null;}

4、回到刚才ConcurrentMap初始化 tab = initTable()

  private transient volatile int sizeCtl; // 注意此处sizeCtl为volatileprivate final Node<K,V>[] initTable() {Node<K,V>[] tab; int sc;while ((tab = table) == null || tab.length == 0) {// 第一次进来sizeCtl值默认为0,那么其他线程过来时发现sizeCtl已经变成了-1，// 直接Thread.yield()让出cpu时间片。故此就实现了只有一个线程去初始化数组if ((sc = sizeCtl) < 0)Thread.yield(); // lost initialization race; just spinelse if (U.compareAndSwapInt(this, SIZECTL, sc, -1)) {// 将sc的值置为-1try {if ((tab = table) == null || tab.length == 0) {int n = (sc > 0) ? sc : DEFAULT_CAPACITY;@SuppressWarnings("unchecked")Node<K,V>[] nt = (Node<K,V>[])new Node<?,?>[n];table = tab = nt;sc = n - (n >>> 2);}} finally {// 将sizeCtl置为 -1sizeCtl = sc;}break;}}return tab;}

5、ConcurrentMap扩容注意点

ConcurrentMap扩容时，其他线程不能再往map中添加/删除数据了，那么难道其他线程在那边等待,直到扩容完成？当然不是了，其他线程会去帮助扩容线程一起进行扩容，每个线程都会去领取属于自己的任务

    final Node<K,V>[] helpTransfer(Node<K,V>[] tab, Node<K,V> f) {Node<K,V>[] nextTab; int sc;if (tab != null && (f instanceof ForwardingNode) &&(nextTab = ((ForwardingNode<K,V>)f).nextTable) != null) {int rs = resizeStamp(tab.length);while (nextTab == nextTable && table == tab &&(sc = sizeCtl) < 0) {if ((sc >>> RESIZE_STAMP_SHIFT) != rs || sc == rs + 1 ||sc == rs + MAX_RESIZERS || transferIndex <= 0)break;if (U.compareAndSwapInt(this, SIZECTL, sc, sc + 1)) {transfer(tab, nextTab);break;}}return nextTab;}return table;}