java 1.8 hashmap 源码分析

1.摘要

    java 1.8优化了HashMap代码，主要有hash值计算方法，添加红黑树，resize顺序保证等，下面将一一进行阐述。

2.HashMap简介

    HashMap和HashTable的区别在于hashtable是线程安全的，而且不允许有null值，而HashMap允许一个键为null，多个值为null。但是HashTable为遗留类，如果需要线程安全，可以使用性能更高的ConcurrentHashMap（使用了分段锁）。

    HashMap还有LinkedHashMap和TreeHashMap两个子类，LinkedHashMap保存数据插入顺序，TreeHashMap会对插入的key进行排序。

    HashMap是一个Node数组（哈希桶数组），每个数组的元素是一个链表或者红黑树，默认如果链表的长度超过8会自动转换为红黑树。

3.计算hash值

    本质上为三个计算： 求hash值，高位运算，取模运算。

    代码为：

//求hash值、高位运算(h=key.hashCode()) ^ (h >>> 16)//取模i = h & (length-1)

4. put方法

思路：哈希桶数组是否为空；通过key.hashCode得到数组下标，判断该哈希桶是否为空；判断该key值在哈希桶中是否存在。

1）判断哈希桶数组是否为空，若是，新建Node，直接跳到8；

2）获取下标，判断该哈希桶是否为空，若是，新建Node，直接跳到8；

3）新建缓存变量Node<K, V> e ；

4）判断该桶的key是否相等，将该Node赋值给e；

5）判断该桶是否为红黑树，如果是，插入e；

6）for循环遍历链表

6.1）没有相同的key，末端添加new Node；

6.2）有相同的key，break；

7）如果e非空，保存e中的旧值，赋值新值，并返回旧值；

8）size增加，并判断是否需要resize。

源码如下：

    /**     * 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 <tt>key</tt>, or     *         <tt>null</tt> if there was no mapping for <tt>key</tt>.     *         (A <tt>null</tt> return can also indicate that the map     *         previously associated <tt>null</tt> with <tt>key</tt>.)     */    public V put(K key, V value) {        return putVal(hash(key), key, value, false, true);    }    /**     * Implements Map.put and related methods     *     * @param hash hash for key     * @param key the key     * @param value the value to put     * @param onlyIfAbsent if true, don't change existing value     * @param evict if false, the table is in creation mode.     * @return previous value, or null if none     */    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;    }

5. resize方法

思路：计算newCap，newThr；新建容量为newCap的Node数组；将之前数组的值拷贝到新数组。

要强调的是

    1. java 1.7拷贝数组的方法是挨个计算key的hash值对应的下标i，并将相应的Node放到tab[i]的头，这样会导致存放顺序相反。java 1.8中，因为扩容后newCap为oldCap的两倍，在hash值取模后，可能不一样的地方只在newCap的最高位，所以旧数组下标i的元素只可能被分配到i或者i+oldCap的下标中，故，只要新建两个链表，将对应不同下标的值赋给不同的链表，然后将两个链表在赋值给table[i]和table[i+oldCap]，就保证了Node的顺序一致性；

    2. 在resize往新的链表中插入Node的时候，如果链表的长度大于8，会将链表转为红黑树。

    /**     * Initializes or doubles table size.  If null, allocates in     * accord with initial capacity target held in field threshold.     * Otherwise, because we are using power-of-two expansion, the     * elements from each bin must either stay at same index, or move     * with a power of two offset in the new table.     *     * @return the table     */    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;    }

参考

Java 8系列之重新认识HashMap