源码分析--HashMap（JDK1.8）

　　在JDK1.8中对HashMap的底层实现做了修改。本篇对HashMap源码从核心成员变量到常用方法进行分析。

　　HashMap数据结构如下：

　　先看成员变量：

　　1、底层存放数据的是Node<K,V>[]数组，数组初始化大小为16。

/**

     * The default initial capacity - MUST be a power of two.

     */

    static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16

　　2、Node<K,V>[]数组最大容量

/**

     * The maximum capacity, used if a higher value is implicitly specified

     * by either of the constructors with arguments.

     * MUST be a power of two <= 1<<30.

     */

    static final int MAXIMUM_CAPACITY = 1 << 30;

　　3、负载因子0.75。也就是如果默认初始化，HashMap在size = 16*0.75 = 12时，进行扩容。

/**

     * The load factor used when none specified in constructor.

     */

    static final float DEFAULT_LOAD_FACTOR = 0.75f;

　　4、将链表转化为红黑数的阀值。

 /**

     * The bin count threshold for using a tree rather than list for a

     * bin.  Bins are converted to trees when adding an element to a

     * bin with at least this many nodes. The value must be greater

     * than 2 and should be at least 8 to mesh with assumptions in

     * tree removal about conversion back to plain bins upon

     * shrinkage.

     */

    static final int TREEIFY_THRESHOLD = 8;

　　5、红黑树节点转换为链表的阀值

/**

     * The bin count threshold for untreeifying a (split) bin during a

     * resize operation. Should be less than TREEIFY_THRESHOLD, and at

     * most 6 to mesh with shrinkage detection under removal.

     */

    static final int UNTREEIFY_THRESHOLD = 6;

　　6、转红黑树时，table的最小长度

/**

     * The smallest table capacity for which bins may be treeified.

     * (Otherwise the table is resized if too many nodes in a bin.)

     * Should be at least 4 * TREEIFY_THRESHOLD to avoid conflicts

     * between resizing and treeification thresholds.

     */

    static final int MIN_TREEIFY_CAPACITY = 64;

介绍一下HashMap用hash值定位数组index的过程：

//HahsMap中的静态方法
static final int hash(Object key) {

        int h;

        return (key == null) ?  : (h = key.hashCode()) ^ (h >>> );

    }

//定位计算
int index = (table.length - 1) & hash

先得到key的hashCode值
再将hashCode值与hashCode无符号右移16位的值进行按位异或运算。得到hash值
将(table.length - 1) 与 hash值进行与运算。定位数组index

给一个长度为16的数组，以"TestHash"为key，进行定位的过程实例：

HashMap中Node就是放入的数据节点，代码定义为：

  static class Node<K,V> implements Map.Entry<K,V> {

        final int hash;

        final K key;

        V value;

        Node<K,V> next;

｝

　　Node节点保存key的hash值和K--V，借助next可实现链表。

红黑树封装为TreeNode节点：

static final class TreeNode<K,V> extends LinkedHashMap.Entry<K,V> {

        TreeNode<K,V> parent;  // red-black tree links

        TreeNode<K,V> left;

        TreeNode<K,V> right;

        TreeNode<K,V> prev;    // needed to unlink next upon deletion

        boolean red;

        TreeNode(int hash, K key, V val, Node<K,V> next) {

            super(hash, key, val, next);

        }

介绍get()方法：

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) >  &&

            (first = tab[(n - ) & 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 {

                    if (e.hash == hash &&

                        ((k = e.key) == key || (key != null && key.equals(k))))

                        return e;

                } while ((e = e.next) != null);

            }

        }

        return null;

    }

index定位，得到该索引上的Node节点，赋值给first
对first节点进行判断，如果是要找的元素，直接返回
first节点的next不为空，继续找
如果first节点是红黑树，调用getTreeNode()获取值。
不是红黑树，只能是链表。从头遍历，找到就返回。

　　上面对于红黑树取值的getTreeNode()方法，看一下红黑树的遍历做法：

final TreeNode<K,V> find(int h, Object k, Class<?> kc) {

            TreeNode<K,V> p = this;

            do {

                int ph, dir; K pk;

                TreeNode<K,V> pl = p.left, pr = p.right, q;

                if ((ph = p.hash) > h)

                    p = pl;

                else if (ph < h)

                    p = pr;

                else if ((pk = p.key) == k || (k != null && k.equals(pk)))

                    return p;

                else if (pl == null)

                    p = pr;

                else if (pr == null)

                    p = pl;

                else if ((kc != null ||

                          (kc = comparableClassFor(k)) != null) &&

                         (dir = compareComparables(kc, k, pk)) != )

                    p = (dir < ) ? pl : pr;

                else if ((q = pr.find(h, k, kc)) != null)

                    return q;

                else

                    p = pl;

            } while (p != null);

            return null;

        }

从do-while循环里的第一个if开始。如果当前节点的hash比传入的hash大，往p节点的左边遍历
如果当前节点的hash比传入的hash小，往p节点的右边遍历
如果key值相同，就找到节点了。返回
左节点为空，转到右边遍历
右节点为空，转到左边
如果传入key实现了Comparable接口。就将传入key与p节点key进行比较，根据比较结果选择向左或向右遍历
没有实现接口，直接向右遍历，找到就返回
没找到，向左遍历

介绍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) == )

            n = (tab = resize()).length;

        if ((p = tab[i = (n - ) & 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 = ; ; ++binCount) {

                    if ((e = p.next) == null) {

                        p.next = newNode(hash, key, value, null);

                        if (binCount >= TREEIFY_THRESHOLD - ) // -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;

    }

table为null，初始化
定位到数组index，若该位置为空，直接放
如果该位置上不为空，且hash和key与传入的值相同，说明key重复，直接将该节点赋值给e，结束循环
如果该index上的节点是红黑树，调用putTreeVal()方法
不是红黑树，只能是链表，遍历整个链表
找到最后一个节点，在这个节点后面以k--v新增一个节点。
判断链表长度，binCount达到7，也就是长度达到8。转为红黑树。
遍历过程中，如果找到了相同key，就跳出循环。
如果e不为空，说明遍历结束后存在key重复的节点。做值覆盖
扩容判断

分析红黑树插入方法putTreeVal():

final TreeNode<K,V> putTreeVal(HashMap<K,V> map, Node<K,V>[] tab,

                                       int h, K k, V v) {

            Class<?> kc = null;

            boolean searched = false;

            TreeNode<K,V> root = (parent != null) ? root() : this;

            for (TreeNode<K,V> p = root;;) {

                int dir, ph; K pk;

                if ((ph = p.hash) > h)

                    dir = -;

                else if (ph < h)

                    dir = ;

                else if ((pk = p.key) == k || (k != null && k.equals(pk)))

                    return p;

                else if ((kc == null &&

                          (kc = comparableClassFor(k)) == null) ||

                         (dir = compareComparables(kc, k, pk)) == ) {

                    if (!searched) {

                        TreeNode<K,V> q, ch;

                        searched = true;

                        if (((ch = p.left) != null &&

                             (q = ch.find(h, k, kc)) != null) ||

                            ((ch = p.right) != null &&

                             (q = ch.find(h, k, kc)) != null))

                            return q;

                    }

                    dir = tieBreakOrder(k, pk);

                }

                TreeNode<K,V> xp = p;

                if ((p = (dir <= ) ? p.left : p.right) == null) {

                    Node<K,V> xpn = xp.next;

                    TreeNode<K,V> x = map.newTreeNode(h, k, v, xpn);

                    if (dir <= )

                        xp.left = x;

                    else

                        xp.right = x;

                    xp.next = x;

                    x.parent = x.prev = xp;

                    if (xpn != null)

                        ((TreeNode<K,V>)xpn).prev = x;

                    moveRootToFront(tab, balanceInsertion(root, x));

                    return null;

                }

            }

        }

查找root根节点
从root节点开始遍历
如果当前节点p的hash大于传入的hash值，记dir为-1，代表向左遍历。
小于，记1，代表向右遍历
如果key相同，直接返回
如果key所属的类实现Comparable接口，或者key相等。先从p的左节点、右节点分别调用find()，找到就返回。
没找到，比较p和传入的key值，结果记为dir
根据dir选择向左或向右遍历
依次遍历，直到为null，表示达到最后一个节点，插入新节点
调整位置

分析HashMap扩容方法：

final Node<K,V>[] resize() {

        Node<K,V>[] oldTab = table;

        int oldCap = (oldTab == null) ?  : oldTab.length;

        int oldThr = threshold;

        int newCap, newThr = ;

        if (oldCap > ) {

            if (oldCap >= MAXIMUM_CAPACITY) {

                threshold = Integer.MAX_VALUE;

                return oldTab;

            }

            else if ((newCap = oldCap << ) < MAXIMUM_CAPACITY &&

                     oldCap >= DEFAULT_INITIAL_CAPACITY)

                newThr = oldThr << ; // double threshold

        }

        else if (oldThr > ) // 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 == ) {

            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 = ; j < oldCap; ++j) {

                Node<K,V> e;

                if ((e = oldTab[j]) != null) {

                    oldTab[j] = null;

                    if (e.next == null)

                        newTab[e.hash & (newCap - )] = 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) == ) {

                                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;

    }

通过一系列判断，确认新table的容量
构造一个新容量的Node数组，赋值给table
遍历旧table数组
如果节点是单节点，直接定位到新数组对应的index位置下
如果是红黑树，调用split方法
遍历链表。
如果e的hash值与老数组容量取与运算，值为0。索引位置不变
如果e的hash值与老数组容量取与运算，值为1。这在新数组中索引的位置为老数组索引 + 老数组容量。
链表放置

简要分析多线程下HashMap死循环问题：

　　JDK1.7HashMap扩容时，对于链表位置变化，采用头插法进行操作。多线程下容易形成环形链表，造成死循环。

　　JDK1.8时，会对于链表hash值与容量的计算结果。分成两部分，并改为插入到链表尾部。1.8以后不会再有死循环问题，只是有可能重复放置导致数据丢失。HashMap本身线程不安全的特性并没有改变。

巴特西

源码分析--HashMap（JDK1.8）

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