JDK源码分析 hashmap1.8

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HashMap

只实现了Map接口,没有实现Comparable接口。

hashmap采用数组+链表(可转换成红黑树)的存储结构。

capacity是指数组的容量,threshold是指map中元素个数size的临界值,树化条件:1.数组容量 达到 min_treeify_capcaity2.链表长度达到 treeify_threshold时候会将链表转换成红黑树。

构造方法

无参构造方法采用默认值,而指定了初始容量的构造方法会将初始容量存储在阈值threshold中。(这样做与扩容方法有关)

public HashMap(int initialCapacity, float loadFactor) {
        if (initialCapacity < 0)
            throw new IllegalArgumentException("Illegal initial capacity: " +
                                               initialCapacity);
        if (initialCapacity > MAXIMUM_CAPACITY)
            initialCapacity = MAXIMUM_CAPACITY;
        if (loadFactor <= 0 || Float.isNaN(loadFactor))
            throw new IllegalArgumentException("Illegal load factor: " +
                                               loadFactor);
        this.loadFactor = loadFactor;
        this.threshold = tableSizeFor(initialCapacity);
    }


public HashMap(int initialCapacity) {
        this(initialCapacity, DEFAULT_LOAD_FACTOR);
    }
public HashMap() {
        this.loadFactor = DEFAULT_LOAD_FACTOR; // all other fields defaulted
    }
public HashMap(Map<? extends K, ? extends V> m) {
        this.loadFactor = DEFAULT_LOAD_FACTOR;
        putMapEntries(m, false);
    }

链表容量保证为2^n

static final int tableSizeFor(int cap) {                                                                      
        int n = cap - 1; //
        n |= n >>> 1;
        n |= n >>> 2;
        n |= n >>> 4;
        n |= n >>> 8;
        n |= n >>> 16;
        return (n < 0) ? 1 : (n >= MAXIMUM_CAPACITY) ? MAXIMUM_CAPACITY : n + 1;
    } 

计算元素所在槽

tab[(n - 1) & hash]

计算哈希

static final int hash(Object key) {                                                                          
        int h;
        return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
    }
  1. 为什么不直接用hashcode?而要在hashcode的基础上再一次hash?

    因为hashcode是32位int,而n = tablesize往往很小,如果直接用hashcode&(n-1)则会浪费hashcode的高位部分,导致分布不均匀。所以将高位与低位部分异或。来使得高位与低位都参与计算,降低碰撞

  2. hash & (n-1) 等于hash%n, n 为二进制幂

    假设n等于2^t.则1在第t+1位。因此n-1就是相当于从第一位到t位都是1。即与上n-1等于取hash的1到t位。

public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }
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;
        // p指代当前节点
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
            Node<K,V> e; K k;
            //e 指代应该插入节点的位置。
            //检查头结点 e != null
            if (p.hash == hash &&((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            //红黑树的情况 e根据是否存在key对应的节点返回NULL或者原节点
            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            else {
                for (int binCount = 0; ; ++binCount) {
                    //遍历到末尾了,直接在链表末尾插入当前节点。这里e == null。先插入再树化/扩容
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                            treeifyBin(tab, hash);
                        break;
                    }
                    //如果发现链表中已经存在等待插入的节点, e != null
                    if (e.hash == hash &&
                        ((k = e.key) == key |(key!=null&&key.equals(k))))
                        break;
                    p = e;
                }
            }
            //存在节点,直接更新val值即可。
            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;
    }

红黑树中put。key需要实现Comparable接口

 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;
                //dir表示插入左子树还是右子树
                if ((ph = p.hash) > h)
                    dir = -1;
                else if (ph < h)
                    dir = 1;
                //如果存在该节点,直接返回原节点
                else if ((pk = p.key) == k || (k != null && k.equals(pk)))
                    return p;
                //hash相等,key不等。用KC比
                else if ((kc == null &&
                          (kc = comparableClassFor(k)) == null) ||
                         (dir = compareComparables(kc, k, pk)) == 0) {
                    //从子树里找
                    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;
                    }
                    //子树里也没有。而且当前节点用kc也比不了。就比较类的名字。
                    dir = tieBreakOrder(k, pk);
                }
                //找到了插入的方向dir,开始插入。xp指代x的父母,X 指代新节点
                TreeNode<K,V> xp = p;
                if ((p = (dir <= 0) ? p.left : p.right) == null) {
                    Node<K,V> xpn = xp.next;
                    TreeNode<K,V> x = map.newTreeNode(h, k, v, xpn);
                    if (dir <= 0)
                        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;//表示插入之前红黑树没有该节点
                }
            }
        }
//找出红黑树中对应key的节点,如果不存在返回NULL
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;
               //如果有子树为空,则直接进入另一个子树搜索
               else if (pl == null)
                    p = pr;
                else if (pr == null)
                    p = pl;
                //比较大小,找出下一个搜素的子树
                else if ((kc != null ||
                          (kc = comparableClassFor(k)) != null) &&
                         (dir = compareComparables(kc, k, pk)) != 0)
                    p = (dir < 0) ? pl : pr;
                //如果kc为空。则瞎几把搜索。先搜右子树
                else if ((q = pr.find(h, k, kc)) != null)
                    return q;
                else
                    p = pl;
            } while (p != null);
            return null;
        }

扩容

扩容在put 时候进行,1种是数组为空时候,另一种是size达到了threshold时候的扩容。

在构造方法执行时候,并没有初始化数组,当插入第一个元素时候才会开始初始化(扩容)数组。

final Node<K,V>[] resize() {
        Node<K,V>[] oldTab = table;
        int oldCap = (oldTab == null) ? 0 : oldTab.length;
        int oldThr = threshold;
        // 设置新容量,新阈值
        int newCap, newThr = 0;
        //如果数组存在
        if (oldCap > 0) {
            //老容量已经最大,则只能调阈值了。调完直接退出
            if (oldCap >= MAXIMUM_CAPACITY) {
                threshold = Integer.MAX_VALUE;
                return oldTab;
            }
            //老容量已经超过初始容量的情况且新容量<容量最大值,则新阈值直接翻倍即可。
            else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                     oldCap >= DEFAULT_INITIAL_CAPACITY)
                newThr = oldThr << 1; // double threshold
        }
        //数组不存在,即尚未初始化,对应2种情况,一种是构造方法指定了初始容量
        else if (oldThr > 0) // initial capacity was placed in threshold
            newCap = oldThr;
        //另一种是没有指定初始容量,初始阈值,则采用默认值。
        else {               // zero initial threshold signifies using defaults
            newCap = DEFAULT_INITIAL_CAPACITY;
            newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
        }
        //如果新阈值还没有被赋值。说明要么新容量超过了容量最大值,或者构造方法指定了初始容量,因为构造方法将初始容量存储在阈值中,这时候要重新赋值阈值。
        if (newThr == 0) {
            float ft = (float)newCap * loadFactor;
            newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
                      (int)ft : Integer.MAX_VALUE);
        }
        threshold = newThr;
        // 设置完了新容量,新阈值,开始搬家。
        @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;
                    // 头结点区分链表还是红黑树
                    if (e.next == null)
                        newTab[e.hash & (newCap - 1)] = e;
                    else if (e instanceof TreeNode)
                        ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                    //继续给剩下的链表搬家,保持链表中的顺序
                    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;
                        //搬家的新位置分2种,因为容量是2倍增长,所以链表中的元素的新位置要么在原位置,要么在原位置+oldcap的新位置。分别用lo与hi指代这2个链表。
                            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;
    }

可以看到多线程扩容情况下不会出现1.7版本的死循环问题。但是依然不安全。搞不懂揪着死循环不放干啥,hashmap本来就不是给并发用的,没有死循环也有其他安全问题。

删除

final Node<K,V> removeNode(int hash, Object key, Object value,
                               boolean matchValue, boolean movable) {
        Node<K,V>[] tab; Node<K,V> p; int n, index;
        if ((tab = table) != null && (n = tab.length) > 0 &&
            (p = tab[index = (n - 1) & hash]) != null) {
            Node<K,V> node = null, e; K k; V v;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                node = p;
            else if ((e = p.next) != null) {
                if (p instanceof TreeNode)
                    node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
                else {
                    do {
                        if (e.hash == hash &&
                            ((k = e.key) == key ||
                             (key != null && key.equals(k)))) {
                            node = e;
                            break;
                        }
                        p = e;
                    } while ((e = e.next) != null);
                }
            }
            if (node != null && (!matchValue || (v = node.value) == value ||
                                 (value != null && value.equals(v)))) {
                if (node instanceof TreeNode)
                    ((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
                else if (node == p)
                    tab[index] = node.next;
                else
                    p.next = node.next;
                ++modCount;
                --size;
                afterNodeRemoval(node);
                return node;
            }
        }
        return null; 
    }

链表与红黑树的转换

红黑树节点类,继承了LinkedHashMap.Entry类,which继承了HashMap.Node

链表-> 红黑树:put时候,如果发现放置完当前元素后,链表长度(bincount) == threshold。则调用treefiBin转换成红黑树。而该方法会检查数组容量是否达到MIN_TREEIFY_CAPACITY。如果没达到,则直接扩容返回。put时候,如果发现

final void treeifyBin(Node<K,V>[] tab, int hash) {
        int n, index; Node<K,V> e;
        if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
            resize();
        else if ((e = tab[index = (n - 1) & hash]) != null) {
            TreeNode<K,V> hd = null, tl = null;
            do {
                TreeNode<K,V> p = replacementTreeNode(e, null);
                if (tl == null)
                    hd = p;
                else {
                    p.prev = tl;
                    tl.next = p;
                }
                tl = p;
            } while ((e = e.next) != null);
            if ((tab[index] = hd) != null)
                hd.treeify(tab);
        }
    }
    final void treeify(Node<K,V>[] tab) {
            TreeNode<K,V> root = null;
            for (TreeNode<K,V> x = this, next; x != null; x = next) {
                next = (TreeNode<K,V>)x.next;
                x.left = x.right = null;
                if (root == null) {
                    x.parent = null;
                    x.red = false;
                    root = x;
                }
                else {
                    K k = x.key;
                    int h = x.hash;
                    Class<?> kc = null;
                    for (TreeNode<K,V> p = root;;) {    
                        int dir, ph;
                        K pk = p.key;
                        if ((ph = p.hash) > h)
                            dir = -1;
                        else if (ph < h)
                            dir = 1;
                        else if ((kc == null &&
                                  (kc = comparableClassFor(k)) == null)||
                                (dir = compareComparables(kc, k, pk)) ==0)
                            dir = tieBreakOrder(k, pk);

                        TreeNode<K,V> xp = p;
                        if ((p = (dir <= 0) ? p.left : p.right) == null)                                                                                                                       {
                            x.parent = xp;
                            if (dir <= 0)
                                xp.left = x;
                            else
                                xp.right = x;
                            root = balanceInsertion(root, x);//平衡红黑树
                            break;
                        }
                    }
                }
            }
            moveRootToFront(tab, root);
        }

/**
         * Ensures that the given root is the first node of its bin.
         */
        static <K,V> void moveRootToFront(Node<K,V>[] tab, TreeNode<K,V> root) {
            int n;
            if (root != null && tab != null && (n = tab.length) > 0) {
                int index = (n - 1) & root.hash;
                TreeNode<K,V> first = (TreeNode<K,V>)tab[index];
                if (root != first) {
                    Node<K,V> rn;
                    tab[index] = root;
                    TreeNode<K,V> rp = root.prev;
                    if ((rn = root.next) != null)
                        ((TreeNode<K,V>)rn).prev = rp;
                    if (rp != null)
                        rp.next = rn;
                    if (first != null)
                        first.prev = root;
                    root.next = first;
                    root.prev = null;
                }
                assert checkInvariants(root);
            }
        }

红黑树->链表

红黑树中的节点因为有prev与next 指针,所以拆分还是比较简单的。

在resize的搬家时候如果检测到是红黑树节点则会先拆分成链表,再搬家。理想情况下链表长度会缩短一半。然后等到以后插入发现链表长度达到新threshold才会进行转换成红黑树的操作。

可以看到拆分的时候链表内节点的相对顺序并没有改变。

拆分成链表并没有损坏红黑树结构,因此拆分完就需要调用untreeify来将红黑树节点转换成链表节点。有个优化如果发现hitail == null则说明没有拆成2个链表,树结构不改变,无需untreeify。

    final void split(HashMap<K,V> map, Node<K,V>[] tab, int index, int bit) {
            TreeNode<K,V> b = this;
            // Relink into lo and hi lists, preserving order
            TreeNode<K,V> loHead = null, loTail = null;
            TreeNode<K,V> hiHead = null, hiTail = null;
            int lc = 0, hc = 0;
            for (TreeNode<K,V> e = b, next; e != null; e = next) {
                next = (TreeNode<K,V>)e.next;
                e.next = null;
                if ((e.hash & bit) == 0) {
                    if ((e.prev = loTail) == null)
                        loHead = e;
                    else
                        loTail.next = e;
                    loTail = e;
                    ++lc;
                }
                else {
                    if ((e.prev = hiTail) == null)
                        hiHead = e;
                    else
                        hiTail.next = e;
                    hiTail = e;
                    ++hc;
                }
            }

            if (loHead != null) {
                //
                if (lc <= UNTREEIFY_THRESHOLD)
                    tab[index] = loHead.untreeify(map);
                else {
                    tab[index] = loHead;
                    if (hiHead != null) // (else is already treeified)
                        loHead.treeify(tab);
                }
            }
            if (hiHead != null) {
                if (hc <= UNTREEIFY_THRESHOLD)
                    tab[index + bit] = hiHead.untreeify(map);
                else {
                    tab[index + bit] = hiHead;
                    if (loHead != null)
                        hiHead.treeify(tab);
                }
            }
        }
    final Node<K,V> untreeify(HashMap<K,V> map) {
            Node<K,V> hd = null, tl = null;
            for (Node<K,V> q = this; q != null; q = q.next) {
                Node<K,V> p = map.replacementNode(q, null);
                if (tl == null)
                    hd = p;
                else
                    tl.next = p;
                tl = p;
            }
            return hd;
        }

红黑树节点与链表节点的转换

    TreeNode<K,V> replacementTreeNode(Node<K,V> p, Node<K,V> next) {
        return new TreeNode<>(p.hash, p.key, p.value, next);
    }
    

查找

    public V get(Object key) {
        Node<K,V> e;
        return (e = getNode(hash(key), key)) == null ? null : e.value;
    }
    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) {
                //会调用TreeNode类的find方法
                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;
    }

总结

  1. 对hash的优化方法,使得节点尽可能均匀分布
  2. 数组容量是2的幂次方,简化取模运算(相当于与上capacity-1)
  3. 扩容时候链表选择2倍长度,使得新位置计算更迅速(要么原地,要么偏移old capacity)
  4. 引入红黑树,来缓解当链表长度变长带来的性能下降
  5. 红黑树带来的问题:
    1. 一旦扩容需要拆分红黑树以及untreeify带来的性能消耗
    2. 当不断删除导致map中节点很少时,红黑树查找节点的性能提升抵不上reblance带来的性能下降
  6. 增加MIN_TREEIFY_CAPACITY,当数组较小时优先选择扩容数组而不是树化
  7. 在hashmap的节点有
    1. 实现了Map.Entry接口的HashMapNode节点,含有next链接
    2. 扩展了LinkedHashMap.Entry节点的TreeNode节点,含有parent,left,right,prev链接。ps:LinkedhashMap.Entry节点含有before,after链接,而LinkedHashMap.Entry继承了HashMap.Node。
  8. 下列更改存储数据结构的操作,保留了节点相对顺序(不改变next链接)。事实上,只有TreeNode类中的moveRootToFront操作会改变root节点的相对顺序。包括:
    1. 扩容操作
    2. 树化操作(除了红黑树的根节点,为了保证红黑树根处于桶的头部)
    3. 链表节点删除操作(红黑树节点的删除如果涉及到根节点的变化,则会改变节点相对顺序)
    4. 拆分树操作
    5. 逆树化操作

Ref

  1. blog.csdn.net/weixin_4234…

  2. segmentfault.com/a/119000001…