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  • Java Jdk1.8 HashMap源代码阅读笔记二

    三、源代码阅读

    3、元素包括containsKey(Object key)

     /**
         * Returns <tt>true</tt> if this map contains a mapping for the
         * specified key.
         *
         * @param   key   The key whose presence in this map is to be tested
         * @return <tt>true</tt> if this map contains a mapping for the specified
         * key.
         */
        public boolean containsKey(Object key) {
            return getNode(hash(key), key) != null;
        }
        /**
         * Implements Map.get and related methods
         *
         * @param hash hash for key
         * @param key the key
         * @return the node, or null if none
         */
        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 &&
                //推断tab[hash]位置是否有值
                (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) {
                    if (first instanceof TreeNode)
                        return ((TreeNode<K,V>)first).getTreeNode(hash, key);
                    //遍历寻找 
                    do {
                        if (e.hash == hash &&
                            ((k = e.key) == key || (key != null && key.equals(k))))
                            return e;
                    } while ((e = e.next) != null);
                }
            }
            return null;
        }
         /**
          * Calls find for root node.
          */
        final TreeNode<K,V> getTreeNode(int h, Object k) {
              return ((parent != null) ? root() : this).find(h, k, null);
            }
         /**
             * Returns root of tree containing this node.
             * 获取红黑树的根
             */
            final TreeNode<K,V> root() {
                for (TreeNode<K,V> r = this, p;;) {
                    if ((p = r.parent) == null)
                        return r;
                    r = p;
                }
            }
        /**
             * Finds the node starting at root p with the given hash and key.
             * The kc argument caches comparableClassFor(key) upon first use
             * comparing keys.
             */
            final TreeNode<K,V> find(int h, Object k, Class<?> kc) {// k即key,kc为null
                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)// ph存当前节点hash
                        p = pl;
                    else if (ph < h) // 所查hash比当前节点hash大
                        p = pr;// 查右子树
                    else if ((pk = p.key) == k || (k != null && k.equals(pk)))
                        return p;// hash、key均同样。【找到了!】返回当前节点
                    else if (pl == null)// hash等,key不等,且当前节点的左节点null
                        p = pr;//查右子树
                    else if (pr == null)
                        p = pl;
                   //get->getTreeNode传递的kc为null。

    ||逻辑或,短路运算,有真就可以 // false || (false && ??) else if ((kc != null || (kc = comparableClassFor(k)) != null) && (dir = compareComparables(kc, k, pk)) != 0) p = (dir < 0) ? pl : pr; else if ((q = pr.find(h, k, kc)) != null) return q; else p = pl; } while (p != null); return null; }

    4、get(Object key)

     /**
         *  返回value或null
         */
        public V get(Object key) {
            Node<K,V> e;
            return (e = getNode(hash(key), key)) == null ? null : e.value;
        }
    

    函数getNode在阅读笔记一中已经记录。

    5、移除 remove(Object key)

     /**
         * Removes the mapping for the specified key from this map if present.
         *
         * @param  key key whose mapping is to be removed from the map
         * @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 remove(Object key) {
            Node<K,V> e;
            return (e = removeNode(hash(key), key, null, false, true)) == null ?

    null : e.value; } /** * Implements Map.remove and related methods * * @param hash hash for key * @param key the key * @param value the value to match if matchValue, else ignored * @param matchValue if true only remove if value is equal * @param movable if false do not move other nodes while removing * @return the node, or null if none */ 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 && //先比較内存地址。假设地址不一致。再调用equals进行比較 ((k = p.key) == key || (key != null && key.equals(k)))) node = p; else if ((e = p.next) != null) { //假设是以红黑树处理冲突。则通过getTreeNode查找 if (p instanceof TreeNode) node = ((TreeNode<K,V>)p).getTreeNode(hash, key); else { //假设是以链式的方式处理冲突。则通过遍历链表来寻找节点 do { if (e.hash == hash && ((k = e.key) == key || (key != null && key.equals(k)))) { node = e; break; } p = e; } while ((e = e.next) != null); } } //比对找到的key的value跟要删除的是否匹配 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; //已从结构上改动 此列表的次数 ++modCount; --size; //回调 afterNodeRemoval(node); return node; } } return null; } /** * Removes the given node, that must be present before this call. * This is messier than typical red-black deletion code because we * cannot swap the contents of an interior node with a leaf * successor that is pinned by "next" pointers that are accessible * independently during traversal. So instead we swap the tree * linkages. If the current tree appears to have too few nodes, * the bin is converted back to a plain bin. (The test triggers * somewhere between 2 and 6 nodes, depending on tree structure). */ final void removeTreeNode(HashMap<K,V> map, Node<K,V>[] tab, boolean movable) { int n; if (tab == null || (n = tab.length) == 0) return; int index = (n - 1) & hash; TreeNode<K,V> first = (TreeNode<K,V>)tab[index], root = first, rl; TreeNode<K,V> succ = (TreeNode<K,V>)next, pred = prev; if (pred == null) tab[index] = first = succ; else pred.next = succ; if (succ != null) succ.prev = pred; if (first == null) return; if (root.parent != null) root = root.root(); if (root == null || root.right == null || (rl = root.left) == null || rl.left == null) { tab[index] = first.untreeify(map); // too small return; } TreeNode<K,V> p = this, pl = left, pr = right, replacement; if (pl != null && pr != null) { TreeNode<K,V> s = pr, sl; while ((sl = s.left) != null) // find successor s = sl; boolean c = s.red; s.red = p.red; p.red = c; // swap colors TreeNode<K,V> sr = s.right; TreeNode<K,V> pp = p.parent; if (s == pr) { // p was s's direct parent p.parent = s; s.right = p; } else { TreeNode<K,V> sp = s.parent; if ((p.parent = sp) != null) { if (s == sp.left) sp.left = p; else sp.right = p; } if ((s.right = pr) != null) pr.parent = s; } p.left = null; if ((p.right = sr) != null) sr.parent = p; if ((s.left = pl) != null) pl.parent = s; if ((s.parent = pp) == null) root = s; else if (p == pp.left) pp.left = s; else pp.right = s; if (sr != null) replacement = sr; else replacement = p; } else if (pl != null) replacement = pl; else if (pr != null) replacement = pr; else replacement = p; if (replacement != p) { TreeNode<K,V> pp = replacement.parent = p.parent; if (pp == null) root = replacement; else if (p == pp.left) pp.left = replacement; else pp.right = replacement; p.left = p.right = p.parent = null; } TreeNode<K,V> r = p.red ? root : balanceDeletion(root, replacement); if (replacement == p) { // detach TreeNode<K,V> pp = p.parent; p.parent = null; if (pp != null) { if (p == pp.left) pp.left = null; else if (p == pp.right) pp.right = null; } } if (movable) moveRootToFront(tab, r); }

    方法 final void removeTreeNode临时没有吃透,待兴许补充。

    四、小结

    HashMap高性能须要下面几点:
    1、高效的hash算法
    2、保证hash值到内存地址(数组索引)的映射速度
    3、依据内存地址(数组索引)能够直接得到对应的值

    參考文章:https://yq.aliyun.com/articles/36812?

    spm=5176.8091938.0.0.lGqa1x

    个人微信公众号:
    这里写图片描写叙述

    作者:jiankunking 出处:http://blog.csdn.net/jiankunking

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  • 原文地址:https://www.cnblogs.com/blfbuaa/p/7373404.html
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