Java8-HashMap实现原理<二>

我们接着看

remove 实现

      /**
 * 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) {
      // 找到hash 对应的   数组元素
        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)
            // 链表有 元素，如果是红黑树结构
                node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
            else {
            // 链表有 元素，如果是链表结构
                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)))) {
          // 如果找到元素，则开始 进行移除元素操作
            if (node instanceof TreeNode)
                // 在红黑树上，移除节点
                ((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
            else if (node == p)
                // 在数组上，移除节点
                tab[index] = node.next;
            else
                // 在链表上，移除节点
                p.next = node.next;
            ++modCount;
            --size;
            afterNodeRemoval(node);
            return node;
        }
    }
    return null;
}

可以看到不同的存储结构他采用了不同的查找方式，以及移除方式，我们来看看他的查找方式

getTreeNode实现

          /**
     * Calls find for root node.
     */
    final TreeNode<K,V> getTreeNode(int h, Object k) {
        return ((parent != null) ? root() : this).find(h, k, null);
    }
  /**
     * 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) {
        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)) != 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;
    }

树形结构特点就是查找非常快速，使用了二分查找，对比hash 与 左右分支的节点hash来确认查找的结果，以及接下来去哪个分支上查找。删除的实现，就比较复杂了，当链表的节点数大于8的时候，链表会转化成红黑树，那是不是红黑树的节点数 小于等 8，会还原成 链表呢，带着这个问题，来看看删除的实现：

removeTreeNode实现

  /**
     * 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) {
            // 果然当红黑树的结构太小，就会转化成链表结构
           // 但是转换条件并不是 节点个数的问题，而是树形结构，
           // 这里的判断条件在于红黑树的层级，少于3层树形结构，就会被转化成链表的存储结构
            tab[index] = first.untreeify(map);  // too small
            return;
        }
        // 接下来就是 红黑树删除节点算法了，到时候我们花一篇文章的时间来介绍红黑树结构
      // 以及他的操作算法节点说明
        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);
    }

接下来，我们来分析一下他的扩容操作，扩容大小是多少，以及扩容的时机是什么时候等等

reSize()实现

  /**
 * 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;
    // 存储数组的 长度
    int oldCap = (oldTab == null) ? 0 : oldTab.length;
   // 存储 整个HashMap的 存储数据量
    int oldThr = threshold;
    int newCap, newThr = 0;
    if (oldCap > 0) {
        if (oldCap >= MAXIMUM_CAPACITY) {
            threshold = Integer.MAX_VALUE;
            return oldTab;
        }
      // 如果老的数组长度 不为零，且小于最大数组长度大小 MAXIMUM_CAPACITY 
     // 新的数组长度是原来的2倍大小
        else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                 oldCap >= DEFAULT_INITIAL_CAPACITY)、
            // 如果新的数组长度 小于最大数组长度大小 MAXIMUM_CAPACITY 
            // 如果满足要求，则HashMap容量变成原来的2倍
            newThr = oldThr << 1; // double threshold
    }
    else if (oldThr > 0) // initial capacity was placed in threshold
        newCap = oldThr;
    else {               // zero initial threshold signifies using defaults
        // 如果未初始化，则赋值 初始化数据
        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;
    // 以下就是 在容量发生变化的时候，存储数据进行重新排列赋值操作
    if (oldTab != null) {
        for (int j = 0; j < oldCap; ++j) {
            Node<K,V> e;
            if ((e = oldTab[j]) != null) {
                // 老的数组数据被赋值成空
                oldTab[j] = null;
                if (e.next == null)
                    // 如果当前的桶 中只有一个元素，则直接赋值给新的 数组中
                    newTab[e.hash & (newCap - 1)] = e;
                else if (e instanceof TreeNode)
                    // 如果是红黑树，则使用分离的算法，
                    ((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;
         // 因为 扩容大小是 原有的大小的两倍，使用 e.hash & oldCap 算法计算
         // 计算出为 0 代表当前节点的hash二进制数 与 oldCap的二进制数 不存在相同位为 1的，如果扩容之后重新计算 的话，index 还是会为 j
       // 计算出不为 0 代表当前节点的hash 二进制数  与 oldCap 的二进制数 存在相同位 为 1的，如果扩容之后重新计算的话，index 会变成  j + oldCap
        // 如果为 0 ，则存储在新的数组中的位置  还是为 j
       // 如果不为 0 ，则 存储位置在新的数组中的位置 为 j + oldCap
                        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;
}

bin : 箱子，也可以称作为 桶，对应的就是数组中 index 对应的存储结构了，可以能是链式，可能是红黑树