Java_JDK_HashMap

Posted 2020-07-23 江湖小妞

（二）HashMap

需要注意的无非几点：

• 是什么结构，如何存储的？
• 如何加入元素？既然是hashMap，那么是如何计算hashcode的呢？遇到冲突又是如何解决的呢？
• 如何删除元素？
• 当容量不够时是如何扩容的？

1. 总体的存储结构为一个Node类型的数组：transient Node<K,V>[] table;

其中，Node节点结构为：

   static class Node<K,V> implements Map.Entry<K,V> {
        final int hash;
        final K key;
        V value;
        Node<K,V> next;

        Node(int hash, K key, V value, Node<K,V> next) {
            this.hash = hash;
            this.key = key;
            this.value = value;
            this.next = next;
        }
}

 

可以看到，Node节点中有next指针，说明数组内部是一个指针，指向下一个元素。

故整个HashMap的存储结构如下图所示：

 

2.put()方法：

 

 public V put(K key, V value) {

            return putVal(hash(key), key, value, false, true); //将具体的实现封装在下面的函数中

        }


     /**
         * 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) //如果是空表
                n = (tab = resize()).length;
            if ((p = tab[i = (n - 1) & hash]) == null) //如果(n - 1) & hash处还没有元素（不会产生冲突），则直接放入即可
                tab[i] = newNode(hash, key, value, null);
            else { //否则，产生了冲突，且p记录了应该放入的位置
                Node<K,V> e; K k;
                if (p.hash == hash &&
                    ((k = p.key) == key || (key != null && key.equals(k)))) //如果确实产生了冲突
                    e = p; //用e记录冲突位置
                else if (p instanceof TreeNode) //p是TreeNode ? 不是太理解，先放这里吧。。
                    e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
                else { 
                    for (int binCount = 0; ; ++binCount) {
                        if ((e = p.next) == null) { //如果就放了一个元素，即链表位置为空
                            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; //是否可以直接替换
                    afterNodeAccess(e);
                    return oldValue;
                }
            }
            ++modCount; //修改次数加一
            if (++size > threshold)
                resize(); //超出了限定容量，则扩容
            afterNodeInsertion(evict);
            return null;
        }

可以看到，其中有很多细节处没有吃透，先有个大概的认识，与后面的分析融汇贯通。

 

 3. resize()函数：扩容

下面我们来分析一下这个长长的扩容函数：

    /**
     * Initializes or doubles table size.  If null, allocates in 初始化或者对原来容量扩大两倍
     * 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) { //原来HashMap不空
            if (oldCap >= MAXIMUM_CAPACITY) { //如果原来map的容量就很大，则直接设置到Integer的最大值
                threshold = Integer.MAX_VALUE;
                return oldTab;
            }
            else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                     oldCap >= DEFAULT_INITIAL_CAPACITY)  //如果扩大一倍的新容量仍小于最大容量 && 原始容量大于默认值（16）
                newThr = oldThr << 1; // double threshold，则让容量扩大一倍
        }
        else if (oldThr > 0) // initial capacity was placed in threshold
            newCap = oldThr;
        else {               
            // zero initial threshold signifies using defaults，若oldCap = 0 && oldThr = 0,则用默认值对map初始化
            newCap = DEFAULT_INITIAL_CAPACITY;
            newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
        }
        if (newThr == 0) {//如果上面的初始化没有成功
            float ft = (float)newCap * loadFactor; //loadfactor为负载因子，初始为0.75
            newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
                      (int)ft : Integer.MAX_VALUE); //初始化为0.75倍
        }
        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; //置为null,方便GC进行回收
                    if (e.next == null) //如果j处只有这一个元素（链表长度为1）
                        newTab[e.hash & (newCap - 1)] = e;//将e放入新的位置，为何这样计算呢？
                    else if (e instanceof TreeNode) 
                        ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                    else { // preserve order 如果j处是一个链表，则按照原来的次序进行拷贝
                        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;
    }

 

 

 4. delete函数

/**
 * 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) { //如果map不为空
        Node<K,V> node = null, e; K k; V v;
        //首先寻找要删除的元素，用node记录
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            node = p; //p/node 就是要删除的元素
        else if ((e = p.next) != null) {
            if (p instanceof TreeNode)
                node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
            else { //该位置有很多元素，则找到要删除的那一个
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key ||
                         (key != null && key.equals(k)))) {
                        node = e;
                        break;
                    }
                    p = e;
                } while ((e = e.next) != null);
            }
        }
        //找到用node记录的要删除的元素之后，进行删除操作
        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) //如果p是要删除的节点，直接指向下一个元素或指向null
                tab[index] = node.next;
            else //如果要删除的node在p的一个链表中，则让next指针指向node.next即可
                p.next = node.next;
            ++modCount; //修改次数+1
            --size; //size - 1
            afterNodeRemoval(node);
            return node;
        }
    }
    return null;
}

 

 

5. hashCode

使用了Object类中的hashCode方法，每个对象都会有自己的hashCode, 同时，jdk有自己的根据key值计算hash的方法：

   /**
     * Computes key.hashCode() and spreads (XORs) higher bits of hash
     * to lower.  Because the table uses power-of-two masking, sets of
     * hashes that vary only in bits above the current mask will
     * always collide. (Among known examples are sets of Float keys
     * holding consecutive whole numbers in small tables.)  So we
     * apply a transform that spreads the impact of higher bits
     * downward. There is a tradeoff between speed, utility, and
     * quality of bit-spreading. Because many common sets of hashes
     * are already reasonably distributed (so don\'t benefit from
     * spreading), and because we use trees to handle large sets of
     * collisions in bins, we just XOR some shifted bits in the
     * cheapest possible way to reduce systematic lossage, as well as
     * to incorporate impact of the highest bits that would otherwise
     * never be used in index calculations because of table bounds.
     */
    static final int hash(Object key) {
        int h;
        return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
    }