深入学习JDK1.8之HashMap

HashMap的底层数据结构是一个Node对象数组+链表。

• 数组定义：

transient Node<K,V>[] table;

• Node类

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;//table索引
        this.key = key;
        this.value = value;
        this.next = next;
    }
    public final K getKey()        { return key; }
    public final V getValue()      { return value; }
    public final String toString() { return key + "=" + value; }
    public final int hashCode() {
        return Objects.hashCode(key) ^ Objects.hashCode(value);
    }
    public final V setValue(V newValue) {
       //...省略
    }
    public final boolean equals(Object o) {
        //...省略
    }
}

查找

get(Object key)函数

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

hash(Object key)函数

static final int hash(Object key) {
    int h;
    return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}

hash函数就是用来生成table数组下标的，具体原理参考： JDK 源码中 HashMap 的 hash 方法原理是什么？

---

定义一个Node<K,V>节点对象，然后调用getNode(hash(key), key))，判断是否为null，不为null返回对应的e.value。

getNode(hash(key), key)

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) {
            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;
}

首先定义了Node数组tab，Node节点first、e，数组长度n，键k。

第一个if用来判断数组是否为空，以及赋值操作：

if (tab = table) != null && (n = tab.length) > 0 && (first = tab[(n - 1) & hash]) != null){}

共三个判断条件：

• 赋值tab = table，并判断是否为null；
• 赋值n = tab.length，判断长度是都大于0；
• 首先(n - 1) & hash是计算数组下标或索引，然后取出tab中对应的值赋给first，并判断是否为null。

紧接着判断first是否满足要查找的条件，满足则返回结果，否则继续执行。

因为当Key值存在hash冲突时，Node节点是通过链表连接的，所以接着要做的是进行链表查询，首先判断链表类型是否为TreeNode（当链表长度超过一定限制后，就会用TreeNode存储），如果是，调用TreeNode的查询方法；如果不是，循环遍历链表。

添加

put(K key, V value)

public V put(K key, V value) {
    return putVal(hash(key), key, value, false, true);
}

内部直接调用的putVal()函数。

putVal(...)

我们直接在代码中通过添加注释的方式分析：

final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
               boolean evict) {
    Node<K,V>[] tab; Node<K,V> p; int n, i;
    //判断`tab`是否为`null`
    if ((tab = table) == null || (n = tab.length) == 0)
        //通过resize()扩展容量
        n = (tab = resize()).length;
    if ((p = tab[i = (n - 1) & hash]) == null)
        //如果对应索引下tab值为null，则直接插入
        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)
            //若节点为TreeNode，则调用TreeNode的相应方法
            e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
        else {
            //如不是TreeNode，则依次遍历
            for (int binCount = 0; ; ++binCount) {
                if ((e = p.next) == null) {
                    //插入到链表尾部
                    p.next = newNode(hash, key, value, null);
                    if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                        //当链表长度>=8时，将链表转换为TreeNode
                        treeifyBin(tab, hash);
                    break;
                }
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    break;
                p = e;
            }
        }
        //若存在则替换
        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;
}

resize()

直接在代码中添加注释进行分析：

/**
*初始化或扩容
*/
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) {
        if (oldCap >= MAXIMUM_CAPACITY) {
            threshold = Integer.MAX_VALUE;
            return oldTab;
        }
        else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                 oldCap >= DEFAULT_INITIAL_CAPACITY)
            newThr = oldThr << 1; // double threshold
    }
	//调用HashMap(int initialCapacity)时的初始化
    else if (oldThr > 0) // initial capacity was placed in threshold
        newCap = oldThr;
	//调用HashMap()时的初始化
    else {               // zero initial threshold signifies using defaults
        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;
	//当oldTab不为空时，将旧数组oldTab的数据复制到新数组newTable
    if (oldTab != null) {
        //遍历oldTab数组
        for (int j = 0; j < oldCap; ++j) {
            Node<K,V> e;
            //当oldTab[j]不为空时
            if ((e = oldTab[j]) != null) {
                oldTab[j] = null;//触发GC
                if (e.next == null)//只有一个元素，放入新数组中即可
                    newTab[e.hash & (newCap - 1)] = e;
                else if (e instanceof TreeNode)//当类型为红黑树时
                    ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                else { // preserve order
                    //将原来的链表分成两条链表lo、hi，head代表头，tail代表尾
                    //一条在newTab[j]
                    //另一条在newTab[j + oldCap]
                    Node<K,V> loHead = null, loTail = null;
                    Node<K,V> hiHead = null, hiTail = null;
                    Node<K,V> next;
                    do {
                        next = e.next;
                        //true代表不需要重新分配位置
                        //因为可能存在在原数组对应相同索引的元素在新数组对应不同索引
                        //https://www.jianshu.com/p/3b77eeb4ad18
                        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;
}

remove(Object key)

public V remove(Object key) {
    Node<K,V> e;
    return (e = removeNode(hash(key), key, null, false, true)) == null ?
        null : e.value;
}

removeNode(...)

 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;
}