深入学习JDK1.8之HashMap

刺骨的言语ヽ痛彻心扉 2022-10-01 05:58 92阅读 0赞

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;
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*  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
        }
    }
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## 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;
    }
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### hash(Object key)函数 ###

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

--------------------

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;
    }
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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){}
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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);
    }
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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;
    }
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### resize() ###

直接在代码中添加注释进行分析：

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) {
                //当oldCap大于等最大容量时，无法继续扩容，只能修改threshold
                threshold = Integer.MAX_VALUE;
                return oldTab;
            }
            else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                     oldCap >= DEFAULT_INITIAL_CAPACITY)
                //两倍threshold
                newThr = oldThr << 1; // double threshold
        }
        else if (oldThr > 0) // initial capacity was placed in threshold
            newCap = oldThr;
        else {               // zero initial threshold signifies using defaults
            //当oldCap==0&&oldThr==0时，初始化table
            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;
        if (oldTab != null) {
            //通过for循环对oldTab数组中的重新赋值到newTab数组
            for (int j = 0; j < oldCap; ++j) {
                Node<K,V> e;
                if ((e = oldTab[j]) != null) {
                    //清空，触发GC
                    oldTab[j] = null;
                    if (e.next == null)
                        newTab[e.hash & (newCap - 1)] = e;
                    else if (e instanceof TreeNode)
                        //当链表为红黑树时，调用TreeNode中的相应方法，其内部会判断当长度<=6时，转换为普通的链表存储方式
                        ((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;
                            //这一步是判断是放到原始的index位置，还是index+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;
    }
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### 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;
    }
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### 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;
    }
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转载于:https://juejin.im/post/5cb55b91f265da03a97ae034

[JDK _ HashMap _ hash]: https://link.juejin.im?target=https%3A%2F%2Fwww.zhihu.com%2Fquestion%2F20733617