HashMap

JDK 1.8

构造方法

//最常用的构造方法，默认初始容量是16，懒加载
public HashMap() {
    this.loadFactor = DEFAULT_LOAD_FACTOR; // 0.75
}

put方法

这个是关键！

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

//先看下 hash(key) 里做了什么 👇
static final int hash(Object key) {
    int h;
    //这个hash值后续会参与对数组下标位置散列，即(n-1)%hash
    //而数组容量通常不会很大，这意味着高位参与运算的概率小，为了散列分散更均匀
    //我们将hash右移16位，与原hash相与，参与运算
    return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}

//接下来看最重要的putVal
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
               boolean evict) {
    Node<K,V>[] tab; Node<K,V> p; int n, i;
    //如果第一次put，则创建table数组，懒加载，看① resize() 👇
    if ((tab = table) == null || (n = tab.length) == 0)
        n = (tab = resize()).length;
    //散列求得位置，且该位置为空
    if ((p = tab[i = (n - 1) & hash]) == null)
        tab[i] = newNode(hash, key, value, null);
    //这位置不为空
    else {
        Node<K,V> e; K k;
        //这是key相同的情况，即覆盖情况
        //看看如何判断key相等，先看hash相同吗？不同直接排除，再看地址或equals
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            e = p;
        //红黑树的情况，不谈
        else if (p instanceof TreeNode)
            e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
        //该链表的头节点和key不相同，往下找
        else {
            for (int binCount = 0; ; ++binCount) {
                //找到尾了还没找到，是在链表里新增一个值
                //这里可以看出jdk1.8使用的是尾插法，因为在jdk1.7中使用头插法时，在多线程环境下会引发死链问题
                if ((e = p.next) == null) {
                    p.next = newNode(hash, key, value, null);
                    //这里不一定会树化！看②处代码 👇
                    //所以链表树化的条件是：链表长度达到8且数组长度大于64（默认）
                    if (binCount >= TREEIFY_THRESHOLD - 1) 
                        treeifyBin(tab, hash);
                    break;
                }
                //找到相同的key，是覆盖的情况
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    break;
                p = e;
            }
        }
        //给e赋值，这个e可能是新结点也可能是已存在覆盖的情况
        if (e != null) { // existing mapping for key
            V oldValue = e.value;
            //onlyIfAbsent：key不存在才插入，相当于不能覆盖，只能新增
            if (!onlyIfAbsent || oldValue == null)
                e.value = value;
            afterNodeAccess(e);
            return oldValue;
        }
    }
    //添加修改次数
    ++modCount;
    //如果最后的容量超过阈值了，扩容
    if (++size > threshold)
        resize();
    afterNodeInsertion(evict);
    return null;
}

//① resize() 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) {
        //已经最大了
        if (oldCap >= MAXIMUM_CAPACITY) {
            threshold = Integer.MAX_VALUE;
            return oldTab;
        }
        //扩大成两倍
        else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                 oldCap >= DEFAULT_INITIAL_CAPACITY)
            newThr = oldThr << 1; // double threshold
    }
    //加了初始值的构造函数懒加载
    else if (oldThr > 0) // initial capacity was placed in threshold
        newCap = oldThr;
    //默认构造方法懒加载会到这个分支
    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;
    //rehash过程
    if (oldTab != null) {
        for (int j = 0; j < oldCap; ++j) {
            Node<K,V> e;
            //如果该桶不为空
            if ((e = oldTab[j]) != null) {
                oldTab[j] = null;
                if (e.next == null)
                    newTab[e.hash & (newCap - 1)] = e;
                else if (e instanceof TreeNode)
                    ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                else { // preserve order
                    链表hash...
                }
            }
        }
    }
    return newTab;
    //回去看putVal 👆
}

//② 树化 
final void treeifyBin(Node<K,V>[] tab, int hash) {
    int n, index; Node<K,V> e;
    //如果数组为空或者数组长度大于64，就扩容
    if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
        resize();
    else if ((e = tab[index = (n - 1) & hash]) != null) {
        树化...
    }
}
//回去看putVal 👆

get() 方法

get方法会简单些

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

// 直接看getNode 👇
final Node<K,V> getNode(int hash, Object key) {
    Node<K,V>[] tab; Node<K,V> first, e; int n; K k;
    //数组不为空、长度大于0、hash求的位置上不为空，否则返回null
    if ((tab = table) != null && (n = tab.length) > 0 &&
        (first = tab[(n - 1) & hash]) != null) {
        //和put一样了，先看hash，再看地址或equals
        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;
}

get还是比较简单的

remove() 方法

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

//直接看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;
    //一摸一样的if判断了，试试看能不能看懂
    if ((tab = table) != null && (n = tab.length) > 0 &&
        (p = tab[index = (n - 1) & hash]) != null) {
        //这里的node就是要删除的结点
        Node<K,V> node = null, e; K k; V v;
        //先hash，再地址或equals 🤯
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            node = p;
        //头结点不是，往下查
        else if ((e = p.next) != null) {
            if (p instanceof TreeNode)
                node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
            else {
                //顺着链表查找了
                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);
           	//是头节点
            else if (node == p)
                tab[index] = node.next;
            else
                p.next = node.next;
            ++modCount;
            --size;
            afterNodeRemoval(node);
            return node;
        }
        //没找到要删除的
    }
    return null;
}

Q.E.D.