HashMap源码阅读

核心方法

put方法

可以发现put方法是委托给putVal()方法的，值得注意的方法是hash(key)的实现。hash(key)相当简单, 它可以接受null作为key，这时hash值为0，而正常的hash值是h ^ (h >>> 16)

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

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

看看putVal()的实现，tab[i = (n - 1) & hash]，这一行是选择一个哈希桶，也能说明为什么HashMap的哈希桶数量必须是2的幂次，如初始化大小是16，16-1的二进制是1111，然后1111 ^ hash 则是桶是索引，值肯定是[0,15]中。

    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)
            n = (tab = resize()).length;//第一次put，要进行初始化
        if ((p = tab[i = (n - 1) & hash]) == null)//选择hash桶，如果桶不存在
            tab[i] = newNode(hash, key, value, null);//初始化一个桶
        else {//如果桶存在，就是一个链表操作，包括转换为红黑树
            Node<K,V> e; K k;
            if (p.hash == hash &&//短路操作，hash不同就没有必要用equals()
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;//表示key相同
            else if (p instanceof TreeNode)//新key，增加红黑树节点
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            else {//新key，普通节点
                for (int binCount = 0; ; ++binCount) {//计数，看看是否需要变成红黑树
                    if ((e = p.next) == null) {//找到一个null，插入新节点
                        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;
                afterNodeAccess(e);
                return oldValue;
            }
        }
        ++modCount;//修改次数，用于迭代器fail-fast
        if (++size > threshold)//判断要不要rehash
            resize();
        afterNodeInsertion(evict);//无实现
        return null;
    }


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

resize方法

newCap = oldCap << 1，即也会是2的幂次

    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;
            }
            //将Cap和Thr都扩充一倍
            else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                     oldCap >= DEFAULT_INITIAL_CAPACITY)
                newThr = oldThr << 1; // double threshold
        }
        else if (oldThr > 0) // initial capacity was placed in threshold
            newCap = oldThr;
        else {               // 第一次resize是初始化操作
            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"})
        //创建新的哈希桶
            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;//将原哈希桶清空，防止内存泄漏
                    if (e.next == null)
                        newTab[e.hash & (newCap - 1)] = e;//rehash
                    else if (e instanceof TreeNode)
                        ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                    else { //代表哈希桶是为1个以上
                        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;//这个措施很。。。
                        }
                    }
                }
            }
        }
        return newTab;
    }

get方法

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

    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) {//获取hash桶首节点
            if (first.hash == hash && // always check first node
                ((k = first.key) == key || (key != null && key.equals(k))))
                return first;//命中返回
            if ((e = first.next) != null) {
                if (first instanceof TreeNode)
                    return ((TreeNode<K,V>)first).getTreeNode(hash, key);
                do {//沿着链表往后找
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        return e;
                } while ((e = e.next) != null);
            }
        }
        return null;
    }

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;
    }

    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;
    }

其他有趣的实现

找到比一个数字大的2的幂次

    static final int tableSizeFor(int cap) {
        int n = cap - 1;
        n |= n >>> 1;
        n |= n >>> 2;
        n |= n >>> 4;
        n |= n >>> 8;
        n |= n >>> 16;
        return (n < 0) ? 1 : (n >= MAXIMUM_CAPACITY) ? MAXIMUM_CAPACITY : n + 1;
    }

从别的Map添加到HashMap

else if (s > threshold) resize();可能是先扩容，再添加。

但是问题来了，比如原先是threshold是12，map.size()是300，扩容后threshold是24，添加的时候还是要再次扩容的呀，所以这么做的意义好像不是很大，只是省去一次rehash的时间

    public HashMap(Map<? extends K, ? extends V> m) {
        this.loadFactor = DEFAULT_LOAD_FACTOR;
        putMapEntries(m, false);
    }

    final void putMapEntries(Map<? extends K, ? extends V> m, boolean evict) {
        int s = m.size();
        if (s > 0) {
            if (table == null) { // pre-size
                float ft = ((float)s / loadFactor) + 1.0F;
                int t = ((ft < (float)MAXIMUM_CAPACITY) ?
                         (int)ft : MAXIMUM_CAPACITY);
                if (t > threshold)
                    threshold = tableSizeFor(t);
            }
            else if (s > threshold)
                resize();
            for (Map.Entry<? extends K, ? extends V> e : m.entrySet()) {
                K key = e.getKey();
                V value = e.getValue();
                putVal(hash(key), key, value, false, evict);
            }
        }
    }

clear方法

    public void clear() {
        Node<K,V>[] tab;
        modCount++;
        if ((tab = table) != null && size > 0) {
            size = 0;
            for (int i = 0; i < tab.length; ++i)
                tab[i] = null;
        }
    }

containsValue方法

    public boolean containsValue(Object value) {
        Node<K,V>[] tab; V v;
        if ((tab = table) != null && size > 0) {
            for (Node<K, V> e : tab) {
                for (; e != null; e = e.next) {
                    if ((v = e.value) == value ||
                        (value != null && value.equals(v)))
                        return true;
                }
            }
        }
        return false;
    }

getOrDefault()

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

putIfAbsent()

    @Override
    public V putIfAbsent(K key, V value) {
        return putVal(hash(key), key, value, true, true);
    }

还有大量的函数式方法和大量的红黑树实现代码

函数式代码暂时不感兴趣，而且所有Collection都差不多，写过python或Javascript的，估计很容易看懂。红黑树的代码是特别长又特别难懂，感兴趣看算法，我只要知道是一种二叉平衡树就行了。