人生需要总结

JDK8中的ConcurrentHashMap源码

背景

上文JDK8中的HashMap源码写了HashMap,这次写ConcurrentHashMap

ConcurrentHashMap源码

/**
     * Maps the specified key to the specified value in this table.
     * Neither the key nor the value can be null.
     *
     * <p>The value can be retrieved by calling the {@code get} method
     * with a key that is equal to the original key.
     *
     * @param key key with which the specified value is to be associated
     * @param value value to be associated with the specified key
     * @return the previous value associated with {@code key}, or
     *         {@code null} if there was no mapping for {@code key}
     * @throws NullPointerException if the specified key or value is null
     */
    public V put(K key, V value) {
        return putVal(key, value, false);
    }

    /** Implementation for put and putIfAbsent */
    final V putVal(K key, V value, boolean onlyIfAbsent) {
        if (key == null || value == null) throw new NullPointerException();
        int hash = spread(key.hashCode());
        int binCount = 0;
        for (Node<K,V>[] tab = table;;) {
            Node<K,V> f; int n, i, fh;
            //tab为空,则初始化
            if (tab == null || (n = tab.length) == 0)
                tab = initTable();
            else if ((f = tabAt(tab, i = (n - 1) & hash)) == null) { //该槽为空,则尝试插入
                if (casTabAt(tab, i, null,
                             new Node<K,V>(hash, key, value, null)))
                    break;                   // no lock when adding to empty bin
            }
            else if ((fh = f.hash) == MOVED) //正在移动,
                tab = helpTransfer(tab, f);
            else {
                V oldVal = null;
                synchronized (f) { //对该槽进行加锁
                    if (tabAt(tab, i) == f) {
                        if (fh >= 0) {
                            binCount = 1;
                            for (Node<K,V> e = f;; ++binCount) {
                                K ek;
                                if (e.hash == hash &&
                                    ((ek = e.key) == key ||
                                     (ek != null && key.equals(ek)))) {
                                    oldVal = e.val;
                                    if (!onlyIfAbsent)
                                        e.val = value;
                                    break;
                                }
                                Node<K,V> pred = e;
                                if ((e = e.next) == null) {
                                    pred.next = new Node<K,V>(hash, key,
                                                              value, null);
                                    break;
                                }
                            }
                        }
                        else if (f instanceof TreeBin) {
                            Node<K,V> p;
                            binCount = 2;
                            if ((p = ((TreeBin<K,V>)f).putTreeVal(hash, key,
                                                           value)) != null) {
                                oldVal = p.val;
                                if (!onlyIfAbsent)
                                    p.val = value;
                            }
                        }
                    }
                }
                if (binCount != 0) {
                    if (binCount >= TREEIFY_THRESHOLD)
                        treeifyBin(tab, i);
                    if (oldVal != null)
                        return oldVal;
                    break;
                }
            }
        }
        addCount(1L, binCount);
        return null;
    }
/**
     * Returns the value to which the specified key is mapped,
     * or {@code null} if this map contains no mapping for the key.
     *
     * <p>More formally, if this map contains a mapping from a key
     * {@code k} to a value {@code v} such that {@code key.equals(k)},
     * then this method returns {@code v}; otherwise it returns
     * {@code null}.  (There can be at most one such mapping.)
     *
     * @throws NullPointerException if the specified key is null
     */
    public V get(Object key) {
        Node<K,V>[] tab; Node<K,V> e, p; int n, eh; K ek;
        //获得hash值
        int h = spread(key.hashCode());
        //表非空,且该处不为空
        if ((tab = table) != null && (n = tab.length) > 0 &&
            (e = tabAt(tab, (n - 1) & h)) != null) {
            if ((eh = e.hash) == h) { //判断第1个
                if ((ek = e.key) == key || (ek != null && key.equals(ek)))
                    return e.val;
            }
            else if (eh < 0) //eh<0,找其他的
                return (p = e.find(h, key)) != null ? p.val : null;
            while ((e = e.next) != null) { //遍历
                if (e.hash == h &&
                    ((ek = e.key) == key || (ek != null && key.equals(ek))))
                    return e.val;
            }
        }
        return null;
    }

ConcurrentHashMap代码太多了,粘了好几次粘不上来。只粘几个方法吧。

阅后感

ConcurrentHashMap通过几个原子操作尽量减少加锁操作。

扩容部分没有看太明白,尤其时扩容时进行get操作。后续再继续学习。

 

posted @ 2020-01-01 10:42  水木桶  阅读(405)  评论(0编辑  收藏  举报