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  • 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。

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  • 原文地址:https://www.cnblogs.com/shuimutong/p/12128434.html
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