ConcurrentHashMap学习

官方文档:

ConcurrentHashMap提供了和hashTable同样的线程安全性,但是实现细节却完全不一样。取值操作是不加锁的,而且也不提供方法锁住整个表。

因为get操作不加锁,所以很多更新操作可能会同时发生。取回的值意味着最近完成的结果。对于并集操作,例如 {@code putAll} 和{@code clear}, 并发取值意味着紧紧一部分Node的插入和删除操作。同时,一次只允许一个线程做iterator操作。

重要方法:

get:

可以看到get操作是不经过加锁的。

 /**
     * 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;
        int h = spread(key.hashCode());
        if ((tab = table) != null && (n = tab.length) > 0 &&
            (e = tabAt(tab, (n - 1) & h)) != null) {
            if ((eh = e.hash) == h) {
                if ((ek = e.key) == key || (ek != null && key.equals(ek)))
                    return e.val;
            }
            else if (eh < 0)
                return (p = e.find(h, key)) != null ? p.val : null;
            while ((e = e.next) != null) {
                if (e.hash == h &&
                    ((ek = e.key) == key || (ek != null && key.equals(ek))))
                    return e.val;
            }
        }
        return null;
    }

put:

一个map的存储数据的入口,可以看到put的锁粒度很小,只以Node为锁,所以如果hash没有落到一个Node节点下的话是完全可以并发操作的。

/** Implementation for put and putIfAbsent */
    final V putVal(K key, V value, boolean onlyIfAbsent) {
        //可以看到key和value都不允许为空
        if (key == null || value == null) throw new NullPointerException();
        //计算hash
        int hash = spread(key.hashCode());
        int binCount = 0;
        //开始插入操作
        for (Node<K,V>[] tab = table;;) {
            Node<K,V> f; int n, i, fh;
            //如果当前table为空,初始化
            if (tab == null || (n = tab.length) == 0)
                tab = initTable();
            //尝试直接在对应位置CAS插入,如果成功就返回
            else if ((f = tabAt(tab, i = (n - 1) & hash)) == null) {
                if (casTabAt(tab, i, null,
                             new Node<K,V>(hash, key, value, null)))
                    break;                   // no lock when adding to empty bin
            }
            //如果当前表正处于扩容拷贝阶段,则这个线程也帮着拷贝
            else if ((fh = f.hash) == MOVED)
                tab = helpTransfer(tab, f);
            //找到对应的Node头结点,并以它为锁
            else {
                V oldVal = null;
                synchronized (f) {
                    if (tabAt(tab, i) == f) {
                        if (fh >= 0) {
                            binCount = 1;
                            for (Node<K,V> e = f;; ++binCount) {
                                K ek;
                                //这里的操作是判定当前的key存不存在,根据状态选择不同策略
                                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;
                                }
                            }
                        }
                        //如果是一个树节点,则用树节点的方式更新
                        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;
                            }
                        }
                    }
                }
                //判断一下是否要扩展成红黑树
                if (binCount != 0) {
                    if (binCount >= TREEIFY_THRESHOLD)
                        treeifyBin(tab, i);
                    if (oldVal != null)
                        return oldVal;
                    break;
                }
            }
        }
        //增加当前表的元素个数
        addCount(1L, binCount);
        return null;
    }

remove:

remove操作是通过replaceNode方法实现的。可以看到删除操作其实就是把对应的key的值设置成null。

 /**
     * Removes the key (and its corresponding value) from this map.
     * This method does nothing if the key is not in the map.
     *
     * @param  key the key that needs to be removed
     * @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 is null
     */
    public V remove(Object key) {
        return replaceNode(key, null, null);
    }

    /**
     * Implementation for the four public remove/replace methods:
     * Replaces node value with v, conditional upon match of cv if
     * non-null.  If resulting value is null, delete.
     */
    final V replaceNode(Object key, V value, Object cv) {
        int hash = spread(key.hashCode());
        for (Node<K,V>[] tab = table;;) {
            Node<K,V> f; int n, i, fh;
            //如果不存在表,或当前位置没有头结点,直接返回
            if (tab == null || (n = tab.length) == 0 ||
                (f = tabAt(tab, i = (n - 1) & hash)) == null)
                break;
            //帮助扩容
            else if ((fh = f.hash) == MOVED)
                tab = helpTransfer(tab, f);
            else {
                V oldVal = null;
                boolean validated = false;
                synchronized (f) {
                    if (tabAt(tab, i) == f) {
                        if (fh >= 0) {
                            validated = true;
                            for (Node<K,V> e = f, pred = null;;) {
                                K ek;
                                if (e.hash == hash &&
                                    ((ek = e.key) == key ||
                                     (ek != null && key.equals(ek)))) {
                                    V ev = e.val;
                                    if (cv == null || cv == ev ||
                                        (ev != null && cv.equals(ev))) {
                                        oldVal = ev;
                                        //如果值不为空,替换
                                        if (value != null)
                                            e.val = value;
                                        //值为空把这个节点直接删除
                                        else if (pred != null)
                                            pred.next = e.next;
                                        //没有前驱且值为空,Cas替换头结点的值
                                        else
                                            setTabAt(tab, i, e.next);
                                    }
                                    break;
                                }
                                pred = e;
                                if ((e = e.next) == null)
                                    break;
                            }
                        }
                        else if (f instanceof TreeBin) {
                            validated = true;
                            TreeBin<K,V> t = (TreeBin<K,V>)f;
                            TreeNode<K,V> r, p;
                            if ((r = t.root) != null &&
                                (p = r.findTreeNode(hash, key, null)) != null) {
                                V pv = p.val;
                                if (cv == null || cv == pv ||
                                    (pv != null && cv.equals(pv))) {
                                    oldVal = pv;
                                    if (value != null)
                                        p.val = value;
                                    else if (t.removeTreeNode(p))
                                        setTabAt(tab, i, untreeify(t.first));
                                }
                            }
                        }
                    }
                }
                if (validated) {
                    if (oldVal != null) {
                        if (value == null)
                            addCount(-1L, -1);
                        return oldVal;
                    }
                    break;
                }
            }
        }
        return null;
    }

size:

关于size方法,因为这是一个全局的计数统计,但是要使用这个方法又不能把整个表锁住,所以他是一个估计值,根据baseCount和cas更新失败的计数来决定的一个大概的值。官方也说明要考虑用mappingCount来替换size方法。

 /**
     * {@inheritDoc}
     */
    public int size() {
        long n = sumCount();
        return ((n < 0L) ? 0 :
                (n > (long)Integer.MAX_VALUE) ? Integer.MAX_VALUE :
                (int)n);
    }

 /**
     * A padded cell for distributing counts.  Adapted from LongAdder
     * and Striped64.  See their internal docs for explanation.
     */
    @sun.misc.Contended static final class CounterCell {
        volatile long value;
        CounterCell(long x) { value = x; }
    }

    final long sumCount() {
        CounterCell[] as = counterCells; CounterCell a;
        long sum = baseCount;
        if (as != null) {
            for (int i = 0; i < as.length; ++i) {
                if ((a = as[i]) != null)
                    sum += a.value;
            }
        }
        return sum;
    }

 /**
     * Returns the number of mappings. This method should be used
     * instead of {@link #size} because a ConcurrentHashMap may
     * contain more mappings than can be represented as an int. The
     * value returned is an estimate; the actual count may differ if
     * there are concurrent insertions or removals.
     *
     * @return the number of mappings
     * @since 1.8
     */
    public long mappingCount() {
        long n = sumCount();
        return (n < 0L) ? 0L : n; // ignore transient negative values
    }


Java   编程      Java

本博客所有文章除特别声明外,均采用 CC BY-SA 3.0协议 。转载请注明出处!