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JDK1.8中的ConcurrentHashMap

参考博文

https://www.e-learn.cn/content/qita/1039497

Put操作
    public V put(K key, V value) {
        return putVal(key, value, false);
    }

JDK1.8中segment被取消,锁之间被加在了每个链表的头结点上。

synchronized (f) {
    if (tabAt(tab, i) == f) {
        if (fh >= 0) {
            binCount = 1;
            for (Node<K,V> e = f;; ++binCount) {
                K ek;
                // 键值对存在
                if (e.hash == hash &&
                        ((ek = e.key) == key ||
                                (ek != null && key.equals(ek)))) {
                    oldVal = e.val;
                    if (!onlyIfAbsent)
                        e.val = value;
                    break;
                }
                // 新节点插入到链表的尾部
                Node<K,V> pred = e;
                if ((e = e.next) == null) {
                    pred.next = new Node<K,V>(hash, key,
                            value, null);
                    break;
                }
            }
        }
        // 节点f为树节点,执行红黑树的插入操作
        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;
            }
        }
    }
}
Get操作
    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;
    }
Remove操作
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;
                        if (value != null)
                            e.val = value;
                        else if (pred != null)
                            pred.next = e.next;
                        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));
                }
            }
        }
    }
}