Java 并发编程之 ConcurrentLinkedQueue 非阻塞队列

Java juc 大约 2094 字

介绍

ConcurrentLinkedQueue没有LinkedBlockingQueue类似的puttake等阻塞方法,所以是一个非阻塞的队列。

但和LinkedBlockingQueueoffer/addpoll/remove实现类似,但LinkedBlockingQueue使用ReentrantLock实现锁机制,而ConcurrentLinkedQueue使用CAS锁机制。

offer

public boolean offer(E e) {
    checkNotNull(e);
    final Node<E> newNode = new Node<E>(e);

    for (Node<E> t = tail, p = t;;) {
        Node<E> q = p.next;
        if (q == null) {
            // p is last node
            if (p.casNext(null, newNode)) {
                // Successful CAS is the linearization point
                // for e to become an element of this queue,
                // and for newNode to become "live".
                if (p != t) // hop two nodes at a time
                    casTail(t, newNode);  // Failure is OK.
                return true;
            }
            // Lost CAS race to another thread; re-read next
        }
        else if (p == q)
            // We have fallen off list.  If tail is unchanged, it
            // will also be off-list, in which case we need to
            // jump to head, from which all live nodes are always
            // reachable.  Else the new tail is a better bet.
            p = (t != (t = tail)) ? t : head;
        else
            // Check for tail updates after two hops.
            p = (p != t && t != (t = tail)) ? t : q;
    }
}

poll

public E poll() {
    restartFromHead:
    for (;;) {
        for (Node<E> h = head, p = h, q;;) {
            E item = p.item;

            if (item != null && p.casItem(item, null)) {
                // Successful CAS is the linearization point
                // for item to be removed from this queue.
                if (p != h) // hop two nodes at a time
                    updateHead(h, ((q = p.next) != null) ? q : p);
                return item;
            }
            else if ((q = p.next) == null) {
                updateHead(h, p);
                return null;
            }
            else if (p == q)
                continue restartFromHead;
            else
                p = q;
        }
    }
}
阅读 71 · 发布于 2021-11-08

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