• ConcurrentLinkedQueue是一个基于链接节点的无界线程安全队列。
• 采用FIFO对节点排序
• 采用CAS实现非阻塞

1. ConcurrentLinkedQueue结构

• 由head和tail节点组成
• 每个节点（Node）由节点元素（item）和指向下一个节点的指针（next）组成

2. 入列

• 入列就是将入列节点添加到队列尾部
• 入列源码：

    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;
        }
    }

• 获取p节点的下一个节点

    final Node<E> succ(Node<E> p) {
        Node<E> next = p.next;
        return (p == next) ? head : next;
    }

• 设置入队节点为尾节点
  casNext(null, n)可以将入队节点置为尾节点的next节点，p为null就表示p是队列尾节点了，如果不为空，说明其他线程更新了尾节点，需要重新获取

3. 出列

获取头节点的元素，然后判断头结点是否为空，为空说明其他线程已经获取了该节点，如果不为空，CAS将头结点设置为null，成功返回头结点，不成功重新获取。

    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;
            }
        }
    }