Java并发编程之ConcurrentLinkedQueue解读

private transient volatile Node<E> tail = head;

    public boolean offer(E e) {
        checkNotNull(e);
        // 入队前，创建一个入队节点
        final Node<E> newNode = new Node<E>(e);
        for (Node<E> t = tail, p = t;;) {
            // 创建一个指向tail节点的引用
            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;
        }
    }

/**
*返回 p 的后继节点，或者如果 p.next 已经链接到 self 则返回头节点，这只有在使用现在不在列表*中的陈旧指针遍历时才会为真。 
**/
   final Node<E> succ(Node<E> p) {
        Node<E> next = p.next;
        return (p == next) ? head : next;
    }

public E poll() {
        restartFromHead:
        for (;;) {
            for (Node<E> h = head, p = h, q;;) {
                //入列折腾的tail，那出列折腾的就是head
                E item = p.item;
                //出列判断依据是节点的item=null
                //item ！= null， 并且能将操作节点的item设置null， 表示出列成功
                if (item != null && p.casItem(item, null)) {
                    if (p != h) 
                        //一旦出列成功需要对head进行移动
                        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)
                    //第一轮操作失败，下一轮继续，调回到循环前
                    continue restartFromHead;
                else
                    //推动head节点移动
                    p = q;
            }
        }
    }