1public boolean offer(E e) {
 2    //如果元素为null，那么抛出NPE
 3    checkNotNull(e);
 4    final Node<E> newNode = new Node<E>(e);
 5    for (Node<E> t = tail, p = t;;) {
 6        //volatile读
 7        Node<E> q = p.next;
 8        if (q == null) {
 9            // p is last node
10            if (p.casNext(null, newNode)) {
11                if (p != t) // hop two nodes at a time
12                    casTail(t, newNode);  // Failure is OK.
13                return true;
14            }
15            //casNext竞争失败，再次读取p.next，此时(即q)不为空
16        }
17        else if (p == q)
18            // We have fallen off list.  If tail is unchanged, it
19            // will also be off-list, in which case we need to
20            // jump to head, from which all live nodes are always
21            // reachable.  Else the new tail is a better bet.
22            p = (t != (t = tail)) ? t : head;
23        else
24            //每插入两个新节点时才会重新设置尾节点
25            p = (p != t && t != (t = tail)) ? t : q;
26    }
27}

1p = (p != t && t != (t = tail)) ? t : q;