Java并发容器之ConcurrentLinkedQueue详解

    private static class Node<E> {
        volatile E item;
        volatile Node<E> next;
        //....
    }

public class ConcurrentLinkedQueue<E> extends AbstractQueue<E>
        implements Queue<E>, java.io.Serializable {
        private transient volatile Node<E> head;
        private transient volatile Node<E> tail;  
        //....
}

   public boolean offer(E e) {
        checkNotNull(e);   //为空判断，e为null是抛异常
        final Node<E> newNode = new Node<E>(e); //将e包装成newNode
        for (Node<E> t = tail, p = t;;) {  //循环cas，直至加入成功
            //t = p = tail 
            Node<E> q = p.next;
            if (q == null) {   //判断p是否为尾节点
                //如果是，p.next = newNode
                if (p.casNext(null, newNode)) {
                    //首次添加时，p 等于t，不进行尾节点更新，所以所尾节点存在滞后性  
                    //并发环境，可能存添加/删除，tail就更难保证正确指向最后节点。
                    if (p != t) 
                        //更新尾节点为最新元素
                        casTail(t, newNode);  
                    return true;
                }
            }
            else if (p == q)
                //当tail不执行最后节点时，如果执行出列操作，很有可能将tail也给移除了    
                //此时需要对tail节点进行复位，复位到head节点
                p = (t != (t = tail)) ? t : head;
            else
                //推动tail尾节点往队尾移动
                p = (p != t && t != (t = tail)) ? t : q;
        }
    }

for (Node<E> t = tail, p = t;;)

 private boolean casTail(Node<E> cmp, Node<E> val) {
     return UNSAFE.compareAndSwapObject(this, tailOffset, cmp, val);
 }

boolean casNext(Node<E> cmp, Node<E> val) {
    return UNSAFE.compareAndSwapObject(this, nextOffset, cmp, val);
}

    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;
            }
        }
    }