public boolean offer(E e) { checkNotNull(e); //创建新节点 final Node<E> newNode = new Node<E>(e); //循环开始，tail赋值给t，t赋值给p，此处是死循环，保证该入队操作能够一直重试，直到入队成功 for (Node<E> t = tail, p = t;;) { //p的next赋值给q Node<E> q = p.next; //如果q是null，说明tail的next是null，tail指向的是队列尾部，所以tail的next应该始终是null，此处表示p是最后一个节点 if (q == null) { // p是尾节点，则p的next节点是要入队列的节点 //cas操作，若失败表示其他线程对尾节点进行了修改，重新循环 //将p（p指向tail）的next指向新节点，若成功，进入if语句 if (p.casNext(null, newNode)) { //cas操作成功后，判断p是否等于t，p不等于t，设置newNode为新的尾节点,p不等于t表示什么？(此处有不明白，希望高手帮帮忙。) if (p != t) // hop two nodes at a time casTail(t, newNode); // Failure is OK. return true; } // p和q相等（这部分不是太理解，希望高手帮忙下） } 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; } }
public E poll() { //设置循环标记 restartFromHead: for (;;) { for (Node<E> h = head, p = h, q;;) { E item = p.item; //表头数据不为null，cas操作设置表头数据为null成功进入if if (item != null && p.casItem(item, null)) { //p不等于h，队列头发生了变化，更新队列头为p，返回原来队列头的item if (p != h) // hop two nodes at a time updateHead(h, ((q = p.next) != null) ? q : p); return item; } //队列头下一个节点为null，更新头尾p返回null else if ((q = p.next) == null) { updateHead(h, p); return null; } else if (p == q) continue restartFromHead; else p = q; } } }
peek()方法
返回队列头部的元素，跟poll方法类似
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public E peek() { restartFromHead: for (;;) { for (Node<E> h = head, p = h, q;;) { E item = p.item; if (item != null || (q = p.next) == null) { updateHead(h, p); return item; } else if (p == q) continue restartFromHead; else p = q; } } }
size()方法
需要遍历队列，不推荐使用，尽量使用isEmpty()
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public int size() { int count = 0; for (Node<E> p = first(); p != null; p = succ(p)) if (p.item != null) // Collection.size() spec says to max out if (++count == Integer.MAX_VALUE) break; return count; }