Ejemplo n.º 1
0
  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;
      }
    }
  }
Ejemplo n.º 2
0
 /**
  * Returns the first live (non-deleted) node on list, or null if none. This is yet another variant
  * of poll/peek; here returning the first node, not element. We could make peek() a wrapper around
  * first(), but that would cost an extra volatile read of item, and the need to add a retry loop
  * to deal with the possibility of losing a race to a concurrent poll().
  */
 Node<E> first() {
   restartFromHead:
   for (; ; ) {
     for (Node<E> h = head, p = h, q; ; ) {
       boolean hasItem = (p.item != null);
       if (hasItem || (q = p.next) == null) {
         updateHead(h, p);
         return hasItem ? p : null;
       } else if (p == q) continue restartFromHead;
       else p = q;
     }
   }
 }
Ejemplo n.º 3
0
 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;
     }
   }
 }