问题描述
final boolean acquireQueued(final Node node,int arg) {
boolean Failed = true;
try {
boolean interrupted = false;
for (;;) {
final Node p = node.predecessor();
if (p == head && tryAcquire(arg)) {
setHead(node);
p.next = null; // help GC
Failed = false;
return interrupted;
}
if (shouldParkAfterFailedAcquire(p,node) &&
parkAndCheckInterrupt())
interrupted = true;
}
} finally {
if (Failed)
cancelAcquire(node);
}
}
/**
* Releases in exclusive mode. Implemented by unblocking one or
* more threads if {@link #tryRelease} returns true.
* This method can be used to implement method {@link Lock#unlock}.
*
* @param arg the release argument. This value is conveyed to
* {@link #tryRelease} but is otherwise uninterpreted and
* can represent anything you like.
* @return the value returned from {@link #tryRelease}
*/
public final boolean release(int arg) {
if (tryRelease(arg)) {
Node h = head;
if (h != null && h.waitStatus != 0)
unparkSuccessor(h);
return true;
}
return false;
}
private void unparkSuccessor(Node node) {
/*
* If status is negative (i.e.,possibly needing signal) try
* to clear in anticipation of signalling. It is OK if this
* fails or if status is changed by waiting thread.
*/
int ws = node.waitStatus;
if (ws < 0)
compareAndSetWaitStatus(node,ws,0);
/*
* Thread to unpark is held in successor,which is normally
* just the next node. But if cancelled or apparently null,* traverse backwards from tail to find the actual
* non-cancelled successor.
*/
Node s = node.next;
if (s == null || s.waitStatus > 0) {
s = null;
for (Node t = tail; t != null && t != node; t = t.prev)
if (t.waitStatus <= 0)
s = t;
}
if (s != null)
LockSupport.unpark(s.thread);
}
线程A调用tryRelease方法成功释放锁并唤醒后继线程。如果后继线程此时没有park，但是tryAcquire成功获取了锁，那么线程A会unpark一个正常执行的线程。这样，后续调用就会有不确定因素。 Doug Lea 的考虑是什么？
解决方法
暂无找到可以解决该程序问题的有效方法，小编努力寻找整理中！
如果你已经找到好的解决方法，欢迎将解决方案带上本链接一起发送给小编。
小编邮箱:dio#foxmail.com (将#修改为@）