blocks|key|1205733|text|一般性建议是，在关键部分发生的任何事情都应尽可能简单。特别是，您应该避免嵌套锁。嵌套锁可能是死锁的来源。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1205734|应用于您在“timer”线程中所做的操作时，您可能应该将关键部分与处理分离开来。计时器线程中的IOW只需从公共变量中检索数据，然后执行其余的处理，包括与锁外的UI线程的交互。|1205735|在混合中添加第三个线程不会使您的生活变得更轻松。|1205736|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

The general recommendation is that whatever is happening in critical sections should be as simple as possible. In particular you should avoid nested locks. Nested locks potentially are the source of deadlocks.

As applied to what you are doing in your 'timer' thread you probably should separate the critical sections from processing. IOW in the timer thread just retrieve the data from the common variable(s) and then do the rest of the processing including interactions with UI thread outside the lock.

Adding a third thread to the mix will not make your life any easier

blocks|key|1170719|text|经验法则是使用最轻的锁，使尽可能缩短时间。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|1170720|这里有一篇很好的文章：http://www.moserware.com/2008/09/how-do-locks-lock.html|1170721|entityMap|0|LINK|mutability|MUTABLE|url|http://www.moserware.com/2008/09/how-do-locks-lock.html^0|7|3|D|5|0|B|1J|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@]|D|@$9|V|A|W|1|X]]|E|$]]|$1|H|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

The rule of thumb is to use the lightest lock for the shortest time possible.

There is quite a good article here: <a href="http://www.moserware.com/2008/09/how-do-locks-lock.html" rel="nofollow noreferrer">http://www.moserware.com/2008/09/how-do-locks-lock.html</a>

blocks|key|557537|text|要遵循的一般规则是，在持有锁时尽量减少持有锁的时间，而不调用不拥有和控制的代码(例如事件、虚拟方法或UI线程)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|557538|因此，计时器不应该在保持锁的同时回调到UI中。如果它需要事务访问锁下的数据(读取、调用UI、写)，那么它可能应该设计为回滚和/或重试。|557539|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

A general rule to follow is that you minimize the amount of time that you hold a lock, and you don't call code you don't own and control (such as an event, a virtual method, or a UI thread) while holding a lock.

So the timer should not call back into the UI while holding the lock. If it needs transactional access to the data under the lock (read, call UI, write), then it should probably be designed to roll back and/or retry.

blocks|key|1205809|text|就像mfeingold说的，尽可能避免。在无法避免的情况下，请按照相同的顺序处理两个线程中的锁！我的意思是，如果你有三个锁A，B和C，如果你有下面的锁模式|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1205810|1205811|线程1:+a->B->C|unordered-list-item|1205812|线程2:+A->C->B|1205813|1205814|那么僵局是可能的..。|1205815|嵌套锁通常会降低性能，因为您必须锁定外部锁，即使有时不需要确保在从另一个线程调用该方法时也不需要确保正确的行为。|1205816|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|P|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|Q|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|R|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

Like mfeingold said, avoid it where possible. Where you cannot avoid it take care of taking the locks in both threads in the same order! I mean, if you have three locks A, B and C, if you have following lock patterns

<ul>
<li>Thread 1: A->B->C</li>
<li>Thread 2: A->C->B</li>
</ul>

then a deadlock is possible...

Nesting locks usually costs performance as you have to lock the outer locks even if you would sometimes not need to to ensure correct behavior in the case where the method is called from another thread.

blocks|key|1170819|text|你为什么要这么做？听起来很复杂而且容易出错。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1170820|将计时器线程委托给UI线程的目的是什么？|1170821|这可能是必要的，但我的直接反应是，您可能需要后退一步，并考虑您的设计是否不必要地迫使您陷入复杂的同步问题。|1170822|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

Why are you doing this? It sounds complicated and error prone.

What is the timer thread delegating to the UI thread for?

It may be the case that it's necessary, but my immediate reaction is that you probably need to take a step back and consider whether your design is unnecessarily forcing you into complex synchronization issues.

I have two threads referencing the same variable -- the UI thread and a timer thread. As such, I have wrapped access to it in lock statements in both threads. The timer thread's access has priority -- if it's using the variable and the UI thread also wants access, I want the UI thread's operation to complete, but only after the timer thread's operation completes.

However, the timer thread may delegate back to the UI thread, so the UI thread needs to be free to handle that. To accommodate that, I have the UI thread launching a third thread to handle its operation so that it (that third thread) can wait for the timer operation to complete and the UI thread can be available. The locking happens in that third thread.

What is the correct pattern I should be using for this sort of synchronization?

Correct pattern for multi-thread synchronization? (C#)

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有两个线程引用同一个变量-- UI线程和一个计时器线程。因此，我可以在两个线程中用lock语句包装对它的访问。计时器线程的访问有优先权--如果它使用变量，而UI线程也想要访问，我希望UI线程的操作完成，但只有在计时器线程的操作完成之后。但是，计时器线程可以委托给UI线程，因此UI线程需要自由处理。为了适应这种情况，我让UI线程启动第三个线程来处理它的操作，以便它(第三个线程)可以等待计时器操作的

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