blocks|key|3727743|text|这需要同步，因为如果一个线程将valid计算为false，然后在赋值之前暂停执行，那么另一个线程也会出现，并将valid检查为false，在将valid设置为true之前，您将有两个线程从这里开始运行代码(这可能不是您想要的)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3727744|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

This needs to be synchronized, because if one thread evaluates valid as false, and then suspends execution before the assignment, then another one comes along and also checks valid as false, before setting valid to true, you will have two threads running code following on from here (which presumably you don't want).

blocks|key|5840467|text|使用AtomicBoolean。可以同时检查和设置实例。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5840468|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Use AtomicBoolean. Instances can be simultaneously checked and set.

blocks|key|5840380|text|它不是线程安全的。但如果这是整个代码，那就没什么关系了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5840381|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

It isn't thread-safe. But if this is the whole code, it wouldn't matter.

blocks|key|5840431|text|编辑：一个全能的更好的替代方案是AtomicBoolean，它使用低级操作在没有同步的情况下实现条件更新。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|CODE|entityRanges|data|5840432|有两个单独的(即非原子的)对标志的访问，因此同步是必要的，除非这个线程是唯一一个对标志进行写操作的线程。即使这样，为了确保将来发生变化，最好还是进行同步。|5840433|entityMap|0|LINK|mutability|MUTABLE|url|http://java.sun.com/javase/6/docs/api/java/util/concurrent/atomic/AtomicBoolean.html^0|0|3|G|D|G|D|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|D]]|E|@$9|V|A|W|1|X]]|F|$]]|$1|G|3|H|5|6|7|Y|8|@]|E|@]|F|$]]|$1|I|3|-4|5|6|7|Z|8|@]|E|@]|F|$]]]|J|$K|$5|L|M|N|F|$O|P]]]]

Edit: An all-around superior alternative would be <a href="http://java.sun.com/javase/6/docs/api/java/util/concurrent/atomic/AtomicBoolean.html" rel="nofollow noreferrer"><code>AtomicBoolean</code></a>, which uses low-level operations to achieve conditional updates without synchronization.

There are two separate (i.e. non-atomic) accesses to the flag, thus synchronization is necessary unless this thread is the only one doing write operations on the flag. Even then it would probably be good to have synchronization just to be sure in case of future changes.

blocks|key|3727829|text|您的代码不是threadsafe，但它确实取决于您的其他代码这样做是否安全。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3727830|您是否要求valid+=+true后面的代码只由单个线程执行一次？如果是这样，那么您的代码是不安全的，因为任意数量的线程都可以读取valid的false值，然后将其设置为true。例如：|offset|length|style|CODE|3727831|if+(valid)
++return;
//+Imagine+every+single+one+of+your+threads+stops+and+blocks+here.
//+They+will+all+wake+up+again+and+set+valid+to+true+and+then
//+execute+the+code+to+follow.
valid+=+true;|code-block|syntax|javascript|3727832|但是，如果您只想保证valid+=+true之后的代码最多由任何线程执行一次...那么你所拥有的就很好了。但是如果这是你需要的行为，我会通过其他方法来实现，因为在这种情况下使用volatile看起来就像你不知道你在做什么。例如，您可以不在线程之间共享valid变量，而是允许每个线程只执行一次代码。|3727833|此外，当对同步和易失性进行推理时存在疑问时...只要使用同步即可。它通常更清晰，而且会从使用volatile中得到你想要的一切，除了更容易推断代码是如何工作的。|3727834|entityMap^0|0|5|C|1T|5|1Z|5|2D|4|0|0|A|C|2G|8|3H|5|0|1A|8|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]|$D|W|E|X|F|G]|$D|Y|E|Z|F|G]|$D|10|E|11|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|12|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|13|8|@$D|14|E|15|F|G]|$D|16|E|17|F|G]|$D|18|E|19|F|G]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1A|8|@$D|1B|E|1C|F|G]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|R|$]]

Your code is not threadsafe, but it really depends on your other code whether or not it is safe to do that.

Do you require that the code following <code>valid = true</code> only be executed once by a single thread? If so, then your code is not safe because any number of threads can read a <code>false</code> value of <code>valid</code>, and then end up setting it to <code>true</code>. For example:

<pre><code>if (valid)
 return;
// Imagine every single one of your threads stops and blocks here.
// They will all wake up again and set valid to true and then
// execute the code to follow.
valid = true;
</code></pre>

But if you just want to guarantee that the code after <code>valid = true</code> is executed at most one time by any thread... then what you have is fine. But if that's the behavior you require I would achieve that by other means, because using <code>volatile</code> in that case will just look like you don't know what you're doing. For example, you could just not share the <code>valid</code> variable across threads, and allow each thread to just execute the code once.

Also, when in doubt while reasoning about synchronization and volatile... just use synchronization. It is usually more clear, and will get you everything you wanted from using <code>volatile</code>, except it will be easier to reason how the code works.

blocks|key|2478663|text|线程安全是一个系统范围的属性。您不能孤立地看待一段代码，并将其称为线程安全/不安全。这取决于其他线程如何与它交互，以及一致性要求是什么。也就是说，大多数线程安全设计都使用while()循环，而不是if()块，因此，您的设计很可能是不正确的：)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2478664|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

thread-safety is a system wide property. you cannot look at a piece of code in isolation and call it thread safe/unsafe. it depends on how other threads interact with it and what is the consistency requirement. that being said, most thread-safe design has while() loop instead of if() block, so, your design is most likely incorrect:)

blocks|key|5840571|text|根据Brian+Goetz+https://www.ibm.com/developerworks/java/library/j-jtp06197/的一篇伟大的论文，“你只能在有限的情况下使用易失性变量而不是锁。易失性变量必须满足以下两个标准才能提供所需的线程安全性：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|5840572|5840573|写入该变量不依赖于其当前值。|unordered-list-item|5840574|该变量不会与其他变量一起参与不变量。|5840575|5840576|基本上，这些条件说明可以写入易失性变量的有效值集独立于任何其他程序状态，包括变量的当前状态。“因此，它必须与锁同步。|5840577|entityMap|0|LINK|mutability|MUTABLE|url|https://www.ibm.com/developerworks/java/library/j-jtp06197/^0|E|1N|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]|$1|E|3|F|5|G|7|Z|8|@]|9|@]|C|$]]|$1|H|3|I|5|G|7|10|8|@]|9|@]|C|$]]|$1|J|3|-4|5|6|7|11|8|@]|9|@]|C|$]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|C|$]]|$1|M|3|-4|5|6|7|13|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

According to great paper by Brian Goetz 
<a href="https://www.ibm.com/developerworks/java/library/j-jtp06197/" rel="nofollow">https://www.ibm.com/developerworks/java/library/j-jtp06197/</a> 
"You can use volatile variables instead of locks only under a restricted set of circumstances. Both of the following criteria must be met for volatile variables to provide the desired thread-safety:

<ul>
<li>Writes to the variable do not depend on its current value.</li>
<li>The variable does not participate in invariants with other variables.</li>
</ul>

Basically, these conditions state that the set of valid values that can be written to a volatile variable is independent of any other program state, including the variable's current state."
So it must be synchronized with a lock. It is not thread-safe as is.

If I have a volatile boolean (let's call it valid), is the following piece of code thread-safe in Java?

<pre><code>if (valid)
 return;
valid = true;
</code></pre>

Or, do I need to synchronize since valid is set to true only if it's false (hence set of valid depends on its current value)?

volatile boolean

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如果我有一个易失性布尔值(让我们称之为有效)，下面这段代码在Java中是线程安全的吗？if (valid)  return;valid = true;或者，我是否需要同步，因为只有当valid为false时才将其设置为true (因此，valid的集合取决于其当前值)？

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