blocks|key|2985124|text|这个方法是100%25线程安全的，即使它不是static。当你需要在线程之间共享数据时，线程安全的问题就出现了--你必须注意原子性、可见性等。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2985125|此方法仅对驻留在堆栈上的参数和对堆上不可变对象的引用进行操作。堆栈本质上是线程的本地堆栈，因此永远不会发生数据共享。|BOLD|2985126|不可变对象(在本例中为String)也是线程安全的，因为一旦创建，它们就不能更改，并且所有线程都看到相同的值。另一方面，如果该方法接受(可变的)+Date，那么您可能会遇到问题。两个线程可以同时修改同一个对象实例，从而导致争用条件和可见性问题。|2985127|entityMap^0|K|6|0|C|1A|0|0|3E|B|6|21|4|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|P|8|@$9|Q|A|R|B|H]]|D|@]|E|$]]|$1|I|3|J|5|6|7|S|8|@$9|T|A|U|B|H]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$]]

This method is 100% thread safe, it would be even if it wasn't <code>static</code>. The problem with thread-safety arises when you need to share data between threads - you must take care of atomicity, visibility, etc.

This method only operates on parameters, which reside on stack and references to immutable objects on heap. Stack is inherently local to the thread, so no sharing of data occurs, ever.

Immutable objects (<code>String</code> in this case) are also thread-safe because once created they can't be changed and all threads see the same value. On the other hand if the method was accepting (mutable) <code>Date</code> you could have had a problem. Two threads can simultaneously modify that same object instance, causing race conditions and visibility problems.

blocks|key|3198510|text|当一个方法改变一些共享状态时，它只能是线程不安全的。它是否是静态的是无关紧要的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3198511|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|$]]

A method can only be thread-unsafe when it changes some shared state. Whether it's static or not is irrelevant.

blocks|key|3198527|text|该函数是完全线程安全的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3198528|如果你仔细想想..。假设如果这种情况不同，会发生什么。如果不同步，每个常用函数都会有线程问题，因此JDK中的所有API函数都必须同步，因为它们可能会被多个线程调用。由于大多数时候应用程序都在使用一些API，因此多线程应用程序实际上是不可能的。|3198529|这太荒谬了，不去想它，所以只对你来说:如果有一个明确的原因可能会出现问题，那么方法就不是threadsafe。试着总是想一想，如果我的函数中有多个线程，如果你有一个步骤调试器，会不会一步接着一步地推进第一步……然后第二个线程...也许是第二次...会有问题吗？如果你找到了一个，它就不是线程安全的。|3198530|还请注意，大多数Java1.5集合类都不是线程安全的，除了那些声明的类，比如ConcurrentHashMap。|3198531|如果您真的想深入研究这个问题，可以仔细研究一下volatile关键字及其所有副作用。请看一下信号量()和锁()类，以及它们在java.util.Concurrent中的朋友。阅读类周围的所有API文档。这也是值得学习和满足的。|3198532|对于这个过于复杂的回答，我很抱歉。|3198533|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|P|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|Q|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|R|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|S|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|M|$]]

The function is perfectly thread safe. 

If you think about it... assume what would happen if this were different. Every usual function would have threading problems if not synchronized, so all API functions in the JDK would have to be synchronized, because they could potentially be called by multiple threads. And since most time the app is using some API, multithreaded apps would effectively be impossible.

This is too ridiculous to think about it, so just for you: Methods are not threadsafe if there is a clear reason why there could be problems. Try to always think about what if there were multiple threads in my function, and what if you had a step-debugger and would one step after another advance the first... then the second thread... maybe the second again... would there be problems? If you find one, its not thread safe.

Please be also aware, that most of the Java 1.5 Collection classes are not threadsafe, except those where stated, like ConcurrentHashMap. 

And if you really want to dive into this, have a close look at the volatile keyword and ALL its side effects. Have a look at the Semaphore() and Lock() class, and their friends in java.util.Concurrent. Read all the API docs around the classes. It is worth to learn and satisfying, too.

Sorry for this overly elaborate answer.

blocks|key|275761|text|String对象是不可变的，这是上述线程安全场景的另一个原因。相反，如果使用可变对象(比如makeMutableArray..)那么线程安全肯定会崩溃。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|275762|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|$]]

String objects being immutable is another reason for thread-safe scenario above. Instead if mutable objects are used (say makeMutableArray..) then surely thread-safety will break.

blocks|key|3198566|text|将static关键字与同步的静态方法一起使用可以修改线程之间共享的静态数据。使用static关键字，所有创建的线程都将争用该方法的一个版本。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3198567|将volatile关键字与同步实例方法一起使用将确保每个线程都有自己的共享数据副本，并且线程之间不会泄漏读/写操作。|3198568|entityMap^0|1|6|14|6|0|1|8|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|R|8|@]|D|@]|E|$]]]|I|$]]

Use the <code>static</code> keyword with synchronized static methods to modify static data shared among threads. With the <code>static</code> keyword all the threads created will contend for a single version of the method.

Use the <code>volatile</code> keyword along with synchronized instance methods will guarantee that each thread has its own copy of the shared data and no read/ writes will leak out between threads.

I was wondering if you have a static method that is not synchronised, but does not modify any static variables is it thread-safe? What about if the method creates local variables inside it? For example, is the following code thread-safe?

<pre><code>public static String[] makeStringArray( String a, String b ){
 return new String[]{ a, b };
}
</code></pre>

So if I have two threads calling ths method continously and concurrently, one with dogs (say "great dane" and "bull dog") and the other with cats (say "persian" and "siamese") will I ever get cats and dogs in the same array? Or will the cats and dogs never be inside the same invocation of the method at the same time?

Are non-synchronised static methods thread safe if they don't modify static class variables?

调试器

我想知道你是否有一个静态方法是不是同步的，但是不是修改任何静态变量，它是线程安全的吗？如果该方法在其内部创建局部变量怎么办？例如，下面的代码是线程安全的吗？public static String[] makeStringArray( String a, String b ){    return new String...

问如果非同步静态方法不修改静态类变量，那么它们是线程安全的吗？
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如果非同步静态方法不修改静态类变量，那么它们是线程安全的吗？EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如果非同步静态方法不修改静态类变量，那么它们是线程安全的吗？
EN