C＃中：如果静态方法是从多个线程叫什么？ [英] C# : What if a static method is called from multiple threads?

问题描述

在我的应用我有一个从多个线程在同一时间被称为静态方法。有没有我的数据的任何危险混合了？

In my Application I have a static method that is called from multiple threads at the same time. Is there any danger of my data being mixed up?

在我第一次尝试的方法是不是静态的，我创建了一个类的多个实例。在这种情况下，我的数据得到了某种程度上混淆。我不知道这是如何发生，因为它只是有时会发生。我还在调试。
但现在的方法是静态的对我没有问题为止。也许这只是运气。我不知道。

In my first attempt the method was not static and I was creating multiple instance of the class. In that case my data got mixed up somehow. I am not sure how this happens because it only happens sometimes. I am still debugging. But now the method is static on I have no problems so far. Maybe it's just luck. I don't know for sure.

推荐答案

方法内声明的变量（有可能是个例外抓获的变量）是孤立的，所以你不会得到任何内在问题;但是，如果你的静态方法访问的共享状态，全盘皆输。

Variables declared inside methods (with the possible exception of "captured" variables) are isolated, so you won't get any inherent problems; however, if your static method accesses any shared state, all bets are off.

共享状态的例子是：

• 静态字段


• 从一个共同的高速缓存访​​问的对象（非序列化）

经由输入参数（和状态对这些对象）中获得
• 数据，如果有可能是多个线程都在触摸同一对象（多个）

如果您有共享状态，你必须：

If you have shared state, you must either:

• 要注意不要发生变异的状态，一旦它可以共享（最好：使用不可变对象重新present状态，并采取状态的快照到一个局部变量 - 例如，而不是参照 whatever.SomeData 反反复复，你读 whatever.SomeData 曾经到一个局部变量，然后只使用变量 - 注意，这不仅有利于对不可变状态）


• （所有线程必须同步）同步到数据存取 - 无论是互斥或（更精细）读/写器

• take care not to mutate the state once it can be shared (better: use immutable objects to represent state, and take a snapshot of the state into a local variable - i.e. rather than reference whatever.SomeData repeatedly, you read whatever.SomeData once into a local variable, and then just use the variable - note that this only helps for immutable state!)
• synchronize access to the data (all threads must synchronize) - either mutually exclusive or (more granular) reader/writer

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