为什么编译器认为Environment.Exit可以返回？ [英] Why does the compiler think Environment.Exit can return?

问题描述

示例代码：

switch(something)
{
    case 0:
        System.Environment.Exit(0);
    case 1:
        // blah ...
        break;
}

它不会编译，因为编译器认为执行可以从Exit返回（ ）。编译器显然是错误的。

It won't compile because the compiler thinks that execution can return from Exit(). The compiler is obviously wrong.

没有技巧。 System.Environment.Exit（）是真实的。

No tricks. System.Environment.Exit() is the real one.

对于<$ c来说，这不仅是完全不合逻辑的$ c> System.Environment.Exit（）返回，我跟踪了代码，最终调用了 ExitProcess（exitCode）;

Not only is it utterly illogical for System.Environment.Exit() to return, I traced the code and it eventually calls ExitProcess(exitCode); which can't return.

推荐答案

就语言而言，它可以返回。是的，在现实生活中，该过程将在有机会返回之前终止，但是编译器基于方法签名不知道该过程。

As far as the language is concerned, it can return. Yes, in real-life the process will terminate before it has a chance to return, but the compiler doesn't know that based on the method signature.

您将需要在其中添加 break以使编译器满意。

You'll need to add the "break" in there to make the compiler happy.

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