exit（0）vs return 0 [英] exit(0) vs return 0

问题描述

当exit（0）用于退出程序时，不会调用本地
scoped非静态对象的析构函数。但是析构函数是
，如果使用返回0。注意静态对象​​将
清理，即使我们调用exit（）。

这个逻辑应该有一些原因。我只想知道它是什么？ c>退出（0）， 你正在调用一个函数。你
不期望调用局部变量的析构函数，如果
你正在调用一个函数。编译器不知道，
先验，有任何特殊的 exit（0）。

事实上，这个理由真的只适用于C ++之前的
异常。该标准可以重新定义 exit（）以使用参数抛出
实现定义的异常，并指定
调用 main 被包装在一个try块中，它捕获
这个异常，并将返回的代码传递回系统。
这意味着 exit 在C和C ++中有一个完全不同的
语义;无论如何，在委员会之前没有
的提案来进行此更改。

When exit(0) is used to exit from program, destructors for locally scoped non-static objects are not called. But destructors are called if return 0 is used.Note that static objects will be cleaned up even if we call exit().

There should be some reason behind this logic. i just want to know what it is? Thank you.

解决方案

In the case of exit( 0 ), you're calling a function. You don't expect the destructors of local variables to be called if you're calling a function. And the compiler doesn't know, a priori, that there is anything special about exit( 0 ).

In fact, this rationale really only applies to C++ before exceptions. The standard could redefine exit() to throw an implementation defined exception with the argument, and specify that the call to main is wrapped in a try block which catches this exception, and passes the return code back to the system. This would mean that exit have a completely different semantics in C and in C++, however; at any rate, there's been no proposal before the committee to make this change.

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