c++ - 带有 try-catch-rethrow 的代码是否等同于不带 try-catch 的代码？

Question

以下两种代码在什么情况下不等价？

{
  // some code, may throw and/or have side effects
}

try {
  // same code as above
} catch(...) {
  throw;
}

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edit 澄清一下，我对 (i) 与上述模式的偏差(例如 catch block 中的更多代码)不感兴趣，也不 (ii) 打算邀请有关适当的光顾评论try-catch block 的用法。

我正在寻找有关 C++ 标准的合格答案。这个问题是由 Cheers and hth. - Alf 的评论提示的。至this answer of mine ，声明没有进一步解释上述代码不等价。

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edit 他们确实是不同的。堆栈展开将在后者中完成，但不一定在前者中完成，这取决于在运行时是否找到异常处理程序(一些 catch block 在堆栈更高的位置)。

Answer 1

后者要求堆栈展开，而在前者中，如果堆栈展开则由实现定义。

相关标准引用(均来自N3337):

[except.ctor]/1: As control passes from a throw-expression to a handler, destructors are invoked for all automatic objects constructed since the try block was entered. The automatic objects are destroyed in the reverse order of the completion of their construction.

[except.ctor]/3: The process of calling destructors for automatic objects constructed on the path from a try block to a throw-expression is called “stack unwinding.” [...]

[except.terminate]/2: [When the exception handling mechanism cannot find a handler for a throw exception], std::terminate() is called (18.8.3). In the situation where no matching handler is found, it is implementation-defined whether or not the stack is unwound before std::terminate() is called. [...]

因此，如果您想保证自动对象的析构函数在未处理的异常情况下运行(例如，某些持久性存储必须在销毁时发生突变)，那么 try {/*code*/} catch (...) {throw;} 会这样做，但 {/*code*/} 不会。