什么是单个“投掷”语句do？

问题描述

这些天，我一直在阅读很多，特别是。

These days, I have been reading a lot the C++ F.A.Q and especially this page.

阅读本部分我发现了一个技术，作者称为异常分派器，允许有人在一个方便的函数中将他的所有异常处理分组：

Reading through the section I discovered a "technique" that the author calls "exception dispatcher" that allows someone to group all his exception handling in one handy function:

void handleException()
{
  try {
    throw; // ?!
  }
  catch (MyException& e) {
    //...code to handle MyException...
  }
  catch (YourException& e) {
    //...code to handle YourException...
  }
}

void f()
{
  try {
    //...something that might throw...
  }
  catch (...) {
    handleException();
  }
}

> throw; 语句：如果你考虑给定的例子，那么肯定，它是显而易见的：它重新抛出异常首先捕获在 f（）

What bothers me is the single throw; statement: if you consider the given example then sure, it is obvious what it does: it rethrows the exception first caught in f() and deals with it again.

但是如果我自己调用 handleException（）从 catch（）子句执行它？是否有任何指定的行为？

But what if I call handleException() on its own, directly, without doing it from a catch() clause ? Is there any specified behavior ?

此外，对于奖励积分，是否有任何其他奇怪（可能不是好词）使用 / code>您知道吗？

Additionally for bonus points, is there any other "weird" (probably not the good word) use of throw that you know of ?

谢谢。

推荐答案

如果你自己做一个 throw; ，并且没有当前的异常重新抛出，那么程序突然结束。 （更具体地说， terminate（）被调用。）

If you do a throw; on its own, and there isn't a current exception for it to rethrow, then the program ends abruptly. (More specifically, terminate() is called.)

是重新抛出当前异常的唯一安全方式 - 它不等于

Note that throw; is the only safe way to re-throw the current exception - it's not equivalent to

catch（exception const& e）{throw e; }

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