What does a single "throw;" statement do?

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

Reading through the section I discovered a "technique" that the author calls "exception dispatc开发者_运维技巧her" 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();
  }
}

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.

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

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

Thank you.

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.)

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; }

Yes, it specified behavior, it will call terminate;

15.1, para 8: If no exception is presently being handled, executing a throw expression with no operand calls terminate() (15.5.1).

That's so-called exception handler. It rethrows the "current exception" if any. If there's no exception currently being handled terminate() will be called.