What happens if I use "throw;" without an exception to throw?

Question

Here's the setup.

I have a C++ program which calls several functions, all of which potentially throw the same exception set, and I want the same behaviour for the exceptions in each function (e.g. print error message & reset all the data to the default for exceptionA; simply print for exceptionB; shut-down cleanly for all other exceptions).

It seems like I should be able to set the catch behaviour to call a private function which simply rethrows the error, and performs the catches, like so:

void aFunction()
{
    try{ /* do some stuff that might throw */ }
    catch(...){handle();}
}

void bFunction()
{
    try{ /* do some stuff that might throw */ }
    catch(...){handle();}
}

void handle()
{
    try{throw;}
    catch(anException)
    {
        // common code for both aFunction and bFunction
        // involving the exception they threw
    }
    catch(anotherException)
    {
        // common code for both aFunction and bFunction
        // involving the exception they threw
    }
    catch(...)
    {
        // common code for both aFunction and bFunction
        // involving the exception they threw
    }
}

Now, what happens if "handle" is called outside of the exception class. I'm aware that this should never happen, but I'm wondering if the behaviour is undefined by the C++ standard.

Answer 1

If handle() is called outside the context of an exception, you will throw without an exception being handled. In this case, the standard (see section 15.5.1) specifies that

If no exception is presently being handled, executing a throw-expression with no operand calls terminate().

so your application will terminate. That's probably not what you want here.

Answer 2

If you use throw inside of a catch block, it will rethrow the exception. If you use throw outside of a catch block, it will terminate the application.