Does throw inside a catch ellipsis (...) rethrow the original error in C++?
Solution 1:
Yes. The exception is active until it's caught, where it becomes inactive. But it lives until the scope of the handler ends. From the standard, emphasis mine:
§15.1/4: The memory for the temporary copy of the exception being thrown is allocated in an unspecified way, except as noted in 3.7.4.1. The temporary persists as long as there is a handler being executed for that exception.
That is:
catch(...)
{ // <--
/* ... */
} // <--
Between those arrows, you can re-throw the exception. Only when the handlers scope ends does the exception cease to exist.
In fact, in §15.1/6 the example given is nearly the same as your code:
try {
// ...
}
catch (...) { // catch all exceptions
// respond (partially) to exception <-- ! :D
throw; //pass the exception to some
// other handler
}
Keep in mind if you throw
without an active exception, terminate
will be called. This cannot be the case for you, being in a handler.
If doSomethingElse()
throws and the exception has no corresponding handler, because the original exception is considered handled the new exception will replace it. (As if it had just thrown, begins stack unwinding, etc.)
That is:
void doSomethingElse(void)
{
try
{
throw "this is fine";
}
catch(...)
{
// the previous exception dies, back to
// using the original exception
}
try
{
// rethrow the exception that was
// active when doSomethingElse was called
throw;
}
catch (...)
{
throw; // and let it go again
}
throw "this replaces the old exception";
// this new one takes over, begins stack unwinding
// leaves the catch's scope, old exception is done living,
// and now back to normal exception stuff
}
try
{
throw "original exception";
}
catch (...)
{
doSomethingElse();
throw; // this won't actually be reached,
// the new exception has begun propagating
}
Of course if nothing throws, throw;
will be reached and you'll throw your caught exception as expected.