why try...catch requires EXACT type thrown

Question

I can do this, no problem:

long lngval = 3L;
int i = lngval;

but if I try this:

try {
  throw 3L;
}

catch(int i) {
  cout << "caught " << i << endl;
}

I get an unhandled exception.

This seems inconsistent. what is the reason for no type conversion in this case?

Answer 1

In the first case, the compiler can tell exactly what you want to do: convert a long to an int. In the second case, the compiler has to assume that you might have a construct like this:

try {
  try {
    throw 3L;
  }
  catch (int i) { /* P */ }
}
catch (long l) { /* Q */ }

The idea is that the compiler can never know whether there might be a catch (long l) lurking outside the current context, so it's never safe to just pick the first possible conversion.

This is also why it's common to use a class hierarchy when throwing exceptions rather than random types like int or long: it makes it easy to add more or less specification to your exception handlers in such a way that the compiler can be sure of your intentions (via the is-a relationship).