Throwable and Exception in Java

Question

Suppose I have a class named BugException which extends from RuntimeException And BugException has a copy constructor which takes a Throwable.

This code compiles and typechecks:

Throwable e = ...;
if (e instanceof BugException) {
   throw new BugException(e);
}

Why is it that:

Throwable e = ...;
if (e instanceof BugException) {
   throw e;
}

Does not compile and gives the error message: unhandled exception. java.lang.Throwable. ?

Why is this unnecessary wrapping necessary to satisfy the typechecker?

Testing `instanceof` doesn't change `e`'s static type. – user2357112 May 30 '16 at 02:27 — user2357112, May 30 '16 at 02:27
Or you can simply cast, without the wrapper. – pvg May 30 '16 at 02:41 — pvg, May 30 '16 at 02:41

score 5 · Accepted Answer · answered May 30 '16 at 02:44

At compile time, it is not know what kind of exception e is. It might be a checked Exception, in which case the compiler will need you to either wrap the throw in a try/catch or make the method throw it.

However, if you explicitly cast the unchecked exception, then it will compile.

Throwable e = ...;
if (e instanceof BugException) {
   throw (BugException) e;
}

Throwable and Exception in Java

1 Answers1