Re: Exception in finally block

From:

Thomas Hawtin <usenet@tackline.plus.com>

Newsgroups:

comp.lang.java.programmer

Date:

Mon, 04 Dec 2006 18:27:28 +0000

Message-ID:

<45746845$0$8720$ed2619ec@ptn-nntp-reader02.plus.net>

Red Orchid wrote:

Thomas Hawtin <usenet@tackline.plus.com> wrote or quoted in
Message-ID: <457306ae$0$8759$ed2619ec@ptn-nntp-reader02.plus.net>:

There are different exceptions so that you can diagnose the problem
(possibly). You are going to have fun trying to disentangle multiple
exceptions at once. The most important exception should be the outermost
one.

Do the word "outermost" above mean this ?

By outermost, I mean the one thrown last (from an execution point of
view). There is no retry mechanism in Java (they do not appear to work
well in practice), so exceptions are always thrown outwards.

It's trivial on the acquire side of things, as the code causing the
innermost exception doesn't get to be executed.

try {
   ...
}
catch (XException e) {
   ...
}
catch (YException e) { // <- Outermost
   ...
}

Here both catch blocks would be equally outermost. An exception thrown
from the first catch would not be caught by the second.

try {
    try {
       ...
    }
    catch (XException e) {
       ...
    }
}
catch (YException e) { // <- Outermost
   ...
}

Here if the inner catch block threw a YException, then the YException
would be outermost.

If XException is super class of YException,
YException is unreachable.

And your code wouldn't compile.

Tom Hawtin