Issue2371

When trying to do the same as described in the following stack overflow answer: http://stackoverflow.com/questions/492519/timeout-on-a-python-function-call

jython throws an exception, but does not stop the execution of "loop_forever"

This exception is thrown
Exception in thread "SIGALRM handler" Traceback (most recent call last):
  File "/home/st/svn/stefan/trunk/GEVA-v2.0/GEVA/lib/Jython/jython-standalone-2.7.0.jar/Lib/signal.py", line 117, in handle
  File "<stdin>", line 3, in signal_handler

This one is a little bit tricky. CPython ensures signal delivery to the main thread, which especially works well for single-threaded programs like the one from Stack Overflow. But Java delivers signals to a separate signal handling thread.

Perhaps this is why the author in the SO question you referenced (http://stackoverflow.com/a/494273/423006) states:

> This module doesn't play well with threads (but then, who does?)

The reality is that signal handling can be made to work well on CPython and Jython in a portable fashion. So in practice you should follow the advice in http://snakesthatbite.blogspot.com/2010/09/cpython-threading-interrupting.html, although IMHO it's a bit overkill - Python memory model semantics ensure that even a simple boolean flag would work just fine, not just `threading.Event`.