Message10127
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 |
|
Date |
User |
Action |
Args |
2015-06-22 08:56:58 | the | set | messageid: <1434963418.81.0.719638076278.issue2371@psf.upfronthosting.co.za> |
2015-06-22 08:56:58 | the | set | recipients:
+ the |
2015-06-22 08:56:58 | the | link | issue2371 messages |
2015-06-22 08:56:58 | the | create | |
|