New submission from David Schere <dsch...@arinc.com>: When doing an exit() within a signal handler for an alarm I am seeing something strange:
This code works, it exits within signal handler as expected. You never see the print statement executed. import signal import time import sys import traceback def handler( *args ): sys.exit(-1) #<-- terminate program signal.signal( signal.SIGALRM, handler ) # set alarm to go off in one second that calls handler() signal.alarm( 1 ) time.sleep( 3 ) print 'You should never see this message' This code results in sys.exit() begin ignored and the code inside the “except:” block being executed! import signal import time import sys import traceback def handler( *args ): sys.exit(-1) #<-- terminate program signal.signal( signal.SIGALRM, handler ) # set alarm to go off in one second that calls handler() signal.alarm( 1 ) try: time.sleep( 3 ) except: print 'This message should not be seen' sys.exit() print 'You should never see this message' The work around is to raise an exception inside the handler() function. ---------- messages: 100120 nosy: david_schere severity: normal status: open title: sys.exit() doesn't execute inside a signal handler while blocked inside an try/except type: behavior versions: Python 2.5 _______________________________________ Python tracker <rep...@bugs.python.org> <http://bugs.python.org/issue8021> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: http://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
