[ python-Bugs-1326841 ] SIGALRM alarm signal kills interpreter

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Bugs item #1326841, was opened at 2005-10-14 16:48
Message generated for change (Comment added) made by loewis
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Category: Demos and Tools
Group: None
>Status: Closed
>Resolution: Invalid
Priority: 5
Submitted By: paul rubin (phr)
Assigned to: Nobody/Anonymous (nobody)
Summary: SIGALRM alarm signal kills interpreter

Initial Comment:
This may be similar to #210641.  Example (Python 2.4.1,
Fedora Core 4 GNU/Linux):

sh-3.00$ python
Python 2.4.1 (#1, May 16 2005, 15:19:29)
[GCC 4.0.0 20050512 (Red Hat 4.0.0-5)] on linux2
Type "help", "copyright", "credits" or "license" for
more information.
>>> import signal
>>> signal.alarm(1)      # 1 second passes...
0
>>> Alarm clock
sh-3.00$   # python has exited

Doing the same thing in IDLE results in the subprocess
restarting.

IMO the correct behavior would be to raise an exception
that the outer shell would catch.


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>Comment By: Martin v. Löwis (loewis)
Date: 2005-10-15 17:53

Message:
Logged In: YES 
user_id=21627

This is not a bug. It is documented that way: the default
handler is SIG_DFL, which in turn does the system default
for the signal. For Sigalarm, it is to terminate the process.

If you want an exception raised, you need to install a
signal handler:

>>> def doalarm(signum,frame):
...   raise "alarm"
...
>>> signal.signal(signal.SIGALRM, doalarm)
<function doalalarm at 0xb7d897d4>
>>> signal.alarm(3)
0
>>>
Traceback (most recent call last):
  File "<stdin>", line 0, in ?
  File "<stdin>", line 2, in doalarm
alarm


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