python2.2: signals and exceptions: interrupted system call

[Jakub Moscicki]

Hello,

A small problem: I get a signal during a system call (from xmlrpclib ->
httplib) and an exception "IOError: [Errno 4] Interrupted system call" is
raised (this is system dependant, on other machine it does not raise this
exception). I have my own signal handler so I want to simply ignore this
exception if it occures. But for a reason mysterious to me I cannot catch
this exception in the main's program try block.

Anybody knows what's wrong? Code listings below.

Thanks,

kuba

-------------------
python2.2 client.py
<ServerProxy for xyz.com:8001/RPC2>
Signal handler called with signal 14
I made 51 calls in 3 seconds
Traceback (most recent call last):
File "client.py" , line 31, in ?
server.echo('he llo')
File
"/tmp/DIANE/install/anaphe_top/specific/redhat73/gcc-3.2/PublicDomainPac kages/2.0.0/lib/python2.2/xmlrpclib.py",
line 821, in __call__
return self.__send(sel f.__name, args)
File
"/tmp/DIANE/install/anaphe_top/specific/redhat73/gcc-3.2/PublicDomainPac kages/2.0.0/lib/python2.2/xmlrpclib.py",
line 975, in __request
verbose=self.__ verbose
File
"/tmp/DIANE/install/anaphe_top/specific/redhat73/gcc-3.2/PublicDomainPac kages/2.0.0/lib/python2.2/xmlrpclib.py",
line 842, in request
errcode, errmsg, headers = h.getreply()
File
"/tmp/DIANE/install/anaphe_top/specific/redhat73/gcc-3.2/PublicDomainPac kages/2.0.0/lib/python2.2/httplib.py",
line 752, in getreply
response = self._conn.getr esponse()
File
"/tmp/DIANE/install/anaphe_top/specific/redhat73/gcc-3.2/PublicDomainPac kages/2.0.0/lib/python2.2/httplib.py",
line 595, in getresponse
response.begin( )
File
"/tmp/DIANE/install/anaphe_top/specific/redhat73/gcc-3.2/PublicDomainPac kages/2.0.0/lib/python2.2/httplib.py",
line 119, in begin
line = self.fp.readlin e()
IOError: [Errno 4] Interrupted system call

-------------------
If you want to reproduce the problem try this with python2.2:

--
client.py
--

import xmlrpclib

SERVER = "http://localhost:8001"
server = xmlrpclib.Serve rProxy(SERVER) # local server

timeout = 3

print server

import signal, os

global terminate

def handler(signum, frame):
global terminate
print 'Signal handler called with signal', signum
terminate = 1

signal.signal(s ignal.SIGALRM, handler)
signal.alarm(ti meout)

cnt = 0
terminate = 0
try:
try:
while not terminate:
server.echo('he llo')
cnt += 1
except xmlrpclib.Error , v:
print "ERROR", v
finally:
print "I made %d calls in %d seconds" % (cnt,timeout)
--
server.py
--

import SimpleXMLRPCSer ver

SERVER = 'localhost'
log = 1

server = SimpleXMLRPCSer ver.SimpleXMLRP CServer((SERVER , 8001), logRequests=log )

try:
server.register _function(lambd a x: x, 'echo')
server.serve_fo rever()
finally:
server.socket.c lose()

--
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mow mi KUBA call me KUBA appelle-moi KUBA
-------------------------------------------------------------

[Martin v. Löwis]

Jakub Moscicki wrote:

A small problem: I get a signal during a system call (from xmlrpclib ->
httplib) and an exception "IOError: [Errno 4] Interrupted system call" is
raised (this is system dependant, on other machine it does not raise this
exception). I have my own signal handler so I want to simply ignore this
exception if it occures. But for a reason mysterious to me I cannot catch
this exception in the main's program try block.

I cannot find anything mysterious here: The exception is IOError, but
you try to catch xmlrpclib.error , so you are not catching IOError, so
raising IOError aborts your program.

Regards,
Martin

[Jakub Moscicki]

> > A small problem: I get a signal during a system call (from xmlrpclib ->

httplib) and an exception "IOError: [Errno 4] Interrupted system call" is
raised (this is system dependant, on other machine it does not raise this
exception). I have my own signal handler so I want to simply ignore this
exception if it occures. But for a reason mysterious to me I cannot catch
this exception in the main's program try block.

I cannot find anything mysterious here: The exception is IOError, but
you try to catch xmlrpclib.error , so you are not catching IOError, so
raising IOError aborts your program.

Yes, you are right. Mea culpa. Too many hours against too many
lines of code ;)

kuba

--
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mow mi KUBA call me KUBA appelle-moi KUBA
-------------------------------------------------------------