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From: Aahz on 12 Feb 2010 00:18
In article <mailman.2077.1265524158.28905.python-list(a)python.org>,
Gabriel Genellina <gagsl-py2(a)yahoo.com.ar> wrote:
>
>Strange. With Python 2.6.4 I don't need to do that; I'd say the difference
>is in the OS or antivirus (some AV are known to break the TCP stack).

Perhaps, but I've also found that ctrl-C doesn't work on Windows.
--
Aahz (aahz(a)pythoncraft.com) <*> http://www.pythoncraft.com/

"At Resolver we've found it useful to short-circuit any doubt and just
refer to comments in code as 'lies'. :-)"
From: Frank Millman on 12 Feb 2010 01:59

"Aahz" <aahz(a)pythoncraft.com> wrote in message
news:hl2ob2$3ib$1(a)panix5.panix.com...
> In article <mailman.2077.1265524158.28905.python-list(a)python.org>,
> Gabriel Genellina <gagsl-py2(a)yahoo.com.ar> wrote:
>>
>>Strange. With Python 2.6.4 I don't need to do that; I'd say the difference
>>is in the OS or antivirus (some AV are known to break the TCP stack).
>
> Perhaps, but I've also found that ctrl-C doesn't work on Windows.

I don't know if this is exactly the same, but FWIW ...

I had the following, in a test program -

from wsgiref.simple_server import make_server

myserver = MyServer()
httpd = make_server('', 7789, myserver)
httpd.serve_forever()

try:
while True:
sleep(1)
except KeyboardInterrupt:
httpd.shutdown()

I could not get it to respond to Ctrl-C.

Then I read somewhere that it must run in a separate thread, so I changed it
like this -

- httpd.serve_forever()
+ threading.Thread(target=httpd.serve_forever).start()

Now it does respond to Ctrl-C.

HTH

Frank Millman


From: Aahz on 12 Feb 2010 09:22
In article <mailman.2422.1265961504.28905.python-list(a)python.org>,
Dennis Lee Bieber <wlfraed(a)ix.netcom.com> wrote:
>On 11 Feb 2010 21:18:26 -0800, aahz(a)pythoncraft.com (Aahz) declaimed the
>following in gmane.comp.python.general:
>> In article <mailman.2077.1265524158.28905.python-list(a)python.org>,
>> Gabriel Genellina <gagsl-py2(a)yahoo.com.ar> wrote:
>>>
>>>Strange. With Python 2.6.4 I don't need to do that; I'd say the difference
>>>is in the OS or antivirus (some AV are known to break the TCP stack).
>>
>> Perhaps, but I've also found that ctrl-C doesn't work on Windows.
>
> Unless the running program makes an I/O call to the console, I don't
>think <ctrl-c> gets past the device driver... <G>

That's probably it. It's more annoying for me because I run Windows with
a VM on a Mac, which doesn't have ctrl-break.
--
Aahz (aahz(a)pythoncraft.com) <*> http://www.pythoncraft.com/

"At Resolver we've found it useful to short-circuit any doubt and just
refer to comments in code as 'lies'. :-)"
From: Gabriel Genellina on 12 Feb 2010 20:36
En Fri, 12 Feb 2010 11:22:40 -0300, Aahz <aahz(a)pythoncraft.com> escribi�:

> In article <mailman.2422.1265961504.28905.python-list(a)python.org>,
> Dennis Lee Bieber <wlfraed(a)ix.netcom.com> wrote:
>> On 11 Feb 2010 21:18:26 -0800, aahz(a)pythoncraft.com (Aahz) declaimed the
>> following in gmane.comp.python.general:
>>> In article <mailman.2077.1265524158.28905.python-list(a)python.org>,
>>> Gabriel Genellina <gagsl-py2(a)yahoo.com.ar> wrote:
>>>>
>>>> Strange. With Python 2.6.4 I don't need to do that; I'd say the
>>>> difference
>>>> is in the OS or antivirus (some AV are known to break the TCP stack).
>>>
>>> Perhaps, but I've also found that ctrl-C doesn't work on Windows.
>>
>> Unless the running program makes an I/O call to the console, I don't
>> think <ctrl-c> gets past the device driver... <G>
>
> That's probably it. It's more annoying for me because I run Windows with
> a VM on a Mac, which doesn't have ctrl-break.

On a "real" PC with Windows XP SP3 and Python 2.6.4, with an idle server,
just hitting Ctrl-C was enough:

D:\temp>python -m SimpleXMLRPCServer
Running XML-RPC server on port 8000
Traceback (most recent call last):
File "d:\apps\python26\lib\runpy.py", line 122, in _run_module_as_main
"__main__", fname, loader, pkg_name)
File "d:\apps\python26\lib\runpy.py", line 34, in _run_code
exec code in run_globals
File "d:\apps\python26\lib\SimpleXMLRPCServer.py", line 615, in <module>
server.serve_forever()
File "d:\apps\python26\lib\SocketServer.py", line 224, in serve_forever
r, w, e = select.select([self], [], [], poll_interval)
KeyboardInterrupt

A slightly more realistic example: a busy, single threaded server. I had
to hit Ctrl-C twice: the first time, KeyboardInterrupt was sent as a Fault
response to the client.


import sys

def busy():
x = 50000
y = x**x
return "Ok"

if sys.argv[1]=='server':
import SimpleXMLRPCServer
server = SimpleXMLRPCServer.SimpleXMLRPCServer(("localhost",
8000),logRequests=False)
server.register_function(busy)
server.serve_forever()
else:
import xmlrpclib
proxy = xmlrpclib.ServerProxy("http://localhost:8000/")
while True:
print "client", proxy.busy()


Maybe the OP's code containes a bare 'except:' clause that swallows
KeyboardInterrupt, or the server is so busy that is always executing a
function handler (and all KeyboardInterrupt become Fault and are sent as a
function response).

Mmm, perhaps that's a bug, a KeyboardInterrupt should bubble up to the
server code, not being treated as an error in computing the function
result.

--
Gabriel Genellina

From: News123 on 13 Feb 2010 14:55
Hi Gabriel,




News123 wrote:
> Hi Gabriel,
>
> Gabriel Genellina wrote:
>> En Fri, 05 Feb 2010 20:03:51 -0300, News123 <news123(a)free.fr> escribi�:
>>
>>> I'm using an XMLRPC server under Windows.
>>>
>>> What I wonder is how I could create a server, that can be killed with
>>> CTRL-C
>>>
>>> The server aborts easily with CTRL-BREAK but not with CTRL-C (under
>>> Windows)
>>>
>>> If I press CTRL-C it will only abort when the next RPC call occurs.
>>> It seems it is blocking in the select() call in the handle_request()
>>> function.
>> Python 2.6 and up behaves exactly as you want.
>> On previous versions you may use this:
>>
>> class MyXMLRPCServer(SimpleXMLRPCServer.SimpleXMLRPCServer):
>>
>> ... your methods ...
>>
>> if not hasattr(SimpleXMLRPCServer.SimpleXMLRPCServer, 'shutdown'):
>>
>> # pre 2.6
>> quit = False
>>
>> def serve_forever(self):
>> while not self.quit:
>> self.handle_request()
>>
>> def shutdown(self):
>> self.quit = True
>>
>> def server_bind(self):
>> self.socket.settimeout(1.0)
>> SimpleXMLRPCServer.SimpleXMLRPCServer.server_bind(self)
>>
>
> Overloading server_bind() with your version solved my problem.
>

Overloading server_bind() makes the server killable with ctrl-C.

I noticed however, that I can't use this as solution.
If the bind function sets a time out, then it seems,

that there are regular small time windows, when connecting clients fail.
I don't have the error message any more, but will post it when I capture
the event next time.


bye

N

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