Bugs item #1168427, was opened at 2005-03-22 16:38
Message generated for change (Comment added) made by holo9
You can respond by visiting:
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Category: Windows
Group: Python 2.3
Status: Open
Resolution: None
Priority: 5
Submitted By: holo9 (holo9)
Assigned to: Nobody/Anonymous (nobody)
Summary: Possible windows+python bug
Initial Comment:
This bug is produced on WindowsXP SP1 (OSVer :
5_1_2600) with Python2.3 installed.
Start Python and type (of course x.x.x.x should be
replaced with IP address):
import socket
s=socket.socket(socket.AF_INET,socket.SOCK_RAW,4)
s.sendto("",("x.x.x.x",0))
Press ENTER and your win box should crash immediately.
On my test after restart windows returned BCCode : d1.
By the way, IP protocol 0x04 is "IP over IP", and I
could send such datagrams month ago with Python
(although Microsoft has crippled some protocols).
Now, this is maybe specific to this configuration, or
it could be due to some driver (BCCode: d1 is specific
for drivers related problems). It needs further
testing on different configurations.
Note that the problem doesn't appears when string in
sendto() function is not empty.
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>Comment By: holo9 (holo9)
Date: 2005-03-30 23:15
Message:
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I tested this on Python 2.3.4. But there is no problem in
Python. This bug could be reproduced in C, for example. The
main problem is in Windows built-in firewall. For explanation
take a look at:
http://seclist.org/lists/bugtraq/2005/Mar/0409.html
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Comment By: Terry J. Reedy (tjreedy)
Date: 2005-03-30 21:37
Message:
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Do you get the same problem with either of the current releases
(2.3.5 and/or 2.4.1)?
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Comment By: logistix (logistix)
Date: 2005-03-24 03:18
Message:
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user_id=699438
Running your reproducable on XP SP2 (with real IP) returns
"socket.error: (10022, 'Invalid argument')" for me without a
hard crash.
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