> 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 _______________________________________________________________________ http://www.epi.sk Ekonomicke a pravne informacie - denne aktualizovane. Zakony, sutaze, udaje o firmach, odborne clanky... Najdete na stranke http://www.epi.sk/info1.htm