Path: csiph.com!fu-berlin.de!uni-berlin.de!not-for-mail From: Christian Baier Newsgroups: de.comp.lang.python Subject: [Python-de] Crypto mit Python2-Hausmitteln?. Date: Sun, 30 Jun 2019 22:38:56 +0200 Lines: 78 Message-ID: References: Mime-Version: 1.0 Content-Type: text/plain; charset=utf-8; format=flowed Content-Transfer-Encoding: quoted-printable X-Trace: news.uni-berlin.de rKk85nUzWMvC/kGREuuEBg3y5j/MRuuQ0CdGjK144YYQ== Return-Path: X-Original-To: python-de@python.org Delivered-To: python-de@mail.python.org DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/simple; d=gmx.net; s=badeba3b8450; t=1561927137; bh=rYd5d7PtPOn8oXI723eK81o4HIDkTHW1OzRl9xnJk+I=; h=X-UI-Sender-Class:To:From:Subject:Date; b=diN9T60rpuXX8akiPGJzLAC1sLnEMlpf6YMc8HH8oUB9wc4fmhoyuuncu62BKcduj OzgkCLj9q3g7/6Jpoalvf5+44MsuZMRDFjfVb2lllotrFLuFJhzOZIok6OPpxA2HHY +nKhjsto+Uw5u28b8YExaNHCjM6882lFhuanXx2c= X-UI-Sender-Class: 01bb95c1-4bf8-414a-932a-4f6e2808ef9c User-Agent: Mozilla/5.0 (X11; Linux i686; rv:60.0) Gecko/20100101 Thunderbird/60.7.1 Content-Language: de-DE X-Provags-ID: V03:K1:5zdBID15Zxqt22eqchAhopgncC3S5Eriy6XiLJMI8z2+4JMu9yO qY8cQPsodmxpwTRSynzVif2syttnq3r91NzG3t9wxaTCXthsrmUc4hm4OLDx03gx8esNrk0 fAMxr4fM5nzNpqqhJ8qrR7rrx++T0IWLTJBz0as7BTQKe313PPLTPwz39Kk5Wb4TAD48WHl ZEu8Buj+CpAR4/5q7A2Tw== X-Spam-Flag: NO X-UI-Out-Filterresults: notjunk:1;V03:K0:FCFLBe2g7tI=:7x4PbmXlOT7//7thT0ISn6 qrpAzrQAm30smIfYLmV65XJeJESWqXvlNt8ERS+mzPveqNDOwbSO7SDOYhEWwlpbg/rqkTiCV ZA95gRiIZXscf8sCqTOdnKNha8LhZvjy8gkD4GRYlcnoP06w7cq+pdZvOmHeJzAGnW737t7qK jWyeNuWEDY77ZZtsTPEP6J5Z2fQCISWLSnzZ9clV7d+ySuoiVMbXYWp5kQNPFfu//SkKn2W2p YBZa4t3xWA6vi6XPqGZYly54/YJf+FkMB30cWOgSq/VJeJVVAxDRb57qC3Nrv6P2nB5tAWkFA Ob+9ZuYClfKJT31AEoVqX1XjSRlh11vpEbPgHDaSnyOuiRYUol3i2gEbl5YN6o1MRy8cLTqhD Nexx4VW0tVz7OlogcRKKhQ5Byw7RunAZbjiCeX3SRcHhm0QSS/0YpIAHr/q7EM1a+XEH60ql9 TR8Z/TKkkpu4ljN12WPVaDcuiNyDtqWGpoku9ynWR4SCZe2QQHckFw/FvIX6wQ+nl55wnVuV5 t0cULdfE/jUxfhYYX5Sf7pYLskkBhXSFkKBY/Cb9/vtVNOkYKKvxhSvNymeic/WvJ2T5xFmcG m4JL1PcpEPXj4X2IN084lpu32210ECvEwGS97tj9wGXYX2SAMvhQJroSWpnIEMqTHFeufkdXC N4jY08p5f5dd5WIayRWSq33uWQ/Xry5sQIl0c/4eP82Jhb5iANjLA9Tdbfb8A1Pbcv5cCrST2 WYLfRecQt7gyeY2T1pPwBHp1Ro58ECPqlkq1fZ6BNF0I3JEYPah4vky7sd/qZl9y+F+qFLwHC fbXQUMqlozZm61bqNHeo+cijhv7nDI3Hx7kZEX0tgKINoMyyOSWHr/GpGOrwm6ESrLbcnvYuw YBRaY8hbwnbEETiaXayO7ekNE0ow6YappsYW/Pko5BVSv5L7EnSz/n6b8T32BlBea5nIYmDOS iGnMDZ6HWaAsyFhrWBXxMRfIVkljCApVDOAtM6K0gVtqOC95btlFshYkKEmxcJDzoyIDyrrt0 Ceu3yHxX+Zx0CcM536ot7W6DLmZ6m7a7I3NOVp4EtLyqC0xpg4d1oGncJVEfzNecoUq7hH3BP /GXQvN4a1jD57A2Roemh4yeY/LeGSlPCtKy X-BeenThere: python-de@python.org X-Mailman-Version: 2.1.29 Precedence: list List-Id: Die Deutsche Python Mailingliste List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , X-Mailman-Original-Message-ID: Xref: csiph.com de.comp.lang.python:5521 Hi, gibt es hier Cryptoexperten? Ich habe da einen Rechner, auf dem ich einen Passwortmanager f=C3=BCr (vorgegebene!) Passw=C3=B6rter brauchen k=C3=B6nnte, aber keine fremde Sof= tware ausf=C3=BChren darf, au=C3=9Fer Python2.7-Scripte. Und ich frage mich gerade: Ist das wirklich schon alles was ich brauche, oder habe ich einen Denkfehler?: =2D--8<--- #!/usr/bin/python2 import hashlib, getpass #dictionary pl mit { "Dienstname" : "crypted Text" }: pl =3D { 'test': 'M\xa2X\xdeU-y(\\\xad\xc8YV\x1d\x18\x9d', 'Geheimnis': '>\xcdvA\t\x9c\x82\xb3\xd9\xd8eI\x1amh\xde', } def xor(data, service, masterpassword ): salt =3D ( "g\x9b#\xaeG\xffM\x1a\xa3\t&;\xdb\xbe\xcd6\x04X" + "\xa9z\x03\x86\xb7K\xff\xa1\xd7uR`\xa1\x16\xc5" + "\x91\x02\xa7y;f\x9b\x19%\x86\xea" ) initstr =3D service + salt + masterpassword # zum verlangsamen: * 1000000 h =3D hashlib.sha512( initstr * 1000000 ) key =3D h.digest() # Passwortlaenge verschleiern, mind. 16 data =3D data + " " * ( 16 - len( data ) ) while len( data ) > len( key ): h.update( initstr ) key +=3D h.digest() return "".join( [ chr( ord( chardata ) ^ ord( charkey ) ) for ( chardata, charkey ) in zip( data, key ) ] ) if __name__ =3D=3D '__main__': masterpassword =3D "" while len( masterpassword ) < 8: masterpassword =3D getpass.getpass( """Geben Sie das Master-Passwort ein: """ ) print( "\nGespeicherte Dienste:\n" ) for service in sorted( pl ): print( service ) checked =3D False while True: service =3D raw_input("\nDienst:") cryptedpassword =3D pl.get( service, None ) if cryptedpassword: print( xor( cryptedpassword, service, masterpassword ) ) else: print( """\nDienst noch unbekannt. Bitte 2 x das zu chiffrierende Passwort eingeben.""" ) newpassword =3D "********" passw_check =3D "--------" while passw_check !=3D newpassword: newpassword =3D passw_check passw_check =3D getpass.getpass( "Neues Passwort:" ) if checked or masterpassword =3D=3D getpass.getpass( """Geben Sie noch mal das Master-Passwort ein: """ ): checked =3D True print( """\nDictionary pl im Quelltext erweitern mit:\n %s: %s,\n""" % ( repr( service ), repr( xor( newpassword, service, masterpassword ) ) ) ) =2D-->8--- Habe hier ein schwaches Masterpasswort verwendet, ...falls jemand mein Geheimnis knacken m=C3=B6chte ;-) Freue mich auf R=C3=BCckmeldungen! Gru=C3=9F Christian