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Newsgroups: comp.lang.python
Date: Sun, 11 Mar 2012 03:10:19 -0700 (PDT)
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Subject: Re: How to re-implement the crypt.crypt function?
From: Cosmia Luna <cosmius@gmail.com>
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On Sunday, March 11, 2012 4:16:52 AM UTC+8, Christian Heimes wrote:
> Am 10.03.2012 20:33, schrieb Cosmia Luna:
> > I'm not searching for a full solution and only want to know how to use =
hashlib to create a equivalent string like=20
>=20
> If you chance your mind and choose to use a full solution, then I highly
> recommend passlib [1]. It has an implementation of SHA-512 crypt as
> indicated by the number 6 in the header of your string.
>=20
> By the way "$6$ds41p/9VMA.BHH0U" is *not* the salt. Just
> "ds41p/9VMA.BHH0U" is the salt, 6 is a hash identifier.
>=20
> Christian
>=20
> [1] http://packages.python.org/passlib/

PassLib works for me. What I want is passlib.context, I need it in consider=
ation of forward compatibility.

Thanks a lot.

But I still want to know how it is implemented, I read passlib's source but=
 I found he/she re-implemented hashlib, which I can't understand. Now I kno=
ws that the encoding is hash64 instead of base64, but I can't know that. Pa=
ssLib is too difficult for me. Anyone knows the accurate process?

Cosmia