Re: How to re-implement the crypt.crypt function?

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From	Christian Heimes <lists@cheimes.de>
Subject	Re: How to re-implement the crypt.crypt function?
Date	Sat, 10 Mar 2012 21:16:52 +0100
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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 

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.

By the way "$6$ds41p/9VMA.BHH0U" is *not* the salt. Just
"ds41p/9VMA.BHH0U" is the salt, 6 is a hash identifier.

Christian

[1] http://packages.python.org/passlib/

Thread

How to re-implement the crypt.crypt function? Cosmia Luna <cosmius@gmail.com> - 2012-03-10 11:33 -0800
  Re: How to re-implement the crypt.crypt function? Roy Smith <roy@panix.com> - 2012-03-10 15:15 -0500
    Re: How to re-implement the crypt.crypt function? Christian Heimes <lists@cheimes.de> - 2012-03-10 21:36 +0100
      Re: How to re-implement the crypt.crypt function? Roy Smith <roy@panix.com> - 2012-03-10 15:41 -0500
        Re: How to re-implement the crypt.crypt function? Christian Heimes <lists@cheimes.de> - 2012-03-10 22:07 +0100
  Re: How to re-implement the crypt.crypt function? Christian Heimes <lists@cheimes.de> - 2012-03-10 21:16 +0100
    Re: How to re-implement the crypt.crypt function? Cosmia Luna <cosmius@gmail.com> - 2012-03-11 03:10 -0700
    Re: How to re-implement the crypt.crypt function? Cosmia Luna <cosmius@gmail.com> - 2012-03-11 03:10 -0700

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