How to re-implement the crypt.crypt function?

[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

crypt.crypt('123456', '$6$ds41p/9VMA.BHH0U') returns the string below.

'$6$ds41p/9VMA.BHH0U$yv25s7jLxTRKLDNjIvT0Qc2jbcqdFRi5.PftO3cveTvjK49JhwCarIowOfrrNPD/PpYT3n6oNDIbjAONh8RXt1'

I tried:

from hashlib import sha512
from base64 import b64encode, b64decode

salt='ds41p/9VMA.BHH0U'
pwd='123456'

b64encode( sha512(pwd+salt).digest(), altchars='./' )
b64encode( sha512(salt+pwd).digest(), altchars='./' )
b64encode( sha512( pwd + b64decode(salt, altchars='./') ).digest(), altchars='./')
b64encode( sha512( b64decode(salt, altchars='./') + pwd ).digest(), altchars='./')

of course none of the four returns the value I want, 'yv25s7jLxTRKLDNjIvT0Qc2jbcqdFRi5.PftO3cveTvjK49JhwCarIowOfrrNPD/PpYT3n6oNDIbjAONh8RXt1', how can I get the value? I can't use crypt.crypt because of the consideration of cross-platform.

Thanks,
Cosmia

 

[Roy Smith]

I'm not searching for a full solution and only want to know how to use
hashlib to create a equivalent string like

crypt.crypt('123456', '$6$ds41p/9VMA.BHH0U') returns the string below.

'$6$ds41p/9VMA.BHH0U$yv25s7jLxTRKLDNjIvT0Qc2jbcqdFRi5.PftO3cveTvjK49JhwCarIowO
frrNPD/PpYT3n6oNDIbjAONh8RXt1'
[...]
I can't use crypt.crypt because of the
consideration of cross-platform.

Just out of curiosity, why do you want to do this? The python crypt
module uses the crypt library supplied by the operating system (which is
why it only works on unix). The algorithm implemented is a modification
of DES, i.e. a salt string is used to change some of the tables used in
the DES computation. It goes back to the ancient days of unix.

By today's standards, the algorithm isn't considered very strong. The
only place I'm aware that uses it is unix password files, and even there
many (most?) systems have replaced it with something stronger such as
SHA1. Maybe Apache .htaccess files?

I don't know what your use case is, but unless you're doing something
silly like trying to execute a dictionary attack against a unix password
file, it's almost certain that you'd do better to just use SHA1.

 

[Roy Smith]

Am 10.03.2012 21:15, schrieb Roy Smith:

The algorithm with identifier 6 is a SHA-512 crypt algorithm with a
lengthy salt (IIRC up to 1024 bits) and 40,000 rounds of SHA-512. It's
the default algorithm on modern Linux machines and believed to be very
secure.

The large salt makes a rainbow table attack impossible and the 40,000
rounds require a lot of CPU time, even on modern systems.

But is that what crypt.crypt() does? I though it implemented the
old-style triple-DES.

 

[Cosmia Luna]

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/

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

Thanks a lot.

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

Cosmia

 

[Cosmia Luna]

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/

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

Thanks a lot.

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

Cosmia