[pycrypto] Docs/code mis-match - not specifying IV fails

[Dave Pawson]

What does not make sense is that this group seems to accept this
discrepency between
docs and code.

Does anyone know if the docs maintainer has any interest these days?

regards

On 8 January 2014 09:32, Hugh Macdonald <hugh.macdonald at nvizible.com> wrote:
> Hi Charley,
>
> Thanks for clarifying that - that makes sense to me.
>
>
> Cheers
>
> Hugh
>
>
> On 8 January 2014 00:59, Charley Nibley <cnibley at gmail.com> wrote:
>>
>> Hi Hugh,
>>
>> This behavior you are seeing is expected.  It changed with a fix to
>> prevent an empty string IV passed to the function. IV is always mandatory
>> even if at some point in a previous build it was being passed as all zeros.
>> I believe the doc just needs updating.
>>
>> See line 173:
>> https://github.com/dlitz/pycrypto/commit/411f60f58cea79f7e93476ba0c069b80a2a4c1a0
>>
>> And see the comments here: https://bugs.launchpad.net/pycrypto/+bug/997464
>>
>> Charley Nibley
>>
>>
>> On Tue, Jan 7, 2014 at 9:43 AM, Hugh Macdonald
>> <hugh.macdonald at nvizible.com> wrote:
>>>
>>> Hi,
>>>
>>> I've just been trying to figure out why some code of mine broke after an
>>> update from 2.0.1 to 2.6.1, and found that I hadn't been passing an IV into
>>> AES.new() (using MODE_CFB).
>>>
>>> Previously, it was happy with this, and, according to the docs here
>>> (https://www.dlitz.net/software/pycrypto/api/current/), "It is optional and
>>> when not present it will be given a default value of all zeroes."
>>>
>>> Looking at the code that throws the exception
>>> (https://github.com/dlitz/pycrypto/blob/master/src/block_template.c, line
>>> 163), it doesn't allow for doing what the docs suggest it will.
>>>
>>> I'm not sure whether I've either misunderstood something or which of the
>>> two is wrong, but I thought I'd at least flag it for discussion...
>>>
>>>
>>> Hugh
>>>
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