I just created a PR: https://github.com/pyca/cryptography/pull/1973. It'd
be great to have some feedback on the interface and general approach before
adding some tests and doc.

Thanks!
Dario

On Sat, May 16, 2015 at 1:23 AM, Paul Kehrer <paul.l.keh...@gmail.com>
wrote:

> Help is always appreciated; if you'd like to submit a PR we'll be happy to
> review it!
>
> -Paul
>
> On May 15, 2015 at 3:26:29 PM, Dario Sneidermanis (dari...@gmail.com)
> wrote:
>
> That'd be perfect. If you need a hand I can make a first draft over the
> weekend.
>
> Dario
>
>
> On Thu, May 14, 2015 at 2:45 PM, Alex Gaynor <alex.gay...@gmail.com>
> wrote:
>
>> Perhaps an EllipticCurvePrivateNumbers.from_private_number_and_curve()
>> would be appropriate to automatically compute the public key?
>>
>> Alex
>>
>> On Thu, May 14, 2015 at 12:11 PM, Paul Kehrer <paul.l.keh...@gmail.com>
>> wrote:
>>
>>>  Hi Dario,
>>>
>>> The creation of an EllipticCurvePrivateNumbers object does require you
>>> to do d*G yourself at this time. There is some precedent for us adding
>>> functions to compute other elements (see:
>>> https://cryptography.io/en/latest/hazmat/primitives/asymmetric/rsa/#handling-partial-rsa-private-keys)
>>> so a function to do this makes sense to me.
>>>
>>> -Paul
>>>
>>> On May 14, 2015 at 3:21:29 AM, Dario Sneidermanis (dari...@gmail.com)
>>> wrote:
>>>
>>>    Hi,
>>>
>>> I'm trying to compute an EllipticCurvePrivateKey from a private_value
>>> (d). It seems that the only way to achieve this is to create an
>>> EllipticCurvePrivateNumbers object, but the constructor takes both the
>>> private_value and the public_numbers.
>>>
>>> Why does it need the public_numbers, when it could compute them (knowing
>>> the curve), by doing d*G (G being the curve generator)?
>>>
>>> Is there any way to go from d to the ECPrivKey without doing d*G by
>>> myself?
>>>
>>> Thanks,
>>> Dario
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>>
>>
>>  --
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>> right to say it." -- Evelyn Beatrice Hall (summarizing Voltaire)
>> "The people's good is the highest law." -- Cicero
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