From: Santi Béjar <santi@agolina.net>
Date: Thu, 16 Jul 2009 23:18:14 GMT
Subject: Re: [PATCHv2 2/2] pull: support rebased upstream + fetch + pull --rebase
Message-ID: <adf1fd3d0907161618o61ee4b58of25659f8c36420f7@mail.gmail.com>
In-Reply-To: <7vhbxc8inp.fsf@alter.siamese.dyndns.org>

2009/7/16 Junio C Hamano <gitster@pobox.com>:
> Johannes Schindelin <Johannes.Schindelin@gmx.de> writes:
>
>> How about
>>
>>       oldremoteref="$(git rev-list --boundary HEAD --not \
>>                       $(git rev-list -g $remoteref | sed 's/$/^@/') |
>>               sed -e '/^[^-]/d' -e q)"
>>
>> Explanation: the "git rev-list -g $remoteref" lists the previous commits
>> the remote ref pointed to, and the ^@ appended to them means all their
>> parents.  Now, the outer rev-list says to take everything in HEAD but
>> _not_ in those parents, showing the boundary commits.  The "sed" call
>> lists the first such boundary commit (which must, by construction, be one
>> of the commits shown by the first rev-list).
>
> Hmm, I am not sure about that "(which must..." part.  When you have
>
>          Y---X
>         /
>        B---o---o---o---H
>
> wouldn't "rev-list --boundary H --not X^@" give B, not X nor Y?
>

$git rev-list --boundary H --not X
and
$git rev-list --boundary H --not X^@

return the same output in this case:
o
o
o
-B

In this case the correct command is without ^@, because you want the
commits in the reflog as boundary commits.

In the simpler and usual case, without a rebased upstream:

    z---B---o---o---o---H

B=upstream@{0}

$git rev-list --boundary H --not B^@
o
o
o
B
-z

and:

$git rev-list --boundary H --not B
o
o
o
-B

Also in the rebased upstream case:

      Y---X
     /
    z---B---o---o---o---H


X=upstream@{0}
B=upstream@{1}

$git rev-list --boundary H --not X^@ B^@
o
o
o
B
-z

and:

$git rev-list --boundary H --not X B
o
o
o
-B

HTH,
Santi