2009/7/16 Junio C Hamano <gitster@xxxxxxxxx>:
> Johannes Schindelin <Johannes.Schindelin@xxxxxx> 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
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