> On Fri, May 21, 2010 at 00:09, Andreas Schwab <schwab@linux-m68k.org> wrote:
>> Antriksh Pany <antriksh.pany@gmail.com> writes:
>>
>>> Instead of (what I initially expected):
>>>
>>> A--------o--------o--------o--------o(old B)--------o--------o--------o(old C)
>>>
>>> A2--------o--------o--------o--------B--------o--------o--------C
>>>
>>>
>>> So what I am missing here? Aren't the new commits B~1, B~2, B~3
>>> identical to C~4, C~5, C~6 (respectively) in all ways so as to have
>>> gotten them the same SHA1 and hence appear as what I expected them to
>>> appear?
>>
>> No, they have a different commit time, which is also part of the hash.
>
> Indeed.
>
> To avoid this, you have to:
> - rebase B on top of A2 first,
>
> git rebase --onto A2 A B
>
> - rebase of C on top of the new B.
>
> git rebase --onto B B_old C
> ("git rebase --onto B A C" should work too, as usually git is
> smart enough to see
> that A-B_old is already applied. Use "git rebase --skip" if it isn't)
>
> If A is an ancestor of A2, you can simplify to:
>
> git rebase A2 B
> git rebase B C
>
> (Disclaimer: the examples without --onto I use almost daily, the ones
> with I don't)
>
> Gr{oetje,eeting}s,
>
> Geert
>
> --
> Geert Uytterhoeven -- There's lots of Linux beyond ia32 --
geert@linux-m68k.org
>
> In personal conversations with technical people, I call myself a hacker. But
> when I'm talking to journalists I just say "programmer" or something like that.
> -- Linus Torvalds
>
