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From:   Junio C Hamano <gitster@pobox.com>
To:     Jeff King <peff@peff.net>
Cc:     Vipul Kumar <kumar+git@onenetbeyond.org>, git@vger.kernel.org
Subject: Re: List all commits of a specified file in oldest to newest order
References: <c3932b3c-323a-39d6-26a7-ba0c3d17378b@onenetbeyond.org>
        <YYTorS1DiuTXv0/V@coredump.intra.peff.net>
Date:   Fri, 05 Nov 2021 11:49:03 -0700
In-Reply-To: <YYTorS1DiuTXv0/V@coredump.intra.peff.net> (Jeff King's message
        of "Fri, 5 Nov 2021 04:17:49 -0400")
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Jeff King <peff@peff.net> writes:

> As a workaround, you can get what you want by two separate traversals:
> one to collect the commits via --follow, and then another to actually
> show them (but without doing any further walking). Like:
>
>   git log --follow --format=%H -- $your_file |
>   git log --stdin --no-walk --reverse [--oneline, -p, etc]

We learn new things every day.

Knowing the implementation, it is sort of obvious (we push the
objects into the pending list, populate the revs.commits in
prepare_revision_walk() from the pending list in order,
get_revision() first reverses the revs.commits and then gives out
the elements), but I didn't know the combination of "--no-walk" and
"--reverse" did something sensible ;-)

Thanks.