Re: [PATCH 0/4] Test name-rev with small stack

[Jeff King <peff@peff.net>]

On Fri, Sep 08, 2017 at 02:33:35PM +0200, Michael J Gruber wrote:

> Looking at it more closely, the solution in cbc60b6720 ("git tag
> --contains: avoid stack overflow", 2014-04-24) seems to be a bit "ad
> hoc" to me:
> 
> First of all, there is more than "tag --contains" that may exceed the
> stack by recursion. So I would expect the solution to be more general,
> and not localised and specialised to builtin/tag.c

At the time, tag was the only one using this depth-first contains
algorithm. It's since been adapted to ref-filter.c, but of course the
stack handling went with it.

Most traversals have a date-sorted queue, so are effectively doing a
breadth-first iteration with no recursion.

> Second, this is a walk, so I'm wondering whether our revision walk
> machinery should be the place to add missing functionality (if any).
> That way, everything would benefit from possible or existing
> improvements there. For example, I think some of our "extra walkers"
> don't heed object replacements. (I need to test more.)

It's possible that name-rev could make better use of the existing
traversal machinery. It's often tough to do so, though, because that
machinery gives you a linearized ordering of the commits. Whereas
something like name-rev really cares about the order that it visits the
commits, because it's building up the names.

It's the same for this "tag --contains" traversal. It _used_ to be a
series of merge-base computations. But by doing a custom traversal, we
can cache incremental results through the graph and avoid walking over
the same bits multiple times. There actually is a way to do it with the
regular breadth-first traversal, but you have to store one bit per ref
you're checking for each commit.

I played around with that a bit a while ago, and it did seem to work. I
can dig up the patches if you're interested. But one downside is that
one bit per ref per commit adds up if you have a lot of refs. A large
number of those bitfields will be the same, so you could probably get by
with a copy-on-write scheme, but I never implemented that.

Of course somebody may have a more clever algorithm, too. I don't claim
the above is a proof. ;)

-Peff