Re: [RFC] notes: avoid recommitting identical trees

From: Jeff King <peff@peff.net>
Date: Aug 31, 2010, 18:29 UTC
Message-ID: <20100831182931.GA21489@sigill.intra.peff.net>
In-Reply-To: <4C7D495A.1080806@drmicha.warpmail.net>

On Tue, Aug 31, 2010 at 08:26:34PM +0200, Michael J Gruber wrote:

Show 18 quoted lines

> Junio C Hamano venit, vidit, dixit 31.08.2010 18:44:
> > Michael J Gruber <git@drmicha.warpmail.net> writes:
> > 
> >>>> +	if (!parent || parse_commit(parent->item) || parse_tree(parent->item->tree) ||
> >>>> +		hashcmp(parent->item->tree->object.sha1, tree_sha1)) {
> >>>
> >>> I didn't check, but I can imagine you can drop the parse_tree here. We
> >>> should know the object sha1 once the commit is parsed.
> >>
> >> parse_commit() does a lookup_tree() but I don't think that it parses the
> >> tree, i.e. I don't hink it fills in tree->object.sha1.
> > 
> > Huh?  parse_tree(tree) calls read_sha1_file(tree->object.sha1) to parse
> > the tree.  How can it do without filling tree->object.sha1?
> > 
> 
> Sure parse_tree() does that. That's why I call it. I never claimed it
> doesn't.

I think the claim is that it is already parsed. Look at parse_tree. The first non-declaration lines are:

  if (item->object.parsed)
          return 0;
  buffer = read_sha1_file(item->object.sha1, &type, &size);

So _somebody_ has already filled in item->object.sha1.

-Peff