Re: [RFC] notes: avoid recommitting identical trees

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Jeff King venit, vidit, dixit 31.08.2010 20:29:
> On Tue, Aug 31, 2010 at 08:26:34PM +0200, Michael J Gruber wrote:
> 
>> Junio C Hamano venit, vidit, dixit 31.08.2010 18:44:
>>> Michael J Gruber <git@xxxxxxxxxxxxxxxxxxxx> 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

OK, now I understand Junio's answer...
...and I also see where lookup_tree() fills the sha1, which I failed to
see before.

I guess what happened was that I had the parse_tree without
parse_commit, which segfaulted, and then inserted parse_commit.

In any case, it seems Johan is going with his approach, so you can
forget about the RFC, be it with or without parse_tree.

Michael
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