Re: Please undo "Use git-merge instead of git-resolve in

[Petr Baudis <pasky@suse.cz>]

Dear diary, on Fri, Sep 23, 2005 at 11:34:16AM CEST, I got a letter
where Junio C Hamano <junkio@cox.net> told me that...
> Petr Baudis <pasky@suse.cz> writes:
> 
> > 	* cg-fetch can do with symlinked object database
> 
> Do you mean ".git/object" in the repository you are fetching
> into is a symlink to somewhere, or something else?

Yes, exactly that. You get it by doing cg-clone -l /local/path.  O(1)
cloning, but it has its downsides and dangers which I tried to describe
in the cg-clone documentation.

> > 	* cg-fetch won't do useless fetches when we are already
> > 	  up-to-date
> 
> Care to explain?  Perhaps you are talking about rsync transport
> (in which case I would not be surprised)?

Just that if the reference pointer did not change, you don't try to pull
any objects. Speeds things up somewhat, and now it should always work
(well, cg-fetch has -f which makes it always try objects as well, in
case something went wrong anyway).

> > 	* I'm not sure if I could call git-fetch to do the initial
> > 	  fetch during clone
> 
>     $ git-init-db && git fetch http://kernel.org/pub/scm/git/git.git
> 
> should work.

Good.

> There is another minor "con", if you include git-clone in the
> discussion.  Over http transport, it refuses to run against a
> remote repository that does not have info/refs and info/packs,
> even if it is not packed.  I think cg-clone uses recursive wget
> for directory traversal and does not have to use them.

cg-clone just basically does cg-init && cg-fetch, and I'd really want to
keep it on that. So effectively, cg-clone over http just runs
git-http-fetch and trusts it to do the right thing (reasonably fast).

Recursive wget is used only for fetching tags (and it actually seems to
work nicely, when directory listing is enabled, obviously). Yes, I
should teach cg-fetch about info/refs too.

-- 
				Petr "Pasky" Baudis
Stuff: http://pasky.or.cz/
VI has two modes: the one in which it beeps and the one in which
it doesn't.