Re: naive question

From: Petr Baudis <pasky@ucw.cz>
Date: Apr 19, 2005, 17:15 UTC
Message-ID: <20050419171534.GH12757@pasky.ji.cz>
In-Reply-To: <16997.222.917219.386956@cargo.ozlabs.ibm.com>

Dear diary, on Tue, Apr 19, 2005 at 03:00:14PM CEST, I got a letter where Paul Mackerras <paulus@samba.org> told me that...

Show 8 quoted lines

> Is there a way to check out a tree without changing the mtime of any
> files that you have already checked out and which are the same as the
> version you are checking out?  It seems that checkout-cache -a doesn't
> overwrite any existing files, and checkout-cache -f -a overwrites all
> files and gives them the current mtime.  This is a pain if you are
> using make and your tree is large (like, for instance, the linux
> kernel :), because it means that after a checkout-cache -f -a you get
> to recompile everything.

Actually, to then get sensible show-diff output, you need to also update-cache --refresh to compensate for the changes. I personally really hate update-cache --refresh; sure, 0.1s with hot cache, but easily eats 5 minutes (!) with cold cache.

I'd actually prefer, if:

(i) checkout-cache simply wouldn't touch files whose stat matches with what is in the cache; it updates the cache with the stat informations of touched files

(ii) read-tree would take over the stat information from the matching files in previous cache.

This way, doing update-cache --refresh would become a rather rare event. Stuff would become swifter, faster, less I/O bound and you would get rid of problems as the one described above.

What do you think?

-- 
				Petr "Pasky" Baudis
Stuff: http://pasky.or.cz/
C++: an octopus made by nailing extra legs onto a dog. -- Steve Taylor