Dear diary, on Tue, Apr 19, 2005 at 03:00:14PM CEST, I got a letter
where Paul Mackerras <[EMAIL PROTECTED]> told me that...
> 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
C++: an octopus made by nailing extra legs onto a dog. -- Steve Taylor
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