Subject: Re: distinfo file changes
To: Alistair Crooks <agc@pkgsrc.org>
From: None <mcmahill@mtl.mit.edu>
List: current-users
Date: 04/17/2001 18:14:16
I assume that nuking those directories from the CVS server will make it
impossible to
a) get logs of the delete files
b) do something like checkout pkgsrc as tagged for 1.4 or 1.5.
-Dan
On Tue, 17 Apr 2001, Alistair Crooks wrote:
> On Tue, Apr 17, 2001 at 04:42:14PM +0200, Dr. Rene Hexel wrote:
> > Alistair Crooks wrote:
> >
> > > These changes mean that, in the majority of cases, there will be no
> > > files/ subdirectory in packages now. It also means that we have
> > > reduced the number of files in pkgsrc, which will mean a corresponding
> > > speedup in extraction from the pkgsrc.tar.gz file, and any cvs updates
> > > should also be quicker now.
> >
> > Thanks, Al, this is very much appreciated! As mjl pointed out
> > earlier, for cvs updates to actually gain some speed, somebody with root
> > permissions on cvs.netbsd.org needs to remove the (empty) 'files'
> > subdirectories in the repository. At the moment, a 'cvs update -Pd'
> > creates all the 'files' directories, then detects they all are empty and
> > deletes them again.
> >
> > Cheers
> > ,
> > Rene
>
> cvs will make and then prune the directories, yes, assuming cvs -P is
> specified to update or checkout. However, it also does not have to
> calculate the timestamps and enter them in the CVS/Entries file in the
> files/ subdirectories, so there will be a speed up there.
>
> The extraction from the pkgsrc.tar.gz file should also be quicker,
> since there are now fewer files, and fewer files/ subdirectories.
>
> I was under the understanding that, without the files/ dirs in place
> on the cvs server, anyone who has those directories checked out will
> run into problems. That's why we let cvs prune directories, rather
> then removing them from the server. But I'm obviously past it, so what
> do I know?
>
> Regards,
> Alistair
>
> PS. Tomasz reminded me to say that pkglint is being updated, even as I
> write.
>