Re: Proposal for moving /usr/doc
Hi,
>>"Francesco" == Francesco Potorti` <F.Potorti@cnuce.cnr.it> writes:
Francesco> /usr/doc is big, and I would like to move it to a
Francesco> different partition via a symlink. Unfortunately, some
Francesco> relative links inside it point to other files under the
Francesco> /usr tree, which is okay by Debian policy. I suggest that
Francesco> links links under /usr/doc that point outside of /usr/doc
Francesco> are required to be absolute.
This is a big can of worms. This has been discussed
before. Please read the archives for the rationale behind the
decision.
As a solution to your problem, If your partition is,
say. /disk1; create /disk1/doc and symlink that to /usr/doc and
things shall be fine. Alternately, create a new partition for the
docments, and mount it as /usr/doc.
The problem with declaring this directory and that drectory
have absolute symlinks is where does on draw the line? Does one
totally outlaw relative links? For example, /usr/lib is bigger than
/usr/doc on my machine.
______________________________________________________________________
Filesystem 1024-blocks Used Available Capacity Mounted on
/dev/hda8 1026067 606444 366611 62% /usr/lib
__> du -s /usr/doc
196303 /usr/doc
______________________________________________________________________
I think you can never be sure of not breaking _any_ symlinks
on _any_ system; there are too many different styles of operation out
there.
Throw in AFS and DFS, and who know what you have out there?
I think, though, you shall catch the majority of installations
if you were careful of the top level directories. Oh, I can see
people breaking up /var and /usr (heck, I do that on my home
machine); but unless you mount them at a different level below /, you
should be fine.
So where does one draw the line? I think just disallowing
symbolic links *between* top level directories is enough.
manoj
--
Let no guilty man escape. U.S. Grant
Manoj Srivastava <srivasta@acm.org> <http://www.datasync.com/%7Esrivasta/>
Key C7261095 fingerprint = CB D9 F4 12 68 07 E4 05 CC 2D 27 12 1D F5 E8 6E
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