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Re: [PROPOSED] Change package relations policy to remove references to non-free from main



Hi,
>>"Raul" == Raul Miller <moth@debian.org> writes:

 Raul> Raul Miller <moth@debian.org> writes:
 Raul> That's solvable:  create a virtual package which has a free instance
 Raul> (such as Mozilla) which provides the interface you're taking advantage of.

 Raul> Actually, this particular case has nothing to do with Enhances.

 Raul> Re-read the above text.  It looks to me as if you're saying it's right
 Raul> to say suggest netscape and not suggest a virtual package which both
 Raul> mozilla and netscape provide.

        No. But I am voicing my objection to a method that requires
 serendipitous free equivalantrs of all non-free packages to
 serve as a workaround. 

 Raul> And we still don't have a good example case where "free package suggests
 Raul> non-free package" is better than "non-free package enhances free package"

        Latex2html suggesting non-fee bits which can create gifs. ZIf
 it can create a gif, it can't be free. And a gif creator does not
 exist to ``enhance'' the multitudes of packages that can use it;
 those packages should reasonably suggest a  package that provides
 added functionality. 

        In any case, policy has a long shadow. It extends beyond the
 cases we bring up here, and even beyond cases that exist today.

        We all agree that the presence of Enhances does not fully
 obviate the need for suggests (They are related, overlapping, but
 different sets of relationships). We should ot force a suggests
 relationship into a ``enhances'' relationship.

        As I have said, the solution lies in the user interfaces. Do
 not show people who do not want it *any* non0free packages, and don't
 try to hack and coerce relationship declarations to kludge a solution.

        manoj
-- 
 Vail's Second Axiom: The amount of work to be done increases in
 proportion to the amount of work already completed.
Manoj Srivastava   <srivasta@debian.org>  <http://www.debian.org/%7Esrivasta/>
Key C7261095 fingerprint = CB D9 F4 12 68 07 E4 05  CC 2D 27 12 1D F5 E8 6E


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