Romain d'Alverny a écrit :
On Tue, Jul 12, 2011 at 23:08, Balcaen John<mik...@mageia.org> wrote:
On Tuesday 12 July 2011 16:48:58 andre999 wrote:
[...]
For all these reasons, I think that it is much more appropriate to wait to
be approached by the patent holder.
(If not ourselves, then some other distro.)
I hope you're not serious when writing thoses lines.
Actually, this is what seems to be a standard process in the industry
when you read the headlines.
One doesn't hear about (software) patents when licensing occurs "as
expected", but when someone in the industry challenges (or is
challenged by) some other actor patent. Because that's the only way to
prove a patent to be ineffective (for various reasons) before it
expires.
(I am not proposing one or the other option - and it's going out of
the initial question in this thread - tmb reframed it well).
Romain
Ok.
Part of my last post seems to have been overlooked.
It proposes a solution to the problem presented in this thread ... where to put
packages which qualify for both non-free and "tainted".
Note that all packages in "tainted" contain ".tainted." in the name.
rsync, used to update the mirrors, permits adding the option
--exclude '.tainted.'
to permit excluding such packages if a mirror wants to.
So packages now in the "tainted" repos would be put in core or non-free as
appropriate.
And mirrors in affected countries, who choose to opt out of packages now in
"tainted" would simply add the above rsync option.
Most mirrors, which now carry "tainted", would drop the "tainted" repos. (If
they don't, these repos would simply become empty, so no harm done.)
This would have the effect of making packages now in "tainted" more readily
available, since it wouldn't be necessary to add these repos.
Which should please the packagers that make these packages.
We would have to make a few adjustments to show (or not) the packages tagged as
"tainted", but that shouldn't be difficult.
So we would eliminate 10 respositories, while keeping the same functionality,
as well as solving the non-free vs. tainted problem.
Wouldn't that work ?
--
André
