On Tue, 20 Jun 2006 16:14:08 -0700 Donnie Berkholz <[EMAIL PROTECTED]> wrote:
> Mike Owen wrote:
> > From this user's perspective, simple is better. qt3 and qt4 as use
> > flags are completely and utterly obvious as to what they mean, and
> > there is no confusion about them. Adding a plain qt flag in there
> > brings back the gtk/gtk2 mess that we've presumably been trying to
> > avoid in the future.
>
> That depends on how it's done. If it's done in a simple and obvious
> way (USE=qt means use the best available qt, USE=qt# for other weird
where "available" means "available in the tree for arch", not
"already installed on build system" (just to be clear - correct me if
I'm wrong)
> stuff that most people don't care about and so can ignore), it
> shouldn't be that bad.
So are you suggesting something like:
qt - Add support for QT/include QT GUI
qt3 - build for version 3 of QT
where dependencies might be something like:
qt? ( qt3? ( >=dev-libs/qt-3.3.6
<dev-libs/qt-4 )
!qt3? ( >= dev-libs/qt-4.1 ) )
for a package that can build against either qt3 (3.3.6 or higher) or
qt4 (4.1 or higher) but not both simultaneously, and perhaps:
qt? ( >= dev-libs/qt-4.1
qt3? ( >=dev-libs/qt-3.3.6
<dev-libs/qt-4 ) )
for packages that can build simultaneously for both (is this situation
realistic?). We'd need a qt4 to get just the qt3 package in that case:
qt? ( qt4? ( >= dev-libs/qt-4.1 )
qt3? ( >=dev-libs/qt-3.3.6
<dev-libs/qt-4 ) )
Am I making sense? This looks a lot like the gtk/gtk2 flags, but
inverted; according to use.desc, gtk builds gtk+-1 unless gtk2 is set,
whereas the above builds highest version compatible with the
package unless a lower version is specifically requested through USE.
Ideally we should be consistent in handling this issue (which
presumably isn't limited to just gtk and qt), although it may not be
worth the disruption to rework gtk/gtk2 into gtk/gtk1.
--
Kevin F. Quinn
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