[Buildroot] [PATCH] pkg-infra: allow dumping reverse dependencies of a package

[Arnout Vandecappelle]

On 28-09-16 00:10, Yann E. MORIN wrote:
> Arnout, All,
> 
> On 2016-09-28 00:02 +0200, Arnout Vandecappelle spake thusly:
>> > On 27-09-16 23:32, Yann E. MORIN wrote:
>>> > > On 2016-09-27 22:40 +0200, Arnout Vandecappelle spake thusly:
>>>>> > >> > 
>>>>> > >> > 
>>>>> > >> > On 25-09-16 14:56, Yann E. MORIN wrote:
>>>>>>> > >>> > > Finding the packages that select another one in a specific configuration
>>>>>>> > >>> > > is not very trivial:
>>>>>>> > >>> > > 
>>>>>>> > >>> > >   - when optional, the dependency is not expressed in Kconfig
>>>>>>> > >>> > > 
>>>>>>> > >>> > >   - looking at the .mk files is not very nice.
>>>>> > >> > 
>>>>> > >> >  If I understand correctly, the idea is to answer the question "why is package
>>>>> > >> > foo built?", right?
>>> > > Right. In my use-case, that was "what do I need to disable so I can
>>> > > disable that broken package?".
>>> > > 
>>>>> > >> >  It would make sense to also add a foo-graph-rdepends, because often you'll want
>>>>> > >> > the transitive dependencies as well and it's always nice to have it in a graph.
>>> > > Yup, planned. But not trivial...
>> > 
>> >  No? Isn't it just calling foo-show-rdepends from graph-depends?
> It would seem it would be that easy, but not really.
> 
> I don;t recall the details, but one thing is that we can't easily do an
> rdepends graph for 'all' in the current state: we'd need to be able to
> extract all packages that have no dependencies (i.e. leaf packages) so
> that we could construct the 'all' node for rdepends.

 graph-rdepends would be exactly the same as graph-depends. Only
foo-graph-rdepends is relevant.


 Regards,
 Arnout


> 
> Hmm.. Or do we have to? We could construct that list... Lemme think...
> 
> Of course, we could limit the feature to just graphing the rdeps of a
> single package. That would not be overly complex, as you said.
> 
> But I was trying to mimick graph-depends into graph-rdepends, so they
> would be symetrical...
> 
> Well, anyway, it's not at the top of my TODO list... ;-)

-- 
Arnout Vandecappelle                          arnout at mind be
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