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Re: (propagated) 'inputs' depends on 'outputs'?
From: |
Julien Lepiller |
Subject: |
Re: (propagated) 'inputs' depends on 'outputs'? |
Date: |
Tue, 16 Jun 2020 10:28:06 -0400 |
User-agent: |
K-9 Mail for Android |
Le 16 juin 2020 08:41:58 GMT-04:00, zimoun <zimon.toutoune@gmail.com> a écrit :
>Bonjour Julien,
>
>On Sat, 13 Jun 2020 at 07:38, Julien Lepiller <julien@lepiller.eu>
>wrote:
>
>> Exactly, no. You cannot separate inputs from outputs, because they
>are
>> part of the same derivation. When you build an output, you actually
>> build the complete derivation and there's no way to separate that in
>> "this part builds out" and "this part builds doc", etc.
>
>So it means that I need to build all the outputs even if I am
>interested
>in only one, right?
Absolutely, but you shouldn't thenk of it as building multiple outputs, because
you are really just building one thing: there is only one package object, one
derivation. It happens to create multiple directories in the store, but there's
only one thing you build.
>
>If I run "guix install foo:out --no-substitutes" then I potentially
>build any other "outputs"" of foo, e.g., "doc" i.e., potentially
>download a lot of TeX stuff, or in the case of Git, all the Subversion
>stuff. Right?
Yes, because building foo:out doesn't make sense. You build foo and guix is
nice enough to understand that's wgat you mean :)
>
>Even if at the end, only the references used by "foo:out" will be
>tracked and all the others potentially garbage collected. Right?
>
>
>> It would make sense to only propagate for some outputs: suppose at
>> runtime only foo:bin requires the propagation of bar. Since foo and
>> bar are already built, it should be possible to restrict the
>> propagation behaviour to that output. Foo:out would not bring in bar
>> anymore, reducing the closure size.
>
>Yes, it seems making sense to only propagated if the output needs it.
>Well, if it is not implemented yet maybe it is because it is not really
>necessary. :-)
That cannot be automated because usually we use propagation when there is no
direct reference. It wouldn't be useful otherwise.
>
>
>Thank you for explaining. It is clearer for me now.
>
>Cheers,
>simon