[External] : Re: provides and requires static ... runtime error

[Rob Bygrave]

*> only y and y.inject must be in two separate JARs.*

Does this work without *any* module having a `requires y.inject`?  If so,
how is the y.inject module get included without a `requires y.inject` being
specified? (There is an obvious follow up question if that is the case as
it likely breaks the case where only y module is used).


Cheers, Rob.

On Tue, 25 Apr 2023 at 01:32, Ron Pressler <ron.pressler at oracle.com> wrote:

>
>
> On 24 Apr 2023, at 14:23, Josiah Noel <josiahnoel at gmail.com> wrote:
>
>
>
>> I’m confused. To use the implementation of the `x.Plugin` service
>> (defined in X) that you want Y to `provides` an implementation of, some
>> module *must* have a `uses x.Plugin`. The runtime has to know that that
>> service is needed by the program. Maybe the application `uses` it, but X
>> itself could also declare that it `uses` an implementation of a service
>> interface that it, itself, declares. So perhaps it is X that `uses
>> X.plugin`, but someone has to actually say they want to use the service.
>>
>
> I believe I'm picking up what you're putting down, but let's review it
> again to be sure.
>
>
> In artifact X we have:
>
>
>>     exports x.spi;
>>     uses x.spi.Plugin;
>>
>
> in artifact Y we have two modules: y and y.inject.
>
>     exports y.stuff;
>>
>
> And
>
>>     requires x;
>>     provides x.spi.Plugin with y.inject.PluginImpl;
>>
>
> With this in mind, we have two situations we want to support. we have an
> application Z that wants to use the Y artifact alone and a case where Z
> wants to use X and Y.
>
> With Y alone, the z module should look like this correct?
>
>     requires y;
>>
>
>  And in the case where Z wants both X and Y (with X doing the
> service-loading), we do:
>
>     requires x;
>>     requires y;
>>
>
> Am I right to understand that X will service load the y.inject plugin
> correctly?
>
>
> Yes.
> This is also how it works today, only y and y.inject must be in two
> separate JARs.
>
> — Ron
>
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