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'public' function exposes its 'public/*package*/' parameter type SolarEdgeException

I have the following Java and Kotlin classes:

package nl.friesoft.solaredgenotifier;

class SolarEdgeException extends Exception {
    public SolarEdgeException(String s)  {
        super(s);
    }
}



package nl.friesoft.solaredgenotifier

class SiteStorage(context: Context): ISolarEdgeListener {
    override fun onError(site: Site?, exception: SolarEdgeException?) {
    }
}

And I get this compilation error:

'public' function exposes its 'public/package/' parameter type SolarEdgeException

Both classes are in the same package, so why doesn't this compile?

like image 269
Bart Friederichs Avatar asked Mar 09 '19 10:03

Bart Friederichs


1 Answers

Classes by default in Kotlin are public, so every member/function is exposed to the outer world. Contrarily, in Java the default visibility, that is omitting a visibility keyword such as public, protected or private, is package-private.

SolarEdgeException is package-private, while its user SiteStorage is entirely public.

That means a user of SiteStorage cannot catch or use your exception at all.


Unfortunately Kotlin doesn't have the concept of package-private, because packages aren't managed the same way as in Java.

The best you can have is internal, which means the definition is visibile to the entire module.
I'd say, in this case, make SolarEdgeException a public class, maintaining the constructor as package-private.

There is a forum thread on the package-private visibility for Kotlin, it's an interesting read.
kotlin-to-support-package-protected-visibility

like image 160
LppEdd Avatar answered Nov 10 '22 20:11

LppEdd