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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?

Bart Friederichs
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1 Answers1

14

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

LppEdd
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    How to fix this then? How can I make `SiteStorage` package private? – Bart Friederichs Mar 09 '19 at 10:49
  • I am not sure if it is still true; but historically the Java spec didn't mandate that package layout match directory layout, this was a convention. – Gavin Mar 09 '19 at 11:30
  • @Gavin you're right Gavin, I got too much used to that IDE inspection that my brain assimilated it as a compiler error. I will remove that piece of text. – LppEdd Mar 09 '19 at 11:34