CODEWITHSUNDEEP
javascala
Christian Schlichtherle
Christian Schlichtherle

Reputation: 3155

How to access a field of a super-class which is implemented in Java which also has an equal-named method?

I want to extend a class which is implemented in Java. Here is an example:

public interface JavaInterface {
    /** Computes foo. */
    public String foo();
}

public class ThisIsJava implements JavaInterface {
    protected String foo = "foo";

    /** Returns value of property foo. */
    public String foo() { return foo; }
}

But how do I disambiguate between the field and the method foo when I extend this class in Scala?

class ThisIsScala extends ThisIsJava {
  def fooBar = super.foo + "/bar" // doesn't compile
}

Upvotes: 3

Views: 724

Answers (1)

tibtof
tibtof

Reputation: 7957

You cannot access the foo field because it's private. You can only access it by super.foo().

Update: I think you have this problem because in Scala there is no distinction between fields and methods, so the only option option would be to have an adapter for ThisIsJava, and extend that adapter in Scala.

Upvotes: 7

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