如何在 Scala 中实现扩展 Comparable 且没有特定类型的 Java 接口？

Question

I am using a Java library in my Scala project which has an interface that looks like this:

public interface A extends Comparable {
    String getSomething();
}

Note how the library implementation does not specify any specific type for Java's Comparable<T> interface.

In my Scala code I want to create an instance of A like so:

object AFactory {
  def getA: A = new A {
    override def getSomething: String = "test"
    override def compareTo(o: ???): Int = -1
  }
}

Unfortunately, I am not sure what type to specify for ??? . Does anyone know how can I implement this interface and get it to compile?

Answer 1

There is no direct way to do it. One way to make you code run against pre-java5 code, is to have Java bridge methods:

public abstract class B implements A {
    @Override
    public int compareTo(Object o) {
        return 0;
    }
}

object AFactory {
  def getA: A = new B {
    override def getSomething: String = "test"
  }
}

If you not want to have the compareTo implemented in Java. You could may be introduce a bridge method and implement the bridge method in scala:

public abstract class B implements A {
    @Override
    public int compareTo(Object o) {
        return compareToBridge(o);
    }

    protected abstract int compareToBridge(Object o);
}

object AFactory {
  def getA: A = new B {
    override def getSomething: String = "test"

    override def compareToBridge(o: Any): Int = ???
  }
}

Without using bridge methods, I dont think there is a nicer way of solving this problem.