Q: Declaratively Implement a Java Interface, How?

Question

Hi Again,
I want to be able to Declaratively Implement 'Response' in a JavaFX (very impressive stuff if I can do this!). Follow the example code below :)
Java Interface:
public interface Response {
   public String doResponse(String input);
}The JavaFX Class:
public class MyResponse extends Response {
  override public-init var doResponse: function (input:String):String; //doesn't work!
}JavaFX Declarative Implementation (the goal):
var responder:MyResponse {
    doResponse:  function (input:String):String {
        return "JavaFX Rocks!"
}So the above should allow whoever is creating "MyResponse" to implement the doResponse functionality of the class.
Alternatively, I can just have to do the following.....
public class MyResponse extends Response {
  public-init var doResponseImpl:function (input:String):String;
  public override function doResponse(input:String):String{
      return doResponseImpl(input);
}Either way works... but the alternative is not so nearly as clean :)
Thanks in advance...

Answer 1

I don't see a way to do this. You want to transform a function to a variable. Even if that's a variable holding a function, these are incompatible.
But you can just override the function in the class instantiation:
var responder = Response {
    override function doResponse(input: String): String {
        return "JavaFX Rocks!"
}