在Android中使用接口作为类型的java

Question

这是oracle在“将接口用作类型”中的示例：

public Object findLargest(Object object1, Object object2) {
   Relatable obj1 = (Relatable)object1;
   Relatable obj2 = (Relatable)object2;
   if ((obj1).isLargerThan(obj2) > 0)
      return object1;
   else 
      return object2;
}

所以一切都有意义，干净漂亮！然后我看到了创建片段的代码，但看不到该方法的逻辑： public void updateDetail() 侦听器正在使用一个空的方法。有人能帮我清理一下吗

public class MyListFragment extends Fragment {

  private OnItemSelectedListener listener;

  @Override
  public View onCreateView(LayoutInflater inflater, ViewGroup container,
      Bundle savedInstanceState) {
    View view = inflater.inflate(R.layout.fragment_rsslist_overview,
        container, false);
    Button button = (Button) view.findViewById(R.id.button1);
    button.setOnClickListener(new View.OnClickListener() {
      @Override
      public void onClick(View v) {
        updateDetail();
      }
    });
    return view;
  }

  public interface OnItemSelectedListener {
    public void onRssItemSelected(String link);
  }

  @Override
  public void onAttach(Activity activity) {
    super.onAttach(activity);
    if (activity instanceof OnItemSelectedListener) {
      listener = (OnItemSelectedListener) activity;
    } else {
      throw new ClassCastException(activity.toString()
          + " must implemenet MyListFragment.OnItemSelectedListener");
    }
  }

  @Override
  public void onDetach() {
    super.onDetach();
    listener = null;
  }


  public void updateDetail() {
    String newTime = String.valueOf(System.currentTimeMillis());

    listener.onRssItemSelected(newTime);
  }
}