How to set where PopupMenu appears on the screen?

Question

Android

You need to set the coordinates of the pop-up menu, because by itself, it appears somewhere in the back of the interface, namely in the upper left corner. In the vastness of the world, I found an example of such a solution:

import android.support.v7.widget.ListPopupWindow;
import android.support.v7.widget.ListPopupWindow.ForwardingListener;
import android.support.v7.widget.PopupMenu;
import android.support.v7.widget.PopupMenu.OnMenuItemClickListener;

........

final Button menuButton = (Button) vView.findViewById(R.id.OptionsButton);
    final PopupMenu popupMenu = new PopupMenu(mainAct, vView, Gravity.CENTER);
    popupMenu.inflate(R.menu.list_manager_context_menu);
    popupMenu.setOnMenuItemClickListener(new OnMenuItemClickListener(){
                menuButton.setOnClickListener( new OnClickListener() {
      @Override
      public void onClick(View view) {
          
                 ListPopupWindow.ForwardingListener  listener =(ForwardingListener)               
                 popupMenu.getDragToOpenListener();
    		         listener.getPopup().setVerticalOffset(-menuButton.getHeight());

             popupMenu.show();
      }
    });
    return vView;
  }

But my listener.getPopup() returns null.

Answer 1

It turned out that everything is quite simple - just pass the button to which I want to bind the menu to the PopupMenu constructor:

final Button menuButton = (Button) vView.findViewById(R.id.OptionsButton);
final PopupMenu popupMenu = new PopupMenu(mainAct, menuButton , Gravity.CENTER);

Answer 2

The alignment is set in the PopupMenu constructor. You need to look at the markup, or rather the binding of elements in XML