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我在页面上显示了一个弹出窗口，我试图获取它的（x，y）坐标，即使弹出窗口在屏幕上可见，我也得到了异常。

  private void showPopup_Click(object sender, RoutedEventArgs e)
{
      Popup p = new Popup();
      StackPanel panel1 = new StackPanel();
    // Create some content to show in the popup. Typically you would 
    // create a user control.
    Border border = new Border();
    border.BorderBrush = new SolidColorBrush(Colors.Black);
    border.BorderThickness = new Thickness(5.0);


    panel1.Height = 200;
    panel1.Width = 100;
    panel1.Background = new SolidColorBrush(Colors.LightGray);
    p.Height = 200;
    p.Width = 100;

    Button button1 = new Button();
    button1.Content = "Close";
    button1.Margin = new Thickness(5.0);
    button1.Click += new RoutedEventHandler(button1_Click);
    TextBlock textblock1 = new TextBlock();
    textblock1.Text = "The popup control";
    textblock1.Margin = new Thickness(5.0);
    panel1.Children.Add(textblock1);
    panel1.Children.Add(button1);
    border.Child = panel1;

    // Set the Child property of Popup to the border 
    // which contains a stackpanel, textblock and button.
    p.Child = border;

    // Set where the popup will show up on the screen.
    p.VerticalOffset = 100;
    p.HorizontalOffset = 200;

    // Open the popup.
    p.IsOpen = true;

}

</弹出>-->

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