User control with richTextBox, bindable richTextBox - wpf

Question

I try make user control with richTextBox because I need bindable richTextbox.
I found some solution here: Richtextbox wpf binding.
I would like to use solution of Arcturus. Create user control with richTextBox control and use dependency property.
In XAML I have only richTextBox control:
<UserControl x:Class="WpfApplication2.BindableRichTextBoxControl"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
mc:Ignorable="d"
d:DesignHeight="300" d:DesignWidth="300">
<Grid>
<RichTextBox Name="RichTextBox" Grid.Row="0"/>
</Grid>
</UserControl>
In CodeBehind:
public partial class BindableRichTextBoxControl : UserControl
{
public static readonly DependencyProperty DocumentProperty =
DependencyProperty.Register("Document", typeof(FlowDocument), typeof(BindableRichTextBoxControl),
new PropertyMetadata(OnDocumentChanged));
public FlowDocument Document
{
get { return (FlowDocument)GetValue(DocumentProperty); }
set { SetValue(DocumentProperty, value); }
}
private static void OnDocumentChanged(DependencyObject d, DependencyPropertyChangedEventArgs e)
{
var control = (BindableRichTextBoxControl)d;
if (e.NewValue == null)
control.RichTextBox.Document=new FlowDocument();
//?
control.RichTextBox.Document = document;
}
public BindableRichTextBoxControl()
{
InitializeComponent();
}
}
I am little confuse with last line in OnDocumentChanged method.
control.RichTextBox.Document = document;
I can’t identify what is varibale document.

Answer 1

I think he means this:
private static void OnDocumentChanged(DependencyObject d, DependencyPropertyChangedEventArgs e)
{
RichTextBoxControl control = (RichTextBoxControl) d;
if (e.NewValue == null)
control.RTB.Document = new FlowDocument(); //Document is not amused by null :)
else
control.RTB.Document = e.NewValue;
}
but I recommend you leave a comment on his original answer.