Re: property exists (OnPaint followup)

> As a follow on to my painting question, I need to obtain the 'BackColor'
> of the hosting 'canvas' control.  For the VS controls, this is simple
> since there is a BackColor property.  For other controls that changed
> the name of the 'BackColor' property or that have the BackColor property
> another level deep in their object model, it isn't quite as simple.
>
> For example, the infragistics control I want to use as the canvas
> exposes the BackColor property inside an 'Appearance' property.
>
> So for the System.Windows.Forms.Panel object, I get the BackColor from
>
> Color myColor = _canvas.BackColor;
>
> For the Infragistics.WinGroupBox object, I get the BackColor from...
>
> Color myColor = _canvas.Appearance.BackColor;
>
> If the 'Appearance' property exists for the object assigned to _canvas,
> then I will use it.  So what I am asking is, how do I check to see if
> the object '_canvas' has a property called 'Appearance'?

You'll notice when the C# compiler gives an error.

In a programming language like C#, if you have a canvas variable of
type "Control", you can only invoke methods directly supported by the
Control class.

I.e.:
Control _canvas = ...;
Color myColor1 = _canvas.BackColor; // compiles
Color myColor2 = _canvas.Appearance.BackColor; // gives a compilation error

In order to access the Appearance property, the "_canvas" variable
must be of a type which supports that property. In C#, you need to
cast the object to such a type. If you are not sure the cast will
succeed, you can use the "is" or "as" operators to test it:

Control _canvas = ...;
...
Color myColor;
// test whether _canvas implements the InfragisticsControl type
InfragisticsControl _infragisticsControl = _canvas as InfragisticsControl;
if (_infragisticsControl != null)
{
  // yes, so use Appearance
  myColor = _infragisticsControl.Appearance.BackColor;
}
else
{
  // no, use Color
  myColor = _canvas.BackColor;
}

(Note that I don't know Infragistics' controls, so the solution I'm
proposing might not work exactly that way.)

Regards,
Fabian