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You may also check for an implemented interface using the is keyword (also first discussed in 6). If the object in question is not compatible with the specified interface, you are returned the value false. On the other hand, if the type is compatible with the interface in question, you can safely call the members without needing to use try/catch logic. To illustrate, assume we have an array of Shape types containing some members that implement IPointy. Notice how we are able to determine which item in the array supports this interface using the is keyword, as shown in this retrofitted Main() method: static void Main(string[] args) { Console.WriteLine("***** Fun with Interfaces *****\n"); // Make an array of Shapes. Shape[] myShapes = { new Hexagon(), new Circle(), new Triangle("Joe"), new Circle("JoJo")} ;

for(int i = 0; i < myShapes.Length; i++) { // Recall the Shape base class defines an abstract Draw() // member, so all shapes know how to draw themselves. myShapes[i].Draw(); // Who's pointy if(myShapes[i] is IPointy) Console.WriteLine("-> Points: {0}", ((IPointy) myShapes[i]).Points); else Console.WriteLine("-> {0}\'s not pointy!", myShapes[i].PetName); Console.WriteLine(); } Console.ReadLine(); } The output is as follows: ***** Fun with Interfaces ***** Drawing NoName the Hexagon -> Points: 6 Drawing NoName the Circle -> NoName's not pointy! Drawing Joe the Triangle -> Points: 3 Drawing JoJo the Circle -> JoJo's not pointy!

Here s a question for thought: when $form_state['values'] gets passed to the validate and submit functions, will the color field be $form_state['values']['color']['favorite_color'] or $form_state['values']['favorite_color'] In other words, will the value be nested inside the fieldset or not The answer: it depends. By default, the form processor flattens the form values, so that the following function will work correctly: function formexample_nameform_submit($form_id, $form_state) { $name = $form_state['values']['user_name']; $color_key = $form_state['values']['favorite_color']; $color = $form_state['values']['color_options'][$color_key]; drupal_set_message(t('%name loves the color %color!', array('%name' => $name, '%color' => $color))); } The message set by the updated submit handler can be seen in Figure 11-4.

Given that interfaces are valid .NET types, you may construct methods that take interfaces as parameters, as illustrated by the CloneMe() method earlier in this chapter. For the current example, assume you have defined another interface named IDraw3D: // Models the ability to render a type in stunning 3D. public interface IDraw3D { void Draw3D(); } Next, assume that two of your three shapes (ThreeDCircle and Hexagon) have been configured to support this new behavior: // Circle supports IDraw3D. class ThreeDCircle : Circle, IDraw3D {

... public void Draw3D() { Console.WriteLine("Drawing Circle in 3D!"); } } // Hexagon supports IPointy and IDraw3D. class Hexagon : Shape, IPointy, IDraw3D { ... public void Draw3D() { Console.WriteLine("Drawing Hexagon in 3D!"); } } Figure 9-3 presents the updated Visual Studio 2010 class diagram.

Figure 11-4.Message from the submit handler for the form If, however, the #tree property is set to TRUE, the data structure of the form will be reflected in the names of the form values. So, if in our form declaration we had said $form['#tree'] = TRUE;

Figure 9-3. The updated shapes hierarchy If you now define a method taking an IDraw3D interface as a parameter, you can effectively send in any object implementing IDraw3D. (If you attempt to pass in a type not supporting the necessary interface, you receive a compile-time error.) Consider the following method defined within your Program class: // I'll draw anyone supporting IDraw3D. static void DrawIn3D(IDraw3D itf3d) { Console.WriteLine("-> Drawing IDraw3D compatible type"); itf3d.Draw3D(); } We could now test whether an item in the Shape array supports this new interface, and if so, pass it into the DrawIn3D() method for processing: static void Main(string[] args) { Console.WriteLine("***** Fun with Interfaces *****\n"); Shape[] myShapes = { new Hexagon(), new Circle(), new Triangle(), new Circle("JoJo") } ;

for(int i = 0; i < myShapes.Length; i++) { ... // Can I draw you in 3D if(myShapes[i] is IDraw3D) DrawIn3D((IDraw3D)myShapes[i]); } } Here is the output of the updated application. Notice that only the Hexagon object prints out in 3D, as the other members of the Shape array do not implement the IDraw3D interface. ***** Fun with Interfaces ***** Drawing NoName the Hexagon -> Points: 6 -> Drawing IDraw3D compatible type Drawing Hexagon in 3D! Drawing NoName the Circle -> NoName's not pointy! Drawing Joe the Triangle -> Points: 3 Drawing JoJo the Circle -> JoJo's not pointy!