Re: Understanding Implements

Buck,

In the given example, without the interface you would be forcing a class
relationship between GenerateSounds() and Airplane(). Let's look at Joel's
example:

public class GenerateSounds {
    public static void blareItOut( Noiseable obj ) {
        obj.makeSound();

In the context of a method parameter, the Noisable interface above is being
used as a reference data type so that blareItOut() can operate on any object
that provides the method(s) of a Noisable device. Let's see what happen's if
you don't have a noisable interface:

public class GenerateSounds {
    public static void blareItOut( Airplane obj ) {
        obj.makeSound();

Now, the original example will still work, as long as you're passing it an
F-14 Tomcat, but what happens when you want to hear the sound of a 12
cylinder Dodge Viper? See the dilemna? In order to accomplish the cast,
you're going to need to impose an unnecessary object hierarchy upon your
design where Airplane and Automobile are both subclasses of a more
generic --most likely abstract-- "NoiseMaker" object. So now you're right
back to:

public class GenerateSounds {
    public static void blareItOut( NoiseMaker obj ) {
        obj.makeSound();

Except you've lost a lot of flexibility because you've had to force a class
hierarchy.

I'll leave you with one final thought; interfaces are also useful for
exposing an object's programming interface without revealing its class.
These are usually known as anonymous objects, and can be useful when
shipping a package of classes to other developers.


HTH,

John Taylor
Canada

----- Original Message -----
From: "Buck Calabro" <buck.calabro@aptissoftware.com>
To: <JAVA400-L@midrange.com>
Sent: Wednesday, March 21, 2001 13:41
Subject: RE: Understanding Implements


> I'm working toward the "ah-HA!" here, and I deeply appreciate the
discussion
> to date...
>
> I'm missing the advantage of the extra layer of abstraction involved in
> using interface Noisable.  Since every class that implements Noisable will
> be required to have a makeSound() method anyway, what does the empty
> interface bring to the table?
>
> If this is way too "newbie" tell me to go away and read some more.  I will
> do the right thing, I promise!
>
> Buck
>
> > -----Original Message-----
> > From: Joe Teff
> > Sent: Wednesday, March 21, 2001 10:20 AM
> > To: JAVA400-L@midrange.com
> > Subject: RE: Understanding Implements
> >
> > The following is an interface, stored in a .java source file and
compiled
> > the same as classes:
> >
> > public interface Noisable {
> >    public void makeSound();
> > }
> >
> > Notice that the method only contains the signature and no
implementation.
> > Methods in an interface are abstract. The interface only requires that
you
> > supply the method(s) listed. There is no inheritence here.
> >
> > This class will implement that interface:
> >
> > public class Airplane implements Noisable {
> >    // various methods and data
> >    public void makeSound() {
> >       // code to here to make airplane noise
> >    }
> > }
> >
> > This class will not compile unless I code all methods listed in the
> > interface, hence the "contract" reference.
> >
> > Now lets make an application that generates sounds:
> >
> > public class GenerateSounds {
> >    public static void blareItOut( Noiseable obj ) {
> >       obj.makeSound();
> >    }
> > }
> >
> > I can call GenerateSounds.blareItOut(xxx) where xxx is an object created
> > from any class that implements Noisable. The actual code to make the
noise
> > is coded into each class's makeSound() method, thus each one will do it
> > slightly different. blareItOut() doesn't care how the code works, it
just
> > cares that the appropriate method is there. It doesn't even care what
kind
> > of object (class) it even is. The only thing it knows is that it has to
> > have a makeSound() method that it can call.
> >
> > Joe Teff


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