Regarding Polymorphism

[Rick Charles]

How would you explain Compile time and Runtime polymorphism? And also can someone explain me about dynamic dispatch?
[ April 11, 2007: Message edited by: Rick Charles ]

[Jan Cumps]

Polymorphism:
Definition of Richard G. Baldwin (His Introductory Java Programming Tutorial will get you started):
compile-time polymorphism: method overloading
runtime polymorphism: method overriding and inheritance

Dynamic dispatch is explained in wikipedia.

Regards, Jan
[ April 11, 2007: Message edited by: Jan Cumps ]

[Campbell Ritchie]

Sounds more like a beginner's question to me.

I thought polymorphism was only seen at runtime. You will find lots of examples. Imagine you have a superclass, and you have subclasses; they can have methods which have the same signature (overridden) which use different operations. Example:-

public abstract class Shape
{
   protected double size;
 
   public Shape(double size)
   {
      this.size = size;
   }
   . . . 
   public abstract double getArea();
}
. . .
public class Circle extends Shape
{
   public Circle(double radius)
   {
       super(radius);
   }
 
   @Override
   public double getArea()
   {
       return Math.PI * size * size;
   }
}
. . .
public class Square extends Shape
{
  public Circle(double side)
   {
       super(side);Circl
   }
 
   @Override
   public double getArea()
   {
       return size * size;
   }
}
. . .
public class EquilateralTriangle extends Shape
{
  public EquilateralTriangle(double side)
   {
       super(side);
   }
 
   @Override
   public double getArea()
   {
       return size * size * Math.sin(PI / 3.0) / 2.0;
   }
}

At compile time all the compiler will notice is that each of these three concrete classes extends Shape, and has a getArea() method. Also that the return types are compatible (in this case all the same, double).
Now, at runtime, you can create different subclasses of Shape:

. . .
private Shape myShape;
. . .
switch(input)
{
   case 1:
      myShape = new Circle(10.0);
      break;
 
   case 2:
      myShape = new Square(10.0);
      break;
 
   case 3:
      myShape = new EquilateralTriangle(10.0);
      break;
 
   default:
      reportError(input);
      break;
}//end switch
. . .
System.out.printf("Area = %f%n", myShape.getArea());

Your compiler can't tell whether you will enter 1 for Circle or 2 for Square or 3 for Triangle. But each has a getArea() method which the JVM finds and uses at Runtime.

try it