parameter passing

[Paul Salerno]

Case A

class ExSuper
{
   public void func(String p, String s)
   {
      System.out.println(p)
   }
}
public class Example extends ExSuper
{
   public void func(String p, String s)
   {
      System.out.println(p + ":" + s)
   }
   static public void main(String arg[])
   {
      ExSuper e1 = new ExSuper() 
      e1.func("hello", "hi")     // hello
      ExSuper e2 = new Example()
      e2.func("hello2", "hi2")   // hello2, hi2
   }
}

Case B

  
class Super
{
   public int getNumber(int a)
   {
      return 2;  // prints 2
   }
}
public class Subclass extends Super
{
   public int getNumber(int a, char ch)
   {
      return 4;
   }
   public static void main(String[] args)
   {
    System.out.println(new Subclass().getNumber(4)
   }
}

My question is why in case B we are accessing the method in the superclass vs the overridden method in Subclass and is this always the case.

[mahesh zarkar]

Paul,
According to me in case B the method is not overridden but it is overloaded since the no. of parameters are different. also in the main() when u pass only one parameter the program call the appropriate method depending upon the no of parameters passed. Therefore it prints 4 if u pass 2 parameter the 2nd method will be invoked. This is my theory. Other guys please reply to verify my answer.
Mahesh

[Eric Cao]

In case B, you overload the method of the superclass. So when running, it invoking the method getNumber(int a).

[Dirk Schreckmann]

Hello,
Mahesh has it right. To override a method, the method signature of the overriding method must be the same as the method signature of the overridden method. To overload a method, you are just writing new methods with the same name as an existing one, but with different parameters. So, if a parent class has method(int a), it would be overridden by method(int b) in a subclass, while method(String a) in the given subclass would be overloading.
Making sense?
Good Luck,
-Dirk Schreckmann