Super class method

[amarkirt saroay]

    class SuperBase{ void print (SuperBase a){System.out.println("Super");}}
class Base extends SuperBase{  void print(Base b){System.out.println("base");}}
class Derived extends Base{ static void print (Derived d){System.out.println("Derived");}}
 
in a test class:
 
public static void main(String[] args) throws Exception {

SuperBase a1=new SuperBase();
SuperBase b1=new Base();
Base c1=new Derived();
a1.print(new Base());
b1.print(new Derived());
c1.print(new Derived());
}
 
}
 
 
 
 

How is the output as "SuperSuperBase" ? why second invocation calls Super method again?
[ July 24, 2008: Message edited by: amarkirt saroay ]

[Raphael Rabadan]

Because it's and overloading, not an overriding...
The parameters of the methods are of diferent types. So, since its an overloading, its decided on compile time, not run time :-)

See these 2 classes:

class SuperBase {
void print(SuperBase a) {
System.out.println("Super");
}
}
 
class Base extends SuperBase {
void print(Base b) {
System.out.println("base");
}
}

[amarkirt saroay]

Thanks ! got it.