Super ref = new Sub();

[pitambari parekh]

class Employee implements Printable
{
public void print()
{
System.out.println("inside print");
}

public Employee()
{

}

public static void main(String[] args)
{
Printable p1 =new Employee();
Printable p2 =new Employee();
p1.print();
if(p1.equals(p2))
System.out.println("Equal");
System.out.println("Not Equal");
}
}
how is it possible to call "equals" method on Printable reference??? Because with super class/interface reference we can only call methods of super class/interface .. And "Object" class is not a super class for Printable.
Waiting for a reply...

[Campbell Ritchie]

Welcome to the Ranch.

Not sure. There is another problem. The print() method in java.awt.print.Printable takes (Graphics, PageFormat) as parameters, and your print() method doesn't take any parameters, so you are overloading the method, not implementing the interface. You can't call print() from a Printable object. Unless you can get a Graphics object and a PageFormat object from somewhere (or dummy objects) you will never get that class to compile.

But when I tried to compile your class, it didn't say anything about equals().

[Ernest Friedman-Hill]

There is explicit language in the Java Language Spec which states that the methods of Object are available on references of interface types.
[ July 06, 2007: Message edited by: Ernest Friedman-Hill ]

[Campbell Ritchie]

Thank you, Mr Friedmann-Hill. That is why my compiler overlooked "equals". I made a mistake; I failed to notice that print needs an int parameter too.

[pitambari parekh]

Thank you, Mr Friedmann-Hill..

[pitambari parekh]

Thank You Mr. Campbell Ritchie .. But one thing i would like to clarify is that by using Printable I meant simply any interface and not specifically the interface from java.awt.print.Printable..