Overriding Q?

[Vidyavathi saravanan]

class Base {}
 
class Agg extends Base{
        public String getFields(){
         String name =  "Agg";
        return name;
        }
}
 

public class Avf{
public static void main(String argv[]){
        Base a = new Agg();
        //Here
        }
}
 
What code placed after the comment //Here will result in calling the getFields method resulting in the output of the string "Agg"?
 
1) System.out.println(a.getFields());
2) System.out.println(a.name);
3) System.out.println((Base) a.getFields());
4) System.out.println( ((Agg) a).getFields());

Answer is 4.
My Question is why not 1 also.
Since methods are called based on Object , why option1 is not correct?

Anyone Please explain

[Mark Patrick]

Answer 1 will not compile. The compiler checks that the method exist in the reference type, not the object type. At runtime, the JVM will dynamically bind the method that executes based on object type.

Base a = new Agg();  

In this case, at compile time the compiler will confirm that every method called against variable 'a' is defined (at least abstract) in class Base. Attempting to call a.getFields() generates a compile error because getFields() is not defined in class Base.

It's easy in this case to see that variable 'a' will always contain an object of type Agg, but in more complex examples the compiler may not be able to confirm at compile-time the exact object type that the reference will contain...Hence, for safety sake, the compiler must complain.

[Mark Patrick]

PS...You can always try to compile and run the example to see what it does....