Please help with this question

[Kevin Lam]

Hi all, I have just come across this question:

class HelSte{
String nat="dub";
public void jume(){
 
System.out.println("HelSte");

}
 
}
 

public class HarMol extends HelSte{
String nat="uk";
public static void main(String argv[]){
 
HarMol hm = new HarMol();
hm.go();

}
public void jume(){
 
System.out.println("HarMol");
}
 
public void go(){
HelSte hs = new HarMol();
hs.jume();
System.out.println(hs.nat);
 
}
}

The output is MarMol followed by dub...

but why dub?? I thought it would be uk?

Any help is greatly appreciated.

Kev

[Alpana Singh]

Variables are set at complile time.Variables are accessed with the declared time ,which in this case is Helse.

[Kevin Lam]

Still don't quite understand, any more explaination??

Thx

[Mahendar Reddy]

Hi Kevin, The thing is..
Variables and Private methods are binded at the time of compilation.
So it looks for the type of reference u r using, but not for the
type of the object it holding.Got it?

[Kevin Lam]

I c... thx

[Kevin Lam]

So if the reference variable called a private method in the above example, it will call the method of the declared type as well?

[Marcus Green]

This quesiton comes from my site and it includes the following explanation.

"The field displayed is that of the reference type HelSte and the method is from the class that has been instantiated with the call to new which is the version of the method in HarMol."

Is there some way I could clarify or expand this explanation?
Marcus

[Kevin Lam]

I guess your explaination only states as it as fact, but it didn't really explain why is it so....

Kev

[Marcus Green]

What happened when you tried with a private method?

[Tonyo Garcia]

This is how I try to understand it:

Since variables whether static or instance and static methods cannot be overridden, the compiler will not determine these members at runtime and the object denoted by the reference does not matter. They cannot be overridden so why would the compiler care about the object at runtime. So the compiler will just based the determination from the reference type.

Perhaps just remember this, always follow the reference type except on cases when a method is overridden, then follow the object type.

Hope that helps.

PS. Try making jume() static, then it is not overridden but hidden. See if the results will change.