whats happening here!!

[Noel Hinders]

hi all
consider the following code:

public class t{
public static void main(String ar[]){
t1 tins= new t2();
System.out.println(tins.i);
System.out.println(tins.amethod());

}
}
public class t1{
int i=2;
int amethod(){
return i;
}
}
public class t2 extends t1{
int i=3;
int amethod(){
return i;
}
}

please explain me with whats happening
Noel
[This message has been edited by Noel Hinders (edited March 09, 2001).]

[ASHISH KUMAR]

Hi Noel,
I think u 've misquoted your question, coz what i see in your code is nothing is tricky and class in t2 is not getting envolved in any way.
Only thing I see tricky in this case is that all of three classes are declared in a single file, since all three classes are declared public, and a file can't can 've more than 1 public class or interface.
if that's not the case than it should print==>
2
2
'coz u r creating object of t1 in the main of t.
as there is no constructor in t1 , i is getting intialized to 2.
so call to tins.i prints 2
and call to tins.amethod prints 2.
I hope this helps u.
bye
ashish

[Noel Hinders]

Hi ashish,
I have edited the code
t1 tins= new t2();
please consider the example now.
thank you.

[David Basile]

Hello,
If these three classes are created in three separate files, however, the output after you compile and run the t class will be
2
3
I believe this has something to do with either dynamic binding or the subclass t2 hiding t1's amethod. I'm also a bit confused as to why this outputs 3 for tins.amethod call. http://members.spree.com/education/javachina/Cert/FAQ_SCJP3.htm#inh_Q4
may have some useful info, but I'm not sure I understand if it relates to this since this one does not have static methods. My guess is that t2's amethod hides t1's amethod.
Can someone please clarify for us on this? Much thanks.

[Seema Das]

David,
You are right. When you call a method, java sees the class of the object this reference is referencing to. In this case its t2. Thats why it calls method from t2 class and hence i = 3. (Dynamic method binding)
Whereas when you ask for a value of a variable, Java sees the class of the reference and hence value for i is fetched from t1 class and in this case it is equal to 2.
hope this helps.
Thanks

[Jane Griscti]

Hi all,
When you reference instance variables, the declared type of the object reference is used. In the example, the declared type of the object reference variable <code>tins</code> is <code>t1</code> so '2' is printed.
When you call methods, the runtime type is used. In the example, the runtime type is <code>t2</code>. <code>amethod()</code> in class <code>t2</code> is called, and as there is an instance variable <code>i</code> in <code>t2</code>, '3' is printed.
See JLS §15.11 Field Expressions
Hope that helps.
------------------
Jane Griscti
Sun Certified Programmer for the Java� 2 Platform

[David Basile]

Thank you very much, Jane and everyone on this question!
I've seen the references to the JLS in other postings on this site, but never referred to it(boy, do I feel foolish!). The JLS seems very thorough in its explanations and serves as a good complement to the books I've been reading and these types of forums.

[Noel Hinders]

Hi all,
thanks for the info,
I think I should refer to JLS.
I have got it now,
how important is it to gothru jls
any comments???
Noel

[Jane Griscti]

Hi Noel,
IMHO the JLS is a must read; the language is a bit of a pain but it definitely helps to work through the example code it contains.
Hope that helps.
------------------
Jane Griscti
Sun Certified Programmer for the Java� 2 Platform