simple super & this question

[Claire Yang]

In the following code, why super.i & super.s changed in method compare()?

  
class Base {
String s=new String("Base");
int i=3;
}
class test extends Base {
void compare() {
s="changed-again";
System.out.println(super.i); //2
System.out.println(super.s); //changed-again
System.out.println(this.i); //2
System.out.println(this.s); //changed-again
}
public static void main(String[] args) {
Base a=new Base();
test t=new test();
System.out.println(a.i); //3
System.out.println(a.s); //Base
System.out.println(t.i); //3
System.out.println(t.s); //Base
t.i=2;
t.s=new String("changed");
System.out.println(a.i); //3
System.out.println(a.s); //Base
System.out.println(t.i); //2
System.out.println(t.s); //changed
t.compare();
System.out.println(a.i); //3
System.out.println(a.s); //Base
System.out.println(t.i); //2
System.out.println(t.s); //changed-again
}
}

[Maulin Vasavada]

hi Claire
well, in test class we don't have another int i and String s defined right? so, we have ONLY ONE copy of variable i and variable s hence super.i = this.i for test object and super.s = this.s as well.
do u see what i mean?
super.i and this.i is just different mehtods of referring to same variable here (same for super.s and this.s)..
try ,
System.out.println(super.s == this.s ); in compare() method, it will return u "true" meaning both the String referrences are same...
regards
maulin

[Claire Yang]

Maulin,

I see what you mean, I forgot that I created 2 objects in the main method, thanks a lot.