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Reference variable code

Saurabh Verma
Ranch Hand

Joined: Aug 13, 2007
Posts: 38



The output at line 1 is 10.
Can anybody explain the code?
Sivakumar Manikkavasagam
Greenhorn

Joined: Oct 16, 2007
Posts: 15
In Check this Line(Commented) - bb is the new instance of BClass, so it will not pass this 200 to the main program. It will display only 10
Check this program!!!
I think this program will give you the flow of the variable j.
I hope it will be useful for you.


class AClass {
void aMethod(BClass bb) {
bb.j = 10;
System.out.println("bb.j1=" +bb.j);
bb = new BClass();//Check this
bb.j = 200;
System.out.println("bb.j2=" +bb.j);
}
}
class BClass {
int j = 0;
}
public class Test {
public static void main(String[] args) {
int i = 1;
AClass a = new AClass();
BClass b = new BClass();
b.j = 5;
System.out.println("b.j1=" +b.j);
a.aMethod(b);
System.out.println("b.j2=" +b.j);
int j = b.j;
System.out.println("j="+j);//Line 1
}
}

Output
b.j1=5
bb.j1=10
bb.j2=200
b.j2=10
j=10


Thanks & Regards,<br />Sivakumar M
Saurabh Verma
Ranch Hand

Joined: Aug 13, 2007
Posts: 38
Thanks Shivkumar!!
 
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