references and method parameters

[sun par]

What is the output of trying to compile and run the following code?
(Select one correct answer)
-----------------------------------------------------------------------

    public class Test026
    {
        static Test026 t = new Test026();
        String str;
        public static void main(String args[]) {
            Test026 t = new Test026();
            t.method("0");        
        }
        void method(String str) {        
            str = str;
            System.out.println(str);        
            System.out.println(t.str);        
            System.out.println(this.str);
        }
    }

-----------------------------------------------------------------------
The answer is 0 null null
Can someone explain how? I thought this.str should give 0?
Thanks
[Dan changed the titled to something more descriptive and added formatting to the code block.]
[ March 01, 2003: Message edited by: Dan Chisholm ]

[Leandro Oliveira]

hum....
what is the output of attempting to compile the following:
int i=2;
i=i+0;
i=i;// special atention on this!!!
System.out.println(i);
it produces 2
So str=str does not cause any error and any change! since you have an already declared name within the method, class variable str is hiden, the only way to access it is with this.str
hope it helps!!!

[Leandro Oliveira]

oops!!! null is printed because every reference variable declared, which is a class variable, not a local variable, is initialized with null, since it's value did not change, it prints null.

[Sarma Lolla]

Sunitha,
The question is intersting. How ever the method local variable doesn't come in to picture when you call t.str and this.str. As both of them look for the varibale in the object scope.
[ March 01, 2003: Message edited by: Sarma Lolla ]