null instance variable after assigning a value

[Juan Manuel Alberto de los Santos]

class test {
 
    String s;
 
    test(){
 
    }
 
    test(String s){
        this.s = s;
    }
 
        public static void main(String[] args){
            test test1 = new test("Hello");
            test test2 = new test();
            System.out.println(test2.s);
    }
}

why this prints null ?
[ October 02, 2008: Message edited by: Juan Manuel Alberto de los Santos ]

[Tom Johnson]

It wont compile since nc1 is undefined....

BTW....A member variable of a class that is a reference variable (anything that extends Object) will always be initialised to null unless otherwise specified.

tom

[Juan Manuel Alberto de los Santos]

sorry my mistake, obviously i didn't want to ask that ...

the question still remains, why "s" value is "null" instead of "hello" ...

thanks
[ October 02, 2008: Message edited by: Juan Manuel Alberto de los Santos ]

[Henry Wong]

Why should test2.s be set to "hello"? You set test1 to "hello" -- not test2.

Henry

[Juan Manuel Alberto de los Santos]

yes, but when i run the code the two instances executes, and the first one assigns "Hello" to s ...

[fred rosenberger]

each instance of the test class has it's own 's' variable. just like every car has it's own odometer reading.

if you add the line

System.out.println(test1.s);

you'll get your "Hello" printed out.

[Juan Manuel Alberto de los Santos]

ok, but what about this ?

class Fizz {
 
    int x = 5;
 
    public static void main(String[] args) {
        Fizz f1 = new Fizz();
        Fizz f2 = new Fizz(f1);
        System.out.println(f1.x);
    }
 
    Fizz(){}
 
    Fizz (Fizz f) {
        Fizz fizz = f;
        fizz.x = 8;
    }
}

it prints "8"

[Henry Wong]

It prints 8 because f2 goes to f1 and changes the x value to 8. Now can you explain why f2.x is *not* 8?

Henry

[Ravikanth kolli]

hye juan,
for all the instances that you create of a class, will have its own set of instance variables that you define. In the first example you set the String in test1 to "Hello" but not String of test2. so you will be getting null/

In the second example, you are passing the f1 as a parameter to the constructor of f2. Where you are actually setting the value of f1's x to 8.
Hence you get an 8

[Juan Manuel Alberto de los Santos]

the constructor uses f1 object reference and changes the x value to 8

in f2 the reference is simply pointed to the original x, that's why is 5

... i don't know if it's correct, is kind of hard to understand

[Ravikanth kolli]

We are not considering the value of f2's x here in this case.

The new Fizz object that you create in the constructor is actually a reference to the original f1 object. So the initial value for f1's x is 5 and now you are changing the value to 8 with in the constructor of f2.

[Stephen Davies]

Its pretty straight forward.

You have decalred a no-argument constructor and an overloaded constructor which takes a String as an argument. Thus whenever you call new on an instance of the class, the constructor will either run with the no-argument constructor which initializes nothing, or it takes a string which returns initializes the String variable to the value "Hello".

Since you are calling your print function on test2.s, test2 is a new instance of test in which you have not pased a String variable, and thus the instance will initialize the test2 instance using the no-argument constructor. Therefore s has not been initialized by you with any value and thus gets a null from the JVM.

If you had done the same with an int you would get 0, a boolean, false and so on.


[ October 03, 2008: Message edited by: Stephen Davies ]