how does this piece of code work? forward referencing?

[Aakash Goel]

public class test1 {

    private int i = giveJ();
    private int j = 10;

    private int giveJ() {
        return j;
    }

    public static void main(String args[]) {
        System.out.println((new test1()).i);
    }
}

[output : 0]

the question is... when exactly does j get the value of 10? the 'return j' statement returns a 'zero'. I expected it tobe a 10.

Please explain how the code works... i feel quite a lot is going on there which I don't understand. According to me, it should have ben a forward referencing error!

Thanks.

[Pawan Arora]

When you called giveJ() method, the variable was not intialized. So, JVM intialized it with a default value 0. That's why you're getting the result as 0 in the output.

A little modification in the code can sort this problem out. Here's your code with just a mere modification:

public class test1 {


private int j = 10;
private int i = giveJ();
private int giveJ() {
return j;
}

public static void main(String args[]) {
System.out.println((new test1()).i);
}
}

[Pawan Arora]

edit:nvm

[Henry Wong]

the question is... when exactly does j get the value of 10? the 'return j' statement returns a 'zero'. I expected it tobe a 10.

Instance variables (that are not compile time constants) get initialized in the order that they appear in the source (from top to bottom, and left to right). And any forward reference to them, will have them at their default values.

Henry

[Aakash Goel]

thanks. got it!