Thread: Why the value of ' i' is 0;

Hi ! I'm new here.I have a question.I don't konw why the value of ' i' is 0;And the following is my code. Thx.


public class TestDemo {
public static void main(String[] args) {
new Son();
}
}

class Son extends Base {
private int i = 22;
public Son() {
i = 222;
}
public void display() {
System.out.println("i==="+i);
}
}

class Base {
private int i = 2;
public Base() {
this.display();
}
public void display() {
System.out.println("i==="+i);
}
}

public class Son extends Base {
	private int i = 22;    //12
	{
		System.out.println("private int i = "+i);   //13
	}
	public Son() {    //3
		i = 222;    //14
		System.out.println("i: " + i);    //15
	}
 
	public void display() {    //9
		System.out.println("2 Base: " + i);    //10
		System.out.println("i==="+i);    //11
	}
}

public class Base {
	private int i = 2;    //5
	{
		System.out.println("private int i = "+i);    //6
	}
	public Base() {    //4
		System.out.println("I base: " + i);    //7
		this.display();    //8
	}
	public void display() {
		System.out.println("i==="+i);
	}
}

public class Test {
	public static void main(String[] args) {    //1
	 new Son();    //2
	}
}

private int i = 2
I base: 2
2 Base: 0
i===0
private int i = 22
i: 222

When and where was the 'i' defined before printing in "System.out.println("2 Base: " + i);"?
It is a default value.

--- Update ---

Ha,ha.I got it.They have been defined when we creat an instance.But they haven't been initialized.That's why the value of 'i' is 0.Right?