static initializer question from khalid mughal

[Vedhas Pitkar]

This is a question from Khalid Mughal's Book pg.268 about static initializers.
public class MyClass{
public static void main(String args[]){
MyClass ki=new MyClass(l);//l is 0 here,how??
}
static int i=5;
static int l;
int j=7;
int k;
public MyClass(int m){
System.out.println(i+","+j+","+k+","+l+","+m);
}
{j=70;l=20;}// Instance Initializer (initializing l)
static { i=50;} //static Initializer
}
My question is are not the static & instance initializers first run before the constructor body is executed?If so, the value of l sholud be 20 not 0.Please explain.
Also is the Sarga site shifted anywhere else?Coz I simply could not open www.sarga.com/java/jac.htm. IS there any mirror site for this?
Do let me know.
Thanks all,
Vedhas.

[Jane Griscti]

Hi Vedhas,
The output is <code>50,70,0,20,0</code> so 'l' is '20'.
------------------
Jane Griscti
Sun Certified Programmer for the Java� 2 Platform

[Guoqiao Sun]

Hi, Vedhas, remember that order of code execution are as follows:

1. static variable initialization
2. static initialization block execution
3. constructor header(super or this - implicit or explicit)
4. instance variables initialization / instance initializaer block 
5. rest of the constructor

In our case, we use the value "l" before step 3, which means "l" still not initialized at that moment, and thus default value 0 is used for the parameter of the constructor. After that, "l" is initialized to 20 at step 4.
Hope it helps.
Guoqiao

Originally posted by Vedhas Pitkar:

[Vedhas Pitkar]

Thanx Jane @ Sun, u heleped me clear my doubts. I guess reading the question carefully is really important.Thanks,
Vedhas.