Volatile variable

[ankur trapasiya]

source : examlab

public class ThreadClass extends Thread{

     public ThreadClass(int x){
          this.x=x;
     }

     static String s="";
     static volatile int x=0;

     public static void doVIP(){
          s=s+x;
          try{
               Thread.sleep(10);
          }catch(Exception e){
               s=s+x;
          }
     }

     public void run(){
          synchronized(s){
               doVIP();
          }
     }

     public static void doDelay(long g){
          try{Thread.sleep(g);}catch(Exception e){}
     }

     public static void main(String args[]){

          ThreadClass tc1=new ThreadClass(2);
          ThreadClass tc2=new ThreadClass(1);
          ThreadClass tc3=new ThreadClass(0);

          tc1.start(); doDelay(100); System.out.print(s);
          tc2.start(); doDelay(100); System.out.print(s);
          tc3.start(); doDelay(100); System.out.print(s);

     }
}

Here in this program it gives output as 000000.. Now i am not getting why it is printing output like this even if values of the volatile variable is set via the constructor... . .

Any one cal explain me the output ??

Thanks in advance...

[Stephan van Hulst]

Note that volatile is not the only modifier that applies to x. Take another look.

[ankur trapasiya]

I know there is static too. I am setting value of it through constructors but it is always zero. ... I am not getting why this happends...

[Stephan van Hulst]

What does static mean?

[ankur trapasiya]

static means it is a class level variable so it is not based on any instance. It has an independent value...

[Stephan van Hulst]

Yup. So what happens when you set the variable in three different calls of a constructor?

[Leon Omk]

[ankur trapasiya]

ohhhh... what a fool i am !!! ...... i didn't noticed the third constructor....

thanks for reply.. ....

[Stephan van Hulst]