Synchronized method

[Eric Pramono]

Hi all,
I'm kinda new to Threads, so I get a bit confused here.
Please consider the following code:

public class X extends Thread {
    static int i=3;
    public synchronized void changeI (String s) {
        for (int j=0; j<10; j++) {
            for (int k=0; k<9999999; k++);
            i++;
            System.out.println(s+" change i to "+i);
        }
    }
    public void run () {
        i++;
        System.out.println(i);
        String s="";
        if (i==4)
            s="thread #1";
        else
            s="thread #2";
        changeI(s);
    }
    public static void main (String[] args) {
        X o=new X();
        X p=new X();
        o.start();
        p.start();
    }
}

I'm creating two threads here, with both of them accessing the same sychronized method. AFAIK about threads, one thread will access the synchronized method, and the other thread will have to wait until the first thread finishes. (pls correct me if I'm wrong )
But, somehow that's not what happened. The output I'm getting is pretty much the same as a non-synchronized method.
Here's what I get on WindowsNT:

4
thread #1 change i to 5
6
thread #2 change i to 7
thread #1 change i to 8
thread #1 change i to 9
thread #2 change i to 10
thread #2 change i to 11
thread #1 change i to 12
thread #1 change i to 13
thread #2 change i to 14
thread #2 change i to 15
thread #1 change i to 16
thread #1 change i to 17
thread #2 change i to 18
thread #2 change i to 19
thread #1 change i to 20
thread #1 change i to 21
thread #2 change i to 22
thread #2 change i to 23
thread #1 change i to 24
thread #2 change i to 25

Shouldn't those two threads take turns in accessing a synchronized method?
please anyone enlightened me on this..
any help is highly appreciated
thanks in advance
- eric

[William Brogden]

But each Thread has its own object - the Threads are not trying to access the same object.
Bill

[Mini Pilla]

Here 2 different objects p and o using the sync methods.If it is one object then that Thread has a lock for that object and no other Thread can have this lock until the first thread has finished the method.
Hope I was good.
Tahnx
Rajani

[Eric Pramono]

Thank you all,
I've created ten threads of the same object and run as I expected. Please check the following code:

public class X extends Thread {
    static int i=0;
    static int j=0;
    public synchronized void change (String s) {
        for (int k=0; k<10; k++) {
            for (int m=0; m<9999999; m++) { };
            ++i;
            System.out.println(s+" change i to "+i);
        }    
    }
    public void run () {
        ++j;
        String s="thread #"+j;
        change(s);    
    }
    public static void main (String[] args) {
        X o=new X();
        for (int i=0; i<10; i++) {
            new Thread(o).start(); 
            /* o.start(); */
        }
    }
}

My next question would be: why do I get an IllegalThreadStateException if I use the commented line instead?
Please anyone let me know..
Thanks again.
- eric

[Vanitha Sugumaran]

Hi Eric,
The exception is caused because you are trying to start the thread that has already been started.
new Thread(o).start() // you are starting a thread on o
o.start // here you are trying to start it again so you get IllegalStateException
Hope it helps,
Vanitha.

[Eric Pramono]

Thanks Vanitha,
That clears my doubt.
- eric

[Ashik Uzzaman]

Hi Vanitha,
If i comment out new Thread(o).start();
and then use o.start(); there then in runtime the code partially runs and reports the same IllegalThreadStateException. Why?

------------------
azaman

[Vanitha Sugumaran]

Hi Azaman,
Good Catch!

<font size = "+0">
public static void main (String[] args) {   
 //X o=new X(); (1)
System.out.println(Thread.currentThread());
for (int i=0; i<10; i++) {   
   // new Thread(o).start(); (4)
   X o = new X(); (3)
   try
   {  
   System.out.println(Thread.currentThread());
   
            o.start();  (2)
   }
   
   
   catch(IllegalThreadStateException e)
   {
   
   System.out.println(e);
   System.exit(0);
   }      }  

At (1) you create a thread out side the for method and start it at (2), (2) will not create a new thread . But (4) will create a new Thread so you didn't get runtime exception when you use (4).
But if you create a thread inside a method (3) it will create the o everytime and start it, so you won't get exception.
Correct me if I am wrong,
Vanitha.
[This message has been edited by Vanitha Sugumaran (edited August 06, 2001).]
[This message has been edited by Vanitha Sugumaran (edited August 06, 2001).]