Doubt in Thread

[srinibash udayasingh]

Hi Ranchers
please explain the following codes.when i run it gives output 00224466
is it vary or same bcaz 2 thread instance created.

but options are
A 0 2 4 4 6 8 10 6
B 0 2 4 6 8 10 2 4
C 0 2 4 6 8 10 12 14
D 0 0 2 2 4 4 6 6 8 8 10 10 12 12

questions :
class Thread1
{
int x=0;
public class Runner implements Runnable
{
public void run()
{
int cur=0;
for (int i=0;i<4 ;i++ )
{
cur=x;
System.out.println(cur +" ");
x=cur+2;
}
}
};
public static void main(String[] args)
{
(new Thread1()).go();
}
public void go()
{
Runnable r1=new Runner();
(new Thread(r1)).start();
(new Thread(r1)).start();
}
}

[Satya Maheshwari]

I think that the output for this cannot be predicted and would depend on how the 2 threads are scheduled by the OS.

[Gitesh Ramchandani]

Ans is C. I checked running the code.

Why so?

We are printing cur, but look at this:

cur=x;
System.out.println(cur +" ");
x=cur+2;

The value of cur depends on x, which is in fact a class variable whose value keeps on updating every time run() is executed.

So whatever the case may be, at every single execution of for loop (by any thread) x is updated and hence the (local) cur variable is updated printing values 0 2 4 6 8 10 12 14

Hope i'm of some help?

Gitesh

[Thiru Mu]

Yes the Answer is C.
the output is predictable well.

as Gitesh said, we are printing cur, which depends variable x, which is a class variable.

so 2 Threads, 4+4 times the value of x is modified and printed via c.

simple way to understand, comment the //L2 in the below code and run it will run the loop 4 times leaving
0 2 4 6 as output.
then de-comment //L2 ad run you will get the output as option c


class JRThread1
{
int x=0;

public class Runner implements Runnable
{
public void run()
{
int cur=0;
for (int i=0;i<4 ;i++ )
{
cur=x;
System.out.print(cur +" ");
x=cur+2;
}
}
};

public static void main(String[] args)
{
(new JRThread1()).go();
}

public void go()
{
Runnable r1=new Runner();
(new Thread(r1)).start(); //L1
(new Thread(r1)).start(); //L2
}
}

[Satya Maheshwari]

I would still say, the output is unpredictable. Here's my understanding.

1.x is a class variable
2.x is shared between 2 threads.
3.x is being modified in run() through x=cur+2.
4.Let's say

[B]Thread1                   Shared Variable  Thread2                     O/P  CPU                  Time[/B]
 
cur=0                     x=0                                               Executing thread1    1
i=0                       x=0                                               Executing thread1    2
cur=0                     x=0                                               Executing thread1    3
System.out.println(cur);  x=0                                          0    Executing thread1    4
                          x=0              cur=0                            Executing thread2    5
                          x=0              i=0                              Executing thread2    6
                          x=0              cur=0                            Executing thread2    7
                          x=0              System.out.println(cur);    0    Executing thread2    8
x=cur+2                   x=2                                               Executing thread1    9
 
...And so on

So above you see that the output is something like 0 0...
Still in another way it could be 0 2 4 6 8 ...

Hence I do not think it is possible to predict the output.Though your OS may always give the output 0 2 4 6 8 ... (or something else)

I did some minor modification to your code to verify:

class Main {
    int x=0;
    public class Runner implements Runnable {
        public void run() {
            int cur=0;
            for (int i=0;i<4 ;i++ ) {
                cur=x;
                System.out.println(cur +" ");
                try
                {                    
                    Thread.sleep(1000); //Sleep here so that other thread is executed by CPU
                }
                catch(Exception e)
                {
                    e.printStackTrace();
                }
                x=cur+2;
            }
        }
    };
    public static void main(String[] args) {
        (new Main()).go();
    }
    public void go() {
        Runnable r1=new Runner();
        (new Thread(r1)).start();
        (new Thread(r1)).start();
    }
}

And the output was: 0 0 2 2 4 4 6 6
[ August 31, 2007: Message edited by: Satya Maheshwari ]

[srinibash udayasingh]

Hi
still it is confuse to me.in my machine it shows 0 0 2 2 4 4 6 6.

[dolly shah]

I ran this program. I got no. of outputs with different combination. So I think answer is not predictable.

[Gitesh Ramchandani]

i'm sorry i'm also getting output as 00224466.

also got 0246810 once.

working on this question.

Any help from other fellow Ranchers??