Multithreading program output is not expected

Question

multithreading

public class RaceData {

public static void main(String[] args) {

    class UnsafeSequence implements Runnable{
        private int value = 0;

        /** Returns a unique value. */

        public void run(){

            synchronized (this) {
                value = value +1;
                System.out.printf("Thread %s and Count value is %d \n",Thread.currentThread().getName(),value);
            }
        }
    }

    UnsafeSequence s = new UnsafeSequence();
    Thread t1 = new Thread(s);
    t1.setName("t1");
    t1.start();
    Thread t2 = new Thread(s);
    t2.setName("t2");
    t2.start();
    Thread t3 = new Thread(s);
    t3.setName("t3");
    t3.start();
    Thread t4 = new Thread(s);
    t4.setName("t4");
    t4.start();
    Thread t5 = new Thread(s);
    t5.setName("t5");
    t5.start();     
}
}

output is :

Thread t1 and Count value is 1 
Thread t5 and Count value is 2 
Thread t3 and Count value is 3 
Thread t4 and Count value is 4 
Thread t2 and Count value is 5

Why count value is not getting displayed correctly ? sorry i am modifying my question it is not about count variable value, i expected thread t2 getting executed after t1 because i started t2 after t1, here threads are not executed in which they are ordered,though i started one after other.

So, what were you expecting? What would be correct?

Mike Nakis
– Mike Nakis

2012-01-04 00:03:50 +00:00
Commented Jan 4, 2012 at 0:03 — Mike Nakis
– Mike Nakis, Commented Jan 4, 2012 at 0:03
What are you expecting for Count Value?

David Harris
– David Harris

2012-01-04 00:05:12 +00:00
Commented Jan 4, 2012 at 0:05 — David Harris
– David Harris, Commented Jan 4, 2012 at 0:05

millhouse · Accepted Answer · 2012-01-04 00:04:09Z

5

The count is getting displayed correctly. Each time a thread gets access to the synchronized block, count gets incremented.

You cannot expect the threads to get access to the synchronized block in the same order that they were started.

answered Jan 4, 2012 at 0:04

millhouse

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Comments

user684934 · Accepted Answer · 2012-01-04 00:06:12Z

1

They're not executing in the order you're expecting because although the actual printing is done by a synchronized thread, there is no guarantee that the individual threads will arrive at the same time (and wait their turn).

All that the synchronized guarantees is that the block is run by 1 thread at a time.

answered Jan 4, 2012 at 0:06

user684934

1 Comment

Satya Jasti Over a year ago

Thanks bdares i missed the point that threads are created and executed at the dicreation of OS

Chander Shivdasani · Accepted Answer · 2012-01-04 00:13:08Z

1

The order in which you start thread doesnt guarantee the way they will execute. Say, you started 3 threads, T1, T2, T3. Once you call T1.start(), T2.start() and T3.start(), all the threads move to a state called runnable which means, waiting for CPU to execute. CPU randomly selects any one of these threads and executes them. This is one of the perils of concurrent programming, that by default, the order of execution is random.

answered Jan 4, 2012 at 0:13

Chander Shivdasani

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1 Comment

David Schwartz Over a year ago

Or, alternatively, it's one of the major benefits -- you get the most efficient ordering. You only pay the cost of forcing a particular ordering if you force a particular ordering.

Mitch · Accepted Answer · 2012-01-04 00:06:39Z

0

"value" is private to the class UnsafeSequence of which you only have a single instance. Every time run() is executed in the different threads, value is incremented and printed. The output is consistent with what I would expect.

answered Jan 4, 2012 at 0:06

Mitch

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