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Home/ Questions/Q 8708159
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T04:03:36+00:00 2026-06-13T04:03:36+00:00

I have 3 threads 1st printing A 2nd printing B 3rd printing C I

  • 0

I have 3 threads
1st printing A
2nd printing B
3rd printing C

I want to print in sequence A B C A B C A B C and so on…..

So I wrote the program below, but I am not able to achieve the same.
I am aware of the problem that when status=1 at that time say for example B1 and C1 thread are waiting and when I do notifyAll() both waiting thread wake up and depending on CPU allocation it might print B or C.

in this case I want only B to be printed after A.

what modification I need to do.

public class NotifyAllExample {

    int status=1;
    public static void main(String[] args) {

        NotifyAllExample notifyAllExample = new NotifyAllExample();

        A1 a=new A1(notifyAllExample);
        B1 b=new B1(notifyAllExample);
        C1 c=new C1(notifyAllExample);

        a.start();
        b.start();
        c.start();
    }
}

class A1 extends Thread{
    NotifyAllExample notifyAllExample;

    A1(NotifyAllExample notifyAllExample){
        this.notifyAllExample = notifyAllExample;
    }

    @Override
    public void run() {

        try{
            synchronized (notifyAllExample) {

                for (int i = 0; i < 100; i++) {

                    if(notifyAllExample.status!=1){
                        notifyAllExample.wait();
                    }

                    System.out.print("A ");
                    notifyAllExample.status = 2;
                    notifyAllExample.notifyAll();
                }

            }
        }catch (Exception e) {
            System.out.println("Exception 1 :"+e.getMessage());
        }

    }

}

class B1 extends Thread{

    NotifyAllExample notifyAllExample;

    B1(NotifyAllExample notifyAllExample){
        this.notifyAllExample = notifyAllExample;
    }

    @Override
    public void run() {

        try{
            synchronized (notifyAllExample) {

                for (int i = 0; i < 100; i++) {

                    if(notifyAllExample.status!=2){
                        notifyAllExample.wait();
                    }

                    System.out.print("B ");
                    notifyAllExample.status = 3;
                    notifyAllExample.notifyAll();
                }

            }
        }catch (Exception e) {
            System.out.println("Exception 2 :"+e.getMessage());
        }

    }
}


class C1 extends Thread{

    NotifyAllExample notifyAllExample;

    C1(NotifyAllExample notifyAllExample){
        this.notifyAllExample = notifyAllExample;
    }

    @Override
    public void run() {

        try{
            synchronized (notifyAllExample) {

                for (int i = 0; i < 100; i++) {

                    if(notifyAllExample.status!=3){
                        notifyAllExample.wait();
                    }

                    System.out.print("C ");
                    notifyAllExample.status = 1;
                    notifyAllExample.notifyAll();
                }

            }
        }catch (Exception e) {
            System.out.println("Exception 3 :"+e.getMessage());
        }

    }
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-13T04:03:38+00:00Added an answer on June 13, 2026 at 4:03 am

    Convert those IF statements to WHILE statements to get the desired behavior:

    if (notifyAllExample.status != 2){
        notifyAllExample.wait();
    }
    

    to

    while (notifyAllExample.status != 2){
        notifyAllExample.wait();
    }
    

    This will ensure that if a thread is notified, it won’t go out of the while loop until the status value is what it expects.

    Also, mark status as volatile so that the threads won’t have a local copy.

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