How threads are executed in java

Question

I have a code-

public class ThreadOne
{
    public static void main(String[] args)
    {
        Thread1 th=new Thread1();
        Thread1 th2=new Thread1();
        th.start();
        th2.start();
        System.exit(1);
    }
}


class Thread1 extends Thread
{
    public void run()
    {
        for(int i=0;i<10;i++)
        {
            System.out.println(i);
        }
    }
}

What I wanted to know is-

• Why the above code doesn't print anything? Although I have created 2 threads and started them,yet it doesn't print anything.Why?
• If there is no user thread,for e.g. say a normal program consisting of many methods and each method is invoked in main() method,then how many threads are created by default,does it depend on the number of methods we have,or is there a single main thread which is responsible for invoking all methods,and at last who creates the main thread,is it JVM which creates it?

Answer 1

System.exit(1); will terminate the currently running Java Virtual Machine. When your program exit, your threads will also die.

Thread is a part of Process, If Process have exited, then all threads will be destroyed.

Thread.join() will wait until thread run finished.

public class ThreadOne
{
    public static void main(String[] args)
    {
        Thread1 th=new Thread1();
        Thread1 th2=new Thread1();
        th.start();
        th2.start();
        th.join();
        th2.join();
        System.exit(1);
    }
}


class Thread1 extends Thread
{
    public void run()
    {
        for(int i=0;i<10;i++)
        {
            System.out.println(i);
        }
    }
} 

Answer 2

To see the numbers being printed on screen add the following:

try{
       th.join();
       th2.join();
   }catch(Exception e){
}

After th2.start() in main method.

As others have pointed out you will need to give time to the user-defined threads to complete their work and this is done by calling join() on the respective thread objects.join() ensures that the calling thread "Waits for this thread to die" before proceeding ahead.

Answer 3

According to java docs regarding system.exit

Terminates the currently running Java Virtual Machine. The argument serves as a status code; by convention, a nonzero status code indicates abnormal termination.

So above is your answer regarding your first question. When JVM starts usually a single thread is created which calls main and then rest of your methods. Hope it helps.

Answer 4

you can wait till the threads executes by using join.check the following

 public class ThreadOne
 {
public static void main(String[] args)
{
    Thread1 th=new Thread1();
    Thread1 th2=new Thread1();
    th.start();
    th2.start();
    try {
        th.join();
        th2.join();
    } catch (InterruptedException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    System.exit(1);
}

}

  class Thread1 extends Thread
{
public void run()
{
    for(int i=0;i<10;i++)
    {
        System.out.println(i);
    }
}

}

Answer 5

As thread is not Synchronized execution like normal java code. so after calling th.start(); th2.start(); it won't wait for run() to complete thats why System.exit(1); called and you are getting anything.

Answer 6

Since the threads are separate from main: the code in main continues to be executed and System.exit(1); executes without any regard for the Threads and shuts down the program.

Answer 7

your 2 threads are started by the main threads, so there are 3 threads get invoked. but system.exit() kills your main thread causing other 2 threads to terminate before they get any chance to run.

Answer 8

1. Your code does not show anything because you are killing the app with System.exit() as soon as it starts. You should wait for both threads to complete before exiting, by using Thread.join() for example.

2. There is one thread by default, which executes your main() method and it is created by the JVM.