Why my output for this Multi threading program is unordered?

import java.util.concurrent.ExecutionException;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.ScheduledExecutorService;
import java.util.concurrent.ScheduledFuture;
import java.util.concurrent.TimeUnit;
import java.util.concurrent.atomic.AtomicInteger;

public class ScheduledExecuterServicePractice {

  public static void main(String[] args) {
    // TODO Auto-generated method stub
    System.out.println("Started");
    
    NumberManager nm= new NumberManager();
    
    ExecutorService newFixedThreadPool = Executors.newFixedThreadPool(10);
    for(int i=0;i<10;i++) {
    newFixedThreadPool.submit(()-> 
    System.out.println(Thread.currentThread().toString()+" "+nm.incrementCount()));
    }
    newFixedThreadPool.shutdown();
  }

}

class NumberManager{
  //int count =0;
  AtomicInteger count = new AtomicInteger();
  public int incrementCount() {
    
    return count.incrementAndGet();
    
  }
}

The output is

 Started
Thread[pool-2-thread-3,5,main] 1
Thread[pool-2-thread-4,5,main] 4
Thread[pool-2-thread-1,5,main] 3
Thread[pool-2-thread-5,5,main] 5
Thread[pool-2-thread-2,5,main] 2
Thread[pool-2-thread-6,5,main] 6
Thread[pool-2-thread-7,5,main] 8
Thread[pool-2-thread-8,5,main] 7
Thread[pool-2-thread-9,5,main] 9
Thread[pool-2-thread-10,5,main] 10

My question is how the value 4 is printed before 3. I have used atomic class. So only one thread can do the operation at on time.

I have printed the thread name and the count. So if the count is 0 and one thread comes and make it 1 so that should be displayed in the terminal right? then if another thread comes and executes then the value 2 will be printed ? But Here I don’t understand.

>Solution :

The reason is below code

ExecutorService newFixedThreadPool = Executors.newFixedThreadPool(10);
for(int i=0;i<10;i++) {
  newFixedThreadPool.submit(()-> 
  System.out.println(Thread.currentThread().toString()+" "+nm.incrementCount()));
}

The 10 threads are not guarantee to execute in order:

System.out.println and nm.incrementCount() are not guaranteed in the same order

A more details explanation:

1. Thread-1 invoke nm.incrementCount(),the value is 1
2. Thread-2 invoke nm.incrementCount(),the value is 2
3. Thread-2 print value of nm.incrementCount(),the output is 2
4. Thread-1 print value of nm.incrementCount(),the output is 1

If you want to them to print in order,you can use newFixedThreadPool or newSingleThreadExecutor() instead

ExecutorService newFixedThreadPool = Executors.newFixedThreadPool(1);
//ExecutorService newFixedThreadPool = Executors.newSingleThreadExecutor();
for(int i=0;i<10;i++) {
  newFixedThreadPool.submit(()-> 
  System.out.println(Thread.currentThread().getName()+" "+nm.incrementCount()));
}