thread example

There is no guarantee that your threads are executing simultaneously regardless of any trivial example anyone else posts. If your OS only gives the java process one processor to work on, your java threads will still be scheduled for each time slice in a round robin fashion. Meaning, no two will ever be executing simultaneously, but the work they do will be interleaved. You can use monitoring tools like Java's Visual VM (standard in the JDK) to observe the threads executing in a Java process.

Here is a simple example:

ThreadTest.java

public class ThreadTest
{
   public static void main(String [] args)
   {
      MyThread t1 = new MyThread(0, 3, 300);
      MyThread t2 = new MyThread(1, 3, 300);
      MyThread t3 = new MyThread(2, 3, 300);

      t1.start();
      t2.start();
      t3.start();
   }
}

MyThread.java

public class MyThread extends Thread
{
   private int startIdx, nThreads, maxIdx;

   public MyThread(int s, int n, int m)
   {
      this.startIdx = s;
      this.nThreads = n;
      this.maxIdx = m;
   }

   @Override
   public void run()
   {
      for(int i = this.startIdx; i < this.maxIdx; i += this.nThreads)
      {
         System.out.println("[ID " + this.getId() + "] " + i);
      }
   }
}

And some output:

[ID 9] 1
[ID 10] 2
[ID 8] 0
[ID 10] 5
[ID 9] 4
[ID 10] 8
[ID 8] 3
[ID 10] 11
[ID 10] 14
[ID 10] 17
[ID 10] 20
[ID 10] 23

An explanation - Each MyThread object tries to print numbers from 0 to 300, but they are only responsible for certain regions of that range. I chose to split it by indices, with each thread jumping ahead by the number of threads total. So t1 does index 0, 3, 6, 9, etc.

Now, without IO, trivial calculations like this can still look like threads are executing sequentially, which is why I just showed the first part of the output. On my computer, after this output thread with ID 10 finishes all at once, followed by 9, then 8. If you put in a wait or a yield, you can see it better:

MyThread.java

System.out.println("[ID " + this.getId() + "] " + i);
Thread.yield();

And the output:

[ID 8] 0
[ID 9] 1
[ID 10] 2
[ID 8] 3
[ID 9] 4
[ID 8] 6
[ID 10] 5
[ID 9] 7

Now you can see each thread executing, giving up control early, and the next executing.