Overriding the start method of Thread Class

[Dhruv Arya]

class MyThread extends Thread
{
	public void run()
	{
		System.out.println("MyThread: run()");
	}

	public void start()
	{
		System.out.println("MyThread: start()");
	}
}

class MyRunnable implements Runnable
{
	public void run()
	{
		System.out.println("MyRunnable: run()");
	}

	public void start()
	{
		System.out.println("MyRunnable: start()");
	}
}

public class MyTest 
{
	public static void main(String args[])
	{
		MyThread myThread  =  new MyThread();
		MyRunnable myRunnable = new MyRunnable();
		Thread thread  =  new Thread(myRunnable);
		myThread.start();
		thread.start();
	}
}

What happens in the above code ?? I m not able to understand the fact of overriding the start() method of the thread class and still get an output of MyThread: start() followed by MyRunnable:run() Please can some one explain thank you

[Henry Wong]

Dhruv Arya wrote:
What happens in the above code ?? I m not able to understand the fact of overriding the start() method of the thread class and still get an output of MyThread: start() followed by MyRunnable:run() Please can some one explain thank you

When you call, myThread.start(), you will get the "MyThread:start()" message.

When you call, thread.start(), the new thread will call the run() method of your runnable, and you will get the "MyRunnable:run()" message.

Henry

[Byju Joy]

Runnable interface has only one method, public void run(). Even if the runnable object has a public void start() method, it doesn't get into picture. But the if Thread object has a public void start() it does come into picture and affect how thread is submitted for run.

[Dhruv Arya]

Thanks Alot I totally get my mistake you guys made my day Thank You