What will happen when you attempt to compile and run the following code? 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(); } }

  1. Prints : MyThread: start() followed by MyRunnable:run()

  2. Prints : MyThread: run() followed by MyRunnable:start()

  3. Prints : MyThread: start() followed by MyRunnable:start()

  4. Compile time error


Correct Option: A

AI Explanation

To answer this question, let's go through each option to understand why it is correct or incorrect:

Option A) Prints: MyThread: start() followed by MyRunnable: run() - This option is incorrect. The start() method of the MyThread class is overridden to print "MyThread: start()", but it does not start the thread. The run() method of the MyRunnable class is not overridden and it prints "MyRunnable: run()". Since both threads are called using the start() method, both will execute in parallel, and the order of execution is not guaranteed. So it is possible that the output could be "MyRunnable: run()" followed by "MyThread: start()".

Option B) Prints: MyThread: run() followed by MyRunnable: start() - This option is incorrect. As explained in Option A, the start() method of the MyThread class is overridden and does not start the thread. The run() method of the MyRunnable class is not overridden and it prints "MyRunnable: run()". Since both threads are called using the start() method, both will execute in parallel, and the order of execution is not guaranteed. So it is possible that the output could be "MyThread: run()" followed by "MyRunnable: start()".

Option C) Prints: MyThread: start() followed by MyRunnable: start() - This option is incorrect. As explained in Option A, the start() method of the MyThread class is overridden and does not start the thread. The run() method of the MyRunnable class is not overridden and it prints "MyRunnable: run()". Since both threads are called using the start() method, both will execute in parallel, and the order of execution is not guaranteed. So it is possible that the output could be "MyThread: start()" followed by "MyRunnable: start()".

Option D) Compile time error - This option is incorrect. The code will compile without any errors because there are no syntax errors or type mismatches.

The correct answer is A. This option is correct because the start() method of the MyThread class is overridden to print "MyThread: start()", and the run() method of the MyRunnable class is not overridden and it prints "MyRunnable: run()". Since both threads are called using the start() method, both will execute in parallel, and it is possible that the output could be "MyThread: start()" followed by "MyRunnable: run()".

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