Doubt on a thread question from K&B OCP Practice Exams

[Jeffrey Tian]

Below is a question from K&B's OCP practice exams book:

public class Bang extends Thread {
    static Thread t1, t2, t3;

    public static void main(String[] args) throws Exception {
        t1 = new Thread(new Bang());
        t2 = new Thread(new Bang());
        t3 = new Thread(new Bang());
        t1.start();
        t2.start();
        t3.start();
    }

    public void run() {
        for (int i = 0; i < 500; i++) {
            System.out.println(Thread.currentThread().getId() + " ");
            if (i == 250)
                try {
                    System.out.println("**" + t1.getId() + "**");
                    t1.sleep(600);
                } catch (Exception e) {
                }
        }
    }
}

Which are true? (Choose all that apply.)
A. Compilation fails.
B. An exception is thrown at runtime.
C. Bang will execute for a second or two.
D. Bang will execute for at least 10 minutes.
E. Thread t1 will almost certainly be the last thread to finish.
F. Thread t1 will almost certainly be the first thread to finish.
G. It’s difficult to predict which thread will be the last to finish.

Below is answer and explanation from the book:
C and G are correct. The sleep() method’s argument is in milliseconds. E is incorrect
because sleep will be called once on each of the three threads, not on Thread t1 three
times. Remember, the sleep() method is static and operates on whichever thread is
the currently running thread.

But I don't know why C is correct. Could somebody explain it to me?
Thanks.

[nitin sethi]

Jeffrey, could you please why do you think C can't be an answer ?

[Mike Okri]

It depends on how you interpret the answer C. Bang will execute for a second or two. If your interpretation is that there is a possibility that all three threads will successfully execute in one second (1000ms), then C is wrong. If your interpretation is that there is a possibility that all three threads will successfully execute somewhere between one second and two seconds, then C is correct.

[Dan Drillich]

By adding some time print statements, we can see that these threads run at about 672 msc, which makes sense as each runs for the 600 msc of pause + a minor time of running. So this statement of one/two seconds is vague but probably correct as if we round 672 msc we end up with a second.

 

import java.util.Date;
import java.text.DateFormat;
import java.text.SimpleDateFormat;
import java.util.Calendar;



       public void run() { 
        	

        Date date = new Date();
        long ll = date.getTime();
        	
        System.out.println("Start: " +Thread.currentThread().getId()  + " " + ll);
        	
            for (int i = 0; i < 500; i++) {  
                System.out.println(Thread.currentThread().getId() + " ");  
                if (i == 250)  
                    try {  
                        System.out.println("**" + t1.getId() + "**");  
                        t1.sleep(600);  
                    } catch (Exception e) {  
                    }  
            }  
            
            Date date2 = new Date();
            long ll2 = date2.getTime();
            System.out.println("End: " +Thread.currentThread().getId() + " " + ll2);
        } 

Regards,
Dan

[Jeffrey Tian]

Dan Drillich wrote:By adding some time print statements, we can see that these threads run at about 672 msc, which makes sense as each runs for the 600 msc of pause + a minor time of running. So this statement of one/two seconds is vague but probably correct as if we round 672 msc we end up with a second.

 

import java.util.Date;
import java.text.DateFormat;
import java.text.SimpleDateFormat;
import java.util.Calendar;



       public void run() { 
        	

        Date date = new Date();
        long ll = date.getTime();
        	
        System.out.println("Start: " +Thread.currentThread().getId()  + " " + ll);
        	
            for (int i = 0; i < 500; i++) {  
                System.out.println(Thread.currentThread().getId() + " ");  
                if (i == 250)  
                    try {  
                        System.out.println("**" + t1.getId() + "**");  
                        t1.sleep(600);  
                    } catch (Exception e) {  
                    }  
            }  
            
            Date date2 = new Date();
            long ll2 = date2.getTime();
            System.out.println("End: " +Thread.currentThread().getId() + " " + ll2);
        } 

Regards,
Dan

Thank you, Dan.

[Jeffrey Tian]

nitin sethi wrote:Jeffrey, could you please why do you think C can't be an answer ?

My thought is similar to Dan's. It should execute for less than 1 second because these 3 threads don't need to compete for a lock and they can pause simultaneously.

[Dan Drillich]

Absolutely, they practically sleep for 600 concurrently.

Regards,
Dan

[Dan Drillich]

Maybe it's not that simple. It's not clear to me what the relation between the thread and the core-cpu is during these 600 msc. Is the thread bound to the cpu during this time-frame?

Regards,
Dan

[Joey Sanchez]

We can say that the result will be the same if we replace t1.sleep() in run()method with either of t2.sleep(), t3.sleep() or sleep(), don't we?

[gurpeet singh]

Joey Sanchez wrote:We can say that the result will be the same if we replace t1.sleep() in run()method with either of t2.sleep(), t3.sleep() or sleep(), don't we?

yes the result will be same. as said in earlier posts sleep() method is static method of Thread class and each of t1, t2,t3 is of type Thread so each of t1.sleep(), t2.sleep(),t3.sleep() will be replaced internally by Thread.sleep() which means the current running thread will sleep.