JQ+ Mock - Threads - Question ID: 988383703406

[Shanel Jacob]

This question is quite similar to the one I posted here: https://coderanch.com/t/257971/java-programmer-SCJP/certification/JQ-Mock-Threads-ID

Tried to compile and run the code to understand the concepts but I got even more confused as the results seem to contradict my thoughts.

Consider the following code: 
 
public class Test extends Thread
{
 static int x, y;
 public synchronized void run(){  for( ; ; ) { x++; y++; System.out.println(x+" "+y);} }
 public static void main(String[] args)
 {
   new Test().start();
   new Test().start();
 }
}
 
What will the above code print? 
   
Select 1 correct option. 
 
a  It will keep on printin same values for x and y incrementing by 1 on each line.   
               
b  It will keep on printin same values for x and y but they may be incrementing by more than 1 on each line.   
               
c  It may print different values for x and y but both the values will be incrementing by 1 on each line.   
               
d  You cannot say anything about the values.   
       
ANSWER: d
        
General Comments
You may be tempted by the synchronized keyword on the run method. But note that, there are two diferent thread objects and both the threads are accquiring locks for their own thread object. So, in this case synchronized doesn't make any sense. 

I tried running the code and it shows this (below) which really doesn't agree with the explanation given (above). The synchronized method does seem to work.:

Thread-0 1234 1234
Thread-0 1235 1235
Thread-1 1236 1236
Thread-1 1237 1237

[ August 08, 2006: Message edited by: Shanel Jacob ]

[Naseem Khan]

synchronized keyword has nothing to do here as there is two different locks and each thread will acquire its own lock.

Answer is d: Reason is you can't be so sure about the order of execution. Does not matter how many times you got the same output what you posted.


Naseem

[Shanel Jacob]

Still need to clarify the concepts. So we have 2 thread objects and even if the run() is synchronized, it doesn't matter because each thread object calls its own run method? In this scenario, x and y may not be the same.

public synchronized void run()
{  
   for( ; ; ) 
   { 
      x++; y++; System.out.println(x+" "+y);
   } 
}

Now suppose I synchronized by object instead. That means both thread objects (we cannot know which thread) will take turn incrementing x and y, so x and y will always have same values.

public void run()
{  
   synchronized(this)
   {
       for( ; ; ) 
       { 
           x++; y++; System.out.println(x+" "+y);
       }
   } 
}

[Gowher Naik]

if you want to synchronized both threads then check code below.
As you are running two threads with two different instances of Test
class so they will acura two seprate locks on two seprate thread and two seprate locks will not block each other.

public class Test1 extends Thread
{
 static int x, y;
 public void run(){  
        //synchronized (this){//thread level lock (line 1)
 synchronized (Test1.class){ //class level lock (line 1)
  for(int i=0;i<20;i++ ) { 
 x++; y++; 
 System.out.println(Thread.currentThread().getName()+": "+x+" "+y);
 }
  }
}
 public static void main(String[] args)
 {
   new Test1().start();
   new Test1().start();
 }
}

[Shanel Jacob]

Is it ok to summarize the levels of synchronization as follow:


public synchronized void run()
------------------------------
any of the 2 threads can access, x and y might NOT be equal


synchronized (this)
-------------------
any of the 2 threads can access, x and y will always be equal


synchronized (Test1.class)
--------------------------
only 1 thread can access, x and y will always be equal