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  1. What will be the output of the program?

      class s1 implements Runnable
      {    
         int x = 0, y = 0;  
         int addX() {x++; return x;}   
         int aadY() {y++; return y;}      
         public void run() {  
         for(int i = 0; i 10; i++)                     
            system.out.println(addX() + " " + addY());
       }   
          public static void main(String args[])     
          {       
             s1 run1 = new s1();    
             s1 run2 = new s1();             
             Thread t1 = new Thread(run1);        
             Thread t2 = new Thread(run2);          
             t1.start();    
             t2.start();   
       }  
    }
    

  2. A.

    Compile time Error: There is no start() method

    B.

    Will print in this order: 1 1 2 2 3 3 4 4 5 5...

    C.

    Will print but not exactly in an order (e.g: 1 1 2 2 1 1 3 3...)

    D.

    Will print in this order: 1 2 3 4 5 6... 1 2 3 4 5 6...

    View Answer

    Workspace

    Answer : Option C

    Explanation :

    Both threads are operating on different sets of instance variables. If you modify the code of the run() method to print the thread name it will help to clarify the output:

    public void run()
    {
    for(int i = 0; i
    System.out.println(
    Thread.currentThread().getName() + ": " + addX() + " " + addY()
    );

    }


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