tricky thread qn

[cs singh]

class C extends Thread {
  private static String[] sa = new String[]{"Not Done","X","Y","Z"};
  public void run() {
    synchronized (sa) {
      while (!sa[0].equals("Done")) {
        try {sa.wait();} catch (InterruptedException ie) {}
    }}
    System.out.print(sa[1] + sa[2] + sa[3]);
  }
  public static void main (String[] args) {
    start();
    synchronized (sa) {
      sa[0] = "Done";
      sa[1] = "A"; sa[2] = "B"; sa[3] = "C";
      sa.notify();
}}}
 
Which is a possible result of attempting to compile and run the program? 
 
a.  Prints: XYZ 
b.  Prints: AYZ 
c.  Prints: ABZ 
d.  Prints: ABC 
e.  Compile-time error 
f.  Run-time error 
g.  None of the above 
    

answer is e as non static method start() can not be referenced from static context.
but in the following code

 class A extends Thread {
  private int i;
  public void run() {i = 1;}
  public static void main(String[] args) {
    A a = new A();
    a.start();
    System.out.print(a.i);
}}
 
   

it compiles fine, although a.start() has been referenced from static context.

[steven gerrard]

in the second code start is associated with a object unlike the code in 1 where non static method start() is not associated with any object

to call a non static method frm a static method it must have a objkect associated with it

[Peter Petrov]

I think the question would have been much nicer if it was a real Thread-programming question and not one which just aims you to overlook that start() is called illegally. Anyway. What do you think would be the answer if we modify only the main() method like this:

public static void main (String[] args) {    
    C th = new C();
    th.start();    
    synchronized (sa) {      
          sa[0] = "Done";      
          sa[1] = "A"; 
          sa[2] = "B"; 
          sa[3] = "C";      
          sa.notify();
    }
}

[ February 14, 2006: Message edited by: Peter Petrov ]

[jiju ka]

Good question.
Answer will always be 'ABC' because..

The run method waits until sa[0] contains a value 'Done'. If sa[0] gets a value of 'Done' then it is guaranteed that sa[1] + sa[2] + sa[3] is 'ABC'