Java Programming - Threads - Discussion
Discussion Forum : Threads - Finding the output (Q.No. 5)
5.
What will be the output of the program?
public class Q126 implements Runnable
{
private int x;
private int y;
public static void main(String [] args)
{
Q126 that = new Q126();
(new Thread(that)).start( ); /* Line 8 */
(new Thread(that)).start( ); /* Line 9 */
}
public synchronized void run( ) /* Line 11 */
{
for (;;) /* Line 13 */
{
x++;
y++;
System.out.println("x = " + x + "y = " + y);
}
}
}
Answer: Option
Explanation:
The synchronized code is the key to answering this question. Because x and y are both incremented inside the synchronized method they are always incremented together. Also keep in mind that the two threads share the same reference to the Q126 object.
Also note that because of the infinite loop at line 13, only one thread ever gets to execute.
Discussion:
14 comments Page 2 of 2.
Vasavi said:
1 decade ago
What is the use of synchronized keyword? what it work's here?
Vasavi said:
1 decade ago
Why here no error will come? because here we cal start() method 2 times with saMe object.
Alex said:
1 decade ago
Oh, I got it - they are int primitives, zero by default. :)
Alex said:
1 decade ago
How it will work D-way if 'x' and 'y' are never initialized and are private, so there's no access to set them out of this class?
Not to mention comma and space (", ") appeared in your answer in example output.
Thanks, Alex.
Not to mention comma and space (", ") appeared in your answer in example output.
Thanks, Alex.
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