Question

Given:

public class Thread1 {
int x = 0;

public class Runner implements Runnable {

    public  void run() {
        int current = 0;
        for (int i = 0; i < 4; i++) {
            current = x;
            System.out.print(current + " ");
            x = current + 2;
        }
    }
}

public void go() {
    Runnable r1 = new Runner();

    new Thread(r1).start();
    new Thread(r1).start();


}

public static void main(String[] args) {
    new Thread1().go();
}

}

Which two are possible results? (Choose two)

A. 0, 2, 4, 4, 6, 8, 10, 6,

B. 0, 2, 4, 6, 8, 10, 2, 4,

C. 0, 2, 4, 6, 8, 10, 12, 14,

D. 0, 0, 2, 2, 4, 4, 6, 6, 8, 8, 10, 10, 12, 12, 14, 14,

E. 0, 2, 4, 6, 8, 10, 12, 14, 0, 2, 4, 6, 8, 10, 12, 14,

I chose A and B but I’m not certain if that is the correct answer.

Answer 1

First of all D. and E. are out. As i scope local to the function, we know that there is going to be 8 numbers.

Now

B. 0, 2, 4, 6, 8, 10, 2, 4,

is incorrect. Why? In order print the last two positions each thread has to write the variable x at least two times. At this point the minimum value for x is 4. Which means that whatever the race conditions, the last two values should be greater or equal to 4.

EDIT
C is totally possible. Absence of synchronization doesn’t mean it is not possible to have what is called a serial execution (ie as if threads execute one after another). The lesson from concurrency is that you don’t know how threads will be interleaved

    C. 0, 2, 4, 6, 8, 10, 12, 14,

happens for example :

• when the first thread as time to finish updating x before the second start
• when the first thread is suspended before the first current = x;, the second finish executing and the first continue.
• tons of other cases.

As Brian Goetz said:

“Writing correct programs is hard; writing correct concurrent programs is harder.”