Given the code fragment: ``java final List<String> list = new CopyOnWriteArrayList<>(); final AtomicInteger ai = new AtomicInteger(0); final CyclicBarrier barrier = new CyclicBarrier(2, new…

Question

final List<String> list = new CopyOnWriteArrayList<>();
final AtomicInteger ai = new AtomicInteger(0);
final CyclicBarrier barrier = new CyclicBarrier(2, new Runnable() {
 public void run() { System.out.print(list); }
});

Runnable r = new Runnable() {
 public void run() {
 try {
 Thread.sleep(1000 * ai.incrementAndGet());
 list.add("X");
 barrier.await();
 } catch (Exception ex) { }
 }
};

new Thread(r).start();
new Thread(r).start();
new Thread(r).start();
new Thread(r).start();

Options

• A[X, X]
• B[X, X, X]
• C[X]
• D[X, X, X, X]

• A
  76% (51)
• B
  3% (2)
• C
  7% (5)
• D
  13% (9)

Explanation

Option A is correct because the CyclicBarrier is configured with 2 parties, meaning its action fires every time exactly 2 threads call await(). With four threads sleeping for 1, 2, 3, and 4 seconds respectively, the first two threads each add "X" to the list before calling await(), triggering the barrier action which prints [X, X]. Option C ([X]) is impossible because 2 threads must reach the barrier before it trips, guaranteeing at least 2 elements in the list; similarly, B ([X, X, X]) would require parties=3, but the barrier here only needs 2 - it never waits for a third. Option D ([X, X, X, X]) is what gets printed when the barrier fires a second time (threads 3 and 4 arrive later and form a new cycle), not the first observable result. Memory tip: "Parties value = X count per print" - whatever number you pass to CyclicBarrier(n, ...), that's exactly how many threads add their element before each barrier action fires.

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