Java Concurrent Thread Execution Order

Managing concurrent code execution
Answer Correct answer: A — The program prints the values in random order due to unsynchronized concurrent thread execution.

Given the code fragment: and Which is true? - image - image

  1. The program prints t1 : 1 : t2 : 1: t1 : 2 : t2 : 2 : in random order. Correct Answer
  2. The program prints t1 : 1 : t2 : 1: t1 : 2 : t2 : 2 :
  3. The program prints t1 : 1 : t2 : 1: t1 : 1 : t2 : 1 : indefinitely.
  4. The program prints an exception.

Community Votes

A
100%

100% of anonymous learners picked answer A. Votes are pick records left by other test-takers — they are not the verified answer.

Community Insight

The core concept tested is thread safety and scheduling; the common trap is assuming a fixed execution order due to code structure rather than recognizing the lack of synchronization guarantees.

This question tests the unpredictability of thread scheduling in Java concurrent programming. It establishes that without explicit synchronization, the interleaving of operations from multiple threads is non-deterministic.

Candidates often choose Option B, incorrectly assuming that thread t1 will always complete its loop before t2 starts or that they will strictly alternate, ignoring that the OS scheduler decides execution order.

Community Discussion (7 comments)

ASPushkin 👍 1 Selected: A
Only one thread t1 has access to Test.x. Only one thread t2 has access to Test.xObj. Monitor (implicit synchronization) of Instance class Test is not used. There is no thread interference at all here. Even if we remove keyword volatile and exchange optimistic locking AtomicInteger variable regular Integer the thread execution order is generally unpredictable because of not absolute contract with underlying OS in order to perform threads scheduling.
Naoufal18 👍 1 Selected: A
n the given code, two threads (t1 and t2) are running concurrently and printing the values of x and xObj, respectively. The execution order of the threads can vary due to their concurrent nature. Here's the breakdown: Thread 1 (t1) is operating on the volatile int x variable. It increments x from 1 to 2 as part of the while (t.x < 3) loop. Thread 2 (t2) is operating on the AtomicInteger xObj. It increments xObj from 1 to 2 and then to 3 as part of the while (t.xObj.get() < 3) loop.
Uteman 👍 1
A is correct
xplorerpj 👍 1
A is the correct answer. The below gets printed in random Order. t2:1: t2:2: t1:1: t1:2:
minhdev 👍 2
A is correct answer
james2033 👍 2 Selected: A
// Result: // t2 : 1 : t2 : 2 : t1 : 1 : t1 : 2 : // t1 : 1 : t2 : 1 : t2 : 2 : t1 : 2 : // t1 : 1 : t2 : 1 : t1 : 2 : t2 : 2 : // t1 : 1 : t2 : 1 : t1 : 2 : t2 : 2 :
Samps 👍 2
A is correct tested

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Expert Analysis

Why the Answer Is Correct

The correct answer is A because the program involves two threads (t1 and t2) executing concurrently without any synchronization mechanisms like synchronized, Lock, or join. In Java, the thread scheduler determines when each thread runs. Since there is no guarantee of ordering between the increment/print cycles of t1 and t2, the output can appear in various interleaved orders, such as t2:1:t2:2:t1:1:t1:2 or t1:1:t2:1:t1:2:t2:2. The term 'random order' here refers to this non-deterministic interleaving.

Why the Other Options Are Wrong

Option B implies a strict, predictable sequence (e.g., all t1 outputs then all t2, or strict alternation), which is false because thread scheduling is preemptive and dependent on system load and priority. Option C suggests an infinite loop with identical output, which is incorrect because the variables x and xObj are incremented and eventually reach values >= 3, terminating the loops. Option D is incorrect because there are no runtime errors; volatile and AtomicInteger ensure visibility and atomicity of individual reads/writes, preventing exceptions during these simple operations.

Community Comment Notes

Community consensus strongly supports Option A. As user james2033 noted, the output shows varied sequences like t2: 1: t2: 2: t1: 1: t1: 2, confirming the unpredictable nature. ASPushkin highlighted that without monitor usage, there is no thread interference constraint forcing a specific order. Multiple users confirmed that this behavior was observed during testing, validating the non-deterministic result.

Exam Strategy

When analyzing concurrent Java code questions, first look for synchronization keywords (synchronized, wait/notify, Lock). If none are present, assume the execution order is non-deterministic unless the problem explicitly states otherwise (e.g., using join()). Always distinguish between thread visibility (handled by volatile/atomic classes) and thread ordering (requires synchronization).

Frequently Asked Questions

Does volatile ensure thread execution order?

No. Volatile ensures visibility of changes across threads but does not impose any ordering or mutual exclusion on execution flow.

Why doesn't AtomicInteger cause blocking?

AtomicInteger uses CAS (Compare-And-Swap) instructions at the hardware level, allowing non-blocking increments without acquiring locks.

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