Quick Answer
The answer is any value between 1000 and 2000 inclusive. This occurs because `counter++` is a classic example of a non-atomic increment race condition; the operation is actually three separate steps—read the value, increment it, and write it back—and when two threads interleave these steps without synchronization, one thread’s increment can overwrite the other’s, causing lost updates. On the Oracle Java Foundations 1Z0-811 exam, this question tests your understanding of thread safety and the visibility of shared mutable state, often appearing as a trap where students assume the result is always 2000. The key insight is that without atomicity or synchronization, the final count can be as low as 1000 if one thread’s increments are repeatedly overwritten. Memory tip: think of “read-increment-write” as three separate tickets—if two people grab tickets at the same time, one gets lost.
This 1Z0-811 practice question tests your understanding of java basics and syntax. Read the scenario carefully and evaluate each option against the stated constraints before committing to an answer. After answering, compare your reasoning against the explanation and wrong-answer breakdown below. Once you have made your selection, read the full explanation to reinforce the concept and understand why each distractor is designed to mislead on exam day.
A class has a static variable counter initialized to 0. Two threads increment counter 1000 times each using counter++. What is a possible final value of counter?
1000
Why wrong: Too low; both threads perform increments.
0
Why wrong: Impossible because each thread increments counter.
2000
Why wrong: Theoretical maximum but not guaranteed due to race condition.
Any value between 1000 and 2000 inclusive
Actual range due to race condition.
Answer choices
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
Any value between 1000 and 2000 inclusive
Option D is correct because `counter++` is not an atomic operation; it consists of three steps: read, increment, and write. When two threads execute this non-atomic operation concurrently without synchronization, interleaving can cause lost updates, resulting in a final value that is less than 2000 but at least 1000 (since each thread increments at least once). The possible final value is any integer between 1000 and 2000 inclusive.
Key principle: Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Answer analysis
For each option: why learners choose it and why it is or isn't the right answer here.
1000
Why it's wrong here
Too low; both threads perform increments.
0
Why it's wrong here
Impossible because each thread increments counter.
2000
Why it's wrong here
Theoretical maximum but not guaranteed due to race condition.
Any value between 1000 and 2000 inclusive
Why this is correct
Actual range due to race condition.
Related concept
Read the scenario before looking for a memorised answer.
Common exam traps
Oracle often tests the misconception that `counter++` is atomic, leading candidates to incorrectly choose 2000, when in fact the non-atomic nature allows any value between 1000 and 2000 due to lost updates.
Detailed technical explanation
Under the hood, `counter++` compiles to bytecode instructions like `getstatic`, `iconst_1`, `iadd`, and `putstatic`, which are not atomic on the Java Virtual Machine (JVM). The Java Memory Model (JMM) allows threads to cache variables locally, so without `volatile` or `synchronized`, updates may not be visible across threads immediately. In real-world scenarios, this race condition is a classic concurrency bug that can lead to data corruption in multi-threaded counters, inventory systems, or any shared mutable state.
Key takeaway
Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Real-world example
A practitioner preparing for the 1Z0-811 exam encounters this exact type of scenario on the job. The correct answer here is not the most general option — it is the best answer for the specific constraint described. Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option. Real exam questions reward reading the full scenario before eliminating options, because the constraint defines which answer fits.
What to study next
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
Java Basics and Syntax — study guide chapter
Learn the concepts, then practise the questions
Java Basics and Syntax practice questions
Targeted practice on this topic area only
All 1Z0-811 questions
509 questions across all exam domains
Oracle Java Foundations 1Z0-811 study guide
Full concept coverage aligned to exam objectives
1Z0-811 practice test guide
How to use practice tests most effectively before exam day
Related practice questions
Use these pages to review the topic behind this question. This is how one missed question becomes focused revision.
Practise 1Z0-811 questions linked to What is Java.
Practise 1Z0-811 questions linked to Java Basics and Syntax.
Practise 1Z0-811 questions linked to Primitives, Strings and Operators.
Practise 1Z0-811 questions linked to Control Flow and Loops.
Practise 1Z0-811 questions linked to Arrays and Methods.
Practise 1Z0-811 questions linked to Object-Oriented Programming.
Practise 1Z0-811 questions linked to Exception Handling and Development Tools.
Practise 1Z0-811 questions linked to 1Z0-811 fundamentals.
Practise 1Z0-811 questions linked to 1Z0-811 scenario.
Practise 1Z0-811 questions linked to 1Z0-811 troubleshooting.
Practice this exam
Short sessions build daily habit. Longer sessions build exam-day stamina. Try a timed session to simulate real conditions.
FAQ
Java Basics and Syntax — This question tests Java Basics and Syntax — Read the scenario before looking for a memorised answer..
The correct answer is: Any value between 1000 and 2000 inclusive — Option D is correct because `counter++` is not an atomic operation; it consists of three steps: read, increment, and write. When two threads execute this non-atomic operation concurrently without synchronization, interleaving can cause lost updates, resulting in a final value that is less than 2000 but at least 1000 (since each thread increments at least once). The possible final value is any integer between 1000 and 2000 inclusive.
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
Read the scenario before looking for a memorised answer.
About these practice questions
Courseiva creates original exam-style practice questions with explanations and wrong-answer analysis. It does not publish real exam questions, exam dumps, or protected exam content. Learn why practice questions differ from exam dumps →
Last reviewed: Jun 30, 2026
This 1Z0-811 practice question is part of Courseiva's free Oracle certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the 1Z0-811 exam.
Question Discussion
Share a tip, memory trick, or ask about the reasoning behind this question. Do not post real exam questions, leaked content, braindumps, or copyrighted exam material. Comments are moderated and may be removed without notice.
Sign in to join the discussion.