Quick Answer
The answer is {x=10, y=20}. This is correct because `computeIfAbsent` only executes the mapping function when the specified key is absent from the map; since "x" already exists with value 10, the lambda `k -> 30` is ignored and the original value remains unchanged. Conversely, `computeIfPresent` only executes its remapping function when the key is already present; "z" does not exist in the map, so the lambda `(k,v) -> 40` is never invoked and no new entry is added. On the Oracle Certified Professional Java SE 17 Developer 1Z0-829 exam, this question tests your precise understanding of conditional map updates—a common trap is assuming these methods always add or modify entries, when in fact they are conditional on key presence or absence. A reliable memory tip: think of "absent" as "add if absent" and "present" as "process if present"—the method name tells you exactly when the function runs.
This 1Z0-829 practice question tests your understanding of working with arrays and collections. 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.
Given: Map<String, Integer> map = new HashMap<>(); map.put("x", 10); map.put("y", 20); map.computeIfAbsent("x", k -> 30); map.computeIfPresent("z", (k,v) -> 40); System.out.println(map); What is the output?
Compilation fails
Why wrong: The code compiles successfully.
{x=30, y=20}
Why wrong: Incorrect: computeIfAbsent does not replace existing values.
{x=30, y=20, z=40}
Why wrong: Incorrect: neither key is modified.
{x=10, y=20, z=40}
Why wrong: Incorrect: computeIfPresent does not add absent keys.
{x=10, y=20}
Correct: both conditions are false, so map unchanged.
Answer choices
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
{x=10, y=20}
Option E is correct because `computeIfAbsent` only inserts a mapping if the key is absent, and since "x" already exists with value 10, the lambda is not executed and the value remains 10. `computeIfPresent` only modifies a mapping if the key is present, and since "z" is not in the map, the lambda is not executed and no entry is added. Thus the map remains {x=10, y=20}.
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.
Compilation fails
Why it's wrong here
The code compiles successfully.
{x=30, y=20}
Why it's wrong here
Incorrect: computeIfAbsent does not replace existing values.
{x=30, y=20, z=40}
Why it's wrong here
Incorrect: neither key is modified.
{x=10, y=20, z=40}
Why it's wrong here
Incorrect: computeIfPresent does not add absent keys.
{x=10, y=20}
Why this is correct
Correct: both conditions are false, so map unchanged.
Related concept
Read the scenario before looking for a memorised answer.
Common exam traps
The trap here is that candidates often confuse `computeIfAbsent` with `put` or `merge`, assuming it always updates the value, and similarly assume `computeIfPresent` will insert a new key-value pair, when in fact both methods only act conditionally based on key presence.
Detailed technical explanation
`computeIfAbsent` uses the provided mapping function only when the key is not already associated with a value (or is mapped to null). `computeIfPresent` only invokes the remapping function if the key is present and its value is non-null. Under the hood, both methods use internal `Node` structures in HashMap to check existence before applying the lambda, ensuring thread-safe atomicity in ConcurrentHashMap but not in HashMap.
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 1Z0-829 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
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FAQ
Working with Arrays and Collections — This question tests Working with Arrays and Collections — Read the scenario before looking for a memorised answer..
The correct answer is: {x=10, y=20} — Option E is correct because `computeIfAbsent` only inserts a mapping if the key is absent, and since "x" already exists with value 10, the lambda is not executed and the value remains 10. `computeIfPresent` only modifies a mapping if the key is present, and since "z" is not in the map, the lambda is not executed and no entry is added. Thus the map remains {x=10, y=20}.
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.
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Last reviewed: Jun 25, 2026
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