Quick Answer
The correct answer is that the int argument is implicitly cast to double. This happens because Java automatically performs a widening primitive conversion when a method expecting a double parameter receives an int argument, as double can represent all int values without any loss of precision. In the given scenario, the int value 1000 is safely converted to 1000.0 before the calculation, which is why the output remains 50.0 even if the method’s return type changes to int. On the Oracle Java Foundations 1Z0-811 exam, this concept tests your understanding of implicit casting and method invocation with mismatched primitive types. A common trap is assuming that an explicit cast is always required, but Java handles widening conversions automatically. Remember the memory tip: “Widen freely, narrow carefully”—implicit casting only works when moving from a smaller to a larger data type, like int to double.
This 1Z0-811 practice question tests your understanding of primitives, strings and operators. 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 banking application uses a method to calculate interest: double calculateInterest(double balance) { return balance * 0.05; }. The method is called with an int argument: int accountBalance = 1000; double interest = calculateInterest(accountBalance); System.out.println(interest); The output is 50.0, but the expected output is 50.0. However, the developer notices that if the method is changed to return int, the output becomes 50.0 as well. Which statement about implicit casting is true?
The double result is implicitly cast to int.
Why wrong: Narrowing would require explicit cast.
The int argument is implicitly cast to double.
Correct widening conversion.
The code fails to compile because of type mismatch.
Why wrong: Compiles fine.
The multiplication result is automatically rounded.
Why wrong: Not automatic rounding.
Answer choices
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
The int argument is implicitly cast to double.
Option B is correct because when a method expecting a `double` parameter is called with an `int` argument, Java performs implicit widening primitive conversion (casting) from `int` to `double`. This is safe because `double` can represent all `int` values without loss of precision. The `int` value 1000 is automatically converted to `1000.0` before being used in the calculation.
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.
The double result is implicitly cast to int.
Why it's wrong here
Narrowing would require explicit cast.
The int argument is implicitly cast to double.
Why this is correct
Correct widening conversion.
Related concept
Read the scenario before looking for a memorised answer.
The code fails to compile because of type mismatch.
Why it's wrong here
Compiles fine.
The multiplication result is automatically rounded.
Why it's wrong here
Not automatic rounding.
Common exam traps
Oracle often tests the misconception that implicit casting can happen in both directions (widening and narrowing) or that the return type determines the cast direction; the trap here is that candidates may think the `double` result is cast to `int` when the return type changes, but in fact the implicit cast occurs on the argument, not the result.
Detailed technical explanation
Implicit casting (widening primitive conversion) in Java follows the rule that smaller types (byte, short, char, int, long) can be automatically promoted to larger types (float, double) without explicit cast, as per JLS §5.1.2. This is safe because the target type has a wider range and precision. In contrast, narrowing conversions (e.g., double to int) require explicit casting and may lose data. In real-world banking applications, using `double` for monetary calculations can lead to rounding errors; `BigDecimal` is preferred for precision, but the implicit casting behavior remains the same.
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
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FAQ
Primitives, Strings and Operators — This question tests Primitives, Strings and Operators — Read the scenario before looking for a memorised answer..
The correct answer is: The int argument is implicitly cast to double. — Option B is correct because when a method expecting a `double` parameter is called with an `int` argument, Java performs implicit widening primitive conversion (casting) from `int` to `double`. This is safe because `double` can represent all `int` values without loss of precision. The `int` value 1000 is automatically converted to `1000.0` before being used in the calculation.
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 30, 2026
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