Quick Answer
The correct answer is that the result is truncated to int. This occurs because the compound assignment operator += performs an implicit narrowing conversion from double to int when used with an int variable and a double operand, discarding the fractional part rather than rounding. On the Oracle Java Foundations 1Z0-811 exam, this tests your understanding of how compound assignment int double narrowing differs from a simple arithmetic assignment, which would require an explicit cast to compile. A common trap is assuming the operation would cause a compilation error or runtime exception, but compound operators like +=, -=, and *= automatically apply a hidden cast, making the narrowing legal. To remember this, think of the mnemonic “C.A.S.T. for free” — Compound Assignment Saves the cast, Truncating the fractional part.
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 method has parameters: int x, double y. It performs x += y; and returns x. What is the range behavior?
Compilation error: cannot apply += between int and double.
Why wrong: Compound assignment compiles.
Lossy conversion causes runtime exception.
Why wrong: No exception; value is truncated.
Result is truncated to int.
Implicit narrowing cast.
x is promoted to double, result is double.
Why wrong: x remains int due to cast.
Answer choices
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
Result is truncated to int.
Option C is correct because when the compound assignment operator += is used with an int and a double, the right-hand operand (double) is implicitly narrowed to int via a primitive narrowing conversion. This truncates the fractional part of the double value, and the result is stored as an int. The operation compiles without error and does not throw a runtime exception.
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 error: cannot apply += between int and double.
Why it's wrong here
Compound assignment compiles.
Lossy conversion causes runtime exception.
Why it's wrong here
No exception; value is truncated.
Result is truncated to int.
Why this is correct
Implicit narrowing cast.
Related concept
Read the scenario before looking for a memorised answer.
x is promoted to double, result is double.
Why it's wrong here
x remains int due to cast.
Common exam traps
The trap here is that candidates mistakenly believe the result is promoted to double (option D) because they focus on the binary numeric promotion during the addition, forgetting that the compound assignment operator includes an implicit narrowing cast back to the left-hand variable's type.
Detailed technical explanation
Under the hood, the compound assignment operator E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1. This means the double value is first added to the int (with binary numeric promotion promoting the int to double), then the result is explicitly cast to int, truncating the fractional part. This behavior is specified in JLS §15.26.2 and is a common source of precision loss in financial calculations where fractional cents are inadvertently dropped.
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: Result is truncated to int. — Option C is correct because when the compound assignment operator += is used with an int and a double, the right-hand operand (double) is implicitly narrowed to int via a primitive narrowing conversion. This truncates the fractional part of the double value, and the result is stored as an int. The operation compiles without error and does not throw a runtime exception.
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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