The correct answer is that the code will not compile, producing a compilation error. This occurs because the expression `x / y` involves an int and a double, triggering binary numeric promotion, which automatically promotes the int to a double before performing the division; the result is a double value of 2.0. However, the assignment `int result = x / y;` attempts to store that double into an int variable without an explicit cast, and Java does not allow implicit narrowing conversion from double to int, causing the compiler to reject the code. On the Oracle Java Foundations 1Z0-811 exam, this question tests your understanding of type promotion in mixed arithmetic and the strictness of Java’s type safety—a common trap is assuming the result is simply truncated to 2. Instead, remember that any arithmetic involving a double promotes all operands to double, and the result cannot be silently narrowed. A helpful memory tip: “Double dominates—once a double appears, the whole operation floats, and you must cast to land back on int.”
1Z0-811 Java Basics and Syntax Practice Question
This 1Z0-811 practice question tests your understanding of java basics and syntax. The scenario asks you to isolate a root cause — eliminate options that address a different problem before choosing. 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.
Exhibit
public class Test {
public void print(int i) { System.out.println("int"); }
public void print(double d) { System.out.println("double"); }
public static void main(String[] args) {
Test t = new Test();
t.print(10);
}
}
public class Test {
public void print(int i) { System.out.println("int"); }
public void print(double d) { System.out.println("double"); }
public static void main(String[] args) {
Test t = new Test();
t.print(10);
}
}
A
int
The call matches the int version exactly.
B
double
Why wrong: The int version is a better match; no conversion needed.
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
int
The code declares an integer variable `x` with value 5 and a double variable `y` with value 2.5. The expression `x / y` performs division between an int and a double, which triggers binary numeric promotion, converting the int to a double before division. The result is 2.0 (since 5.0 / 2.5 = 2.0), which is a double. However, the assignment `int result = x / y;` attempts to store a double into an int without an explicit cast, causing a compilation error because double cannot be implicitly narrowed to int.
Key principle: Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
✓
int
Why this is correct
The call matches the int version exactly.
Related concept
Read the scenario before looking for a memorised answer.
✗
double
Why it's wrong here
The int version is a better match; no conversion needed.
✗
Runtime error
Why it's wrong here
No runtime error occurs.
✗
Compilation error
Why it's wrong here
The code compiles fine.
Common exam traps
Common exam trap: answer the scenario, not the keyword
Oracle often tests the distinction between implicit widening (allowed) and implicit narrowing (disallowed), tricking candidates into thinking the division result's type matches the left operand or that Java will automatically truncate the double to int.
Detailed technical explanation
How to think about this question
Binary numeric promotion is defined in JLS §5.6.2: when an operator applies to operands of different types, the smaller type is widened to the larger. Here, the int `x` is promoted to double for the division, yielding a double result. The subsequent assignment to an int variable violates Java's strict type-checking rules (JLS §5.2), which require explicit narrowing conversion for double-to-int. This is a common compile-time error that prevents accidental loss of precision.
KKey Concepts to Remember
Read the scenario before looking for a memorised answer.
Find the constraint that changes the correct option.
Eliminate answers that are true in general but not in this case.
TExam Day Tips
→Watch for words such as best, first, most likely and least administrative effort.
→Review why wrong options are wrong, not only why the correct option is correct.
Key takeaway
Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Real-world example
How this comes up in practice
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
Got this wrong? Here's your next step.
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
Java Basics and Syntax — This question tests Java Basics and Syntax — Read the scenario before looking for a memorised answer..
What is the correct answer to this question?
The correct answer is: int — The code declares an integer variable `x` with value 5 and a double variable `y` with value 2.5. The expression `x / y` performs division between an int and a double, which triggers binary numeric promotion, converting the int to a double before division. The result is 2.0 (since 5.0 / 2.5 = 2.0), which is a double. However, the assignment `int result = x / y;` attempts to store a double into an int without an explicit cast, causing a compilation error because double cannot be implicitly narrowed to int.
What should I do if I get this 1Z0-811 question wrong?
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
What is the key concept behind this question?
Read the scenario before looking for a memorised answer.
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Question Discussion
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