1Z0-811 Binary Numeric Promotion Practice Question
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. A key principle to apply: binary Numeric Promotion. 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);
}
}
Refer to the exhibit. What is the output?
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);
}
}
A
int
Correct. The argument is an int due to the cast, so the int version is called, printing 'int'.
B
double
Why wrong: Incorrect. The argument is not a double because the cast makes it an int.
C
Runtime error
Why wrong: Incorrect. The code compiles and runs without error.
D
Compilation error
Why wrong: Incorrect. The code compiles successfully.
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 performs an integer division because the double y is cast to int, making both operands int. The result is an int value of 2. The method typeOf(int) is called, which prints the string 'int'.
Key principle: Binary Numeric Promotion
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
Correct. The argument is an int due to the cast, so the int version is called, printing 'int'.
Related concept
Binary Numeric Promotion
✗
double
Why it's wrong here
Incorrect. The argument is not a double because the cast makes it an int.
✗
Runtime error
Why it's wrong here
Incorrect. The code compiles and runs without error.
✗
Compilation error
Why it's wrong here
Incorrect. The code compiles successfully.
Common exam traps
Common exam trap: answer the scenario, not the keyword
Candidates often overlook the explicit cast and think the division is between int and double, leading to a double result. However, with the cast, the division is integer, and the method overload resolution selects the int version.
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
Binary Numeric Promotion
Method Overloading
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
Binary Numeric Promotion
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. Binary Numeric Promotion 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.
Review binary Numeric Promotion, then practise related 1Z0-811 questions on the same topic to reinforce the concept.
Java Basics and Syntax — This question tests Java Basics and Syntax — Binary Numeric Promotion.
What is the correct answer to this question?
The correct answer is: int — The code performs an integer division because the double y is cast to int, making both operands int. The result is an int value of 2. The method typeOf(int) is called, which prints the string 'int'.
What should I do if I get this 1Z0-811 question wrong?
Review binary Numeric Promotion, then practise related 1Z0-811 questions on the same topic to reinforce the concept.
What is the key concept behind this question?
Binary Numeric Promotion
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Question Discussion
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