1Z0-811 Binary Numeric Promotion Practice Question
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?
⚠ Common exam trap
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.
Answer choices
Why each option matters
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'.
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'.
- ✗
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
A runtime error would occur only if the code attempted an illegal operation, such as accessing an invalid index or dividing by zero, but the exhibit shows no such operation; the loop iterates within array bounds and performs only integer arithmetic. This option is tempting because runtime errors are common when a program tries to use an uninitialised variable or exceed array length, and in a scenario where the loop condition mistakenly allowed an out-of-bounds access, a runtime exception would indeed be the correct answer.
- ✗
Compilation error
Why it's wrong here
Incorrect. The code compiles successfully.
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