1Z0-811 Java Basics and Syntax Practice Question
Given methods: void print(Integer i) { System.out.println("Integer"); } void print(int i) { System.out.println("int"); } What is output of print(10);?
⚠ Common exam trap
Oracle often tests the misconception that autoboxing will always convert a primitive to its wrapper type, leading candidates to incorrectly choose the Integer overload, but the compiler prioritizes exact primitive matches to avoid unnecessary boxing.
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"
When calling print(10) with an integer literal, Java's method overloading resolution prefers the primitive int version over the Integer wrapper type because the literal 10 is a primitive int. The compiler selects the most specific matching method without requiring autoboxing, so the void print(int i) method is invoked, printing "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.
- ✗
Compilation error
Why it's wrong here
Incorrect. Both methods are valid.
- ✓
"int"
Why this is correct
Correct. Exact match with int parameter.
- ✗
"Integer"
Why it's wrong here
Incorrect. Autoboxing is not used when exact match exists.
- ✗
Runtime error
Why it's wrong here
Incorrect. No ambiguity at runtime.
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