Question 256 of 481
1Z0-811 Object-Oriented Programming Practice Question
A developer writes: Object obj = new String("Hello"); System.out.println(obj.length()); What will be the output?
⚠ Common exam trap
Oracle often tests the distinction between compile-time type and runtime type, trapping candidates who assume that because the object is a String, the length() method is automatically available on any reference to it.
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
✓
Compilation error
The code fails to compile because the reference variable 'obj' is of type Object, which does not have a length() method. The length() method is defined in the String class, not in Object. Since the compiler checks the declared type (Object) for method availability, it does not find length() and produces a compilation error.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
null
Why it's wrong here
Incorrect; this does not produce null output.
- ✗
Runtime exception
Why it's wrong here
Incorrect; it fails at compile time, not runtime.
- ✗
5
Why it's wrong here
Incorrect; although the object is a String, the reference type is Object, which does not have length() method.
- ✓
Compilation error
Why this is correct
The code produces a compilation error because the `obj` variable is declared with the static type `Object`. Although the runtime object is a `String`, the Java compiler performs method resolution based solely on the variable's declared type. The `Object` class does not possess a `length()` method. Consequently, attempting to invoke `obj.length()` fails during compilation, as the method signature is not found within the `Object` class's accessible members.
- ✗
0
Why it's wrong here
Incorrect; length() is not called.
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Last reviewed: Jun 30, 2026
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