Consider the following Java program. What is the output of the program?

Question

BlackTom AI · Accepted Answer

Field access in Java is determined by the compile-time type of the reference (field hiding), not by the actual runtime object type.

1. Relevant knowledge points and concepts
- Field hiding: If a subclass declares a field with the same name as a superclass, the subclass field hides the superclass field. Both exist, but which one is accessed depends on the reference type.
- Binding rules: Field access is resolved at compile time using the reference's static type. Method calls use dynamic dispatch (runtime polymorphism), but fields do not.

2. Step-by-step reasoning
- class A declares int x = 10;
- class B extends A and declares int x = 20; (this hides A.x)
- In main: A obj = new B(); — the static (compile-time) type of obj is A, the runtime object is an instance of B.
- System.out.print(obj.x); — because obj is declared as type A, the compiler resolves obj.x to A.x (the field in class A), not B.x.

3. Intermediate state / evaluation
- After A obj = new B(); there exist two fields for the object: A.x == 10 and B.x == 20, but accessed field depends on reference:
  - Using reference declared as A: obj.x → A.x → 10
  - Using reference declared as B (e.g. ((B)obj).x): ((B)obj).x → B.x → 20

4. Why the correct answer is correct
- The language specification requires field access to be based on the compile-time type. Therefore obj.x prints 10.

5. Why other options are incorrect
- a. Compilation error: incorrect — code compiles fine; hiding fields is allowed.
- c. Runtime error: incorrect — no exception occurs.
- d. 20: incorrect — that would assume dynamic dispatch for fields, which Java does not do.

6. Conclusion
The program prints 10. This matches the provided correct answer: b. 10.

Consider the following Java program. What is the output of the program?单项选择题

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