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, not the runtime (actual) object's class. This is the key concept that gives the output.

1) Relevant knowledge points and concepts
- Instance methods in Java are polymorphic (virtual): which method runs depends on the runtime class.
- Instance fields are not polymorphic (they are hidden, not overridden): accessing a field uses the reference type at compile time.
- In the code, A defines int x = 10; B extends A and defines int x = 20; main declares the reference as type A: A obj = new B(); then prints obj.x.

2) Step-by-step reasoning
- Step 1: Class A has a field x with value 10.
- Step 2: Class B extends A and declares its own field x with value 20 (this hides A.x when accessed through a B reference).
- Step 3: In main, the variable obj is declared with compile-time type A but is initialized with a B instance (new B()).
- Step 4: When the compiler sees obj.x, it resolves the field lookup using obj's declared type A, so it binds to A.x.
- Step 5: At runtime, the program accesses the A.x value (10) stored in the object and prints it.

3) Intermediate considerations / examples
- If the code had printed ((B)obj).x it would have accessed B.x and printed 20.
- If x were accessed via a method overridden in B, method dispatch would be dynamic and B's method would run.

4) Why the correct answer is correct
Because field access is compile-time based on reference type A, obj.x refers to A.x whose value is 10.

5) Why other options are incorrect
- a. 20 is wrong because obj is declared as type A, so B.x is not selected.
- b. Compilation error is wrong: the code is legal (fields are accessible and hiding is allowed).
- d. Runtime error is wrong: no exception occurs.

Conclusion: The program prints 10, matching the provided correct answer (c. 10).

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