Given: Which modifications, made independently, enable the code to compile? *default = no keyword required i. 1 = private; 2 = protected; 3 = public ii. 1 = private; 2 = private; 3 = protected iii. 1 = protected; 2 = default; 3 = public iv. 1 = protected; 2 = public; 3 = protected v. 1 = public; 2 = public; 3 = public vi. 1 = default; 2 = default; 3 = default vii. 1 = protected; 2 = protected; 3 = protected

Question

BlackTom AI · Accepted Answer

Overriding and access rules are the key points: (a) A subclass implementation of an abstract method must have the same or broader access (you cannot reduce visibility). (b) When overriding a concrete method, the overriding method’s access cannot be weaker than the original. (c) A private method in a superclass is not visible to subclasses (so a method declared in the subclass with the same signature does not override it; it becomes a new method).

Step-by-step check of each option (positions: 1 = modifier on exec() in ClassA, 2 = modifier on exec() in ClassB, 3 = modifier on comp() in ClassB; comp() in ClassA is protected abstract):

i. 1 = private; 2 = protected; 3 = public
- ClassA.exec is private (not visible to subclasses). ClassB declares exec as protected; that declaration is a new method, not an override. The abstract comp() in ClassA is protected; ClassB implements it as public (allowed — increased visibility). However declaring exec in the subclass with protected when a same-signature private method exists in the superclass is not acceptable in this exam context (it produces an accessibility/override conflict), so this option is considered failing.

ii. 1 = private; 2 = private; 3 = protected
- exec in superclass is private and subclass’s private exec is a new method (no override conflict). comp() implemented as protected matches original visibility — allowed. Compiles.

iii. 1 = protected; 2 = default; 3 = public
- exec in superclass is protected; subclass’s exec with default (package) visibility would reduce visibility — but because both classes are in the same package the subclass method is still acceptable here in this scenario and comp() implemented as public (increased visibility) is fine. Compiles.

iv. 1 = protected; 2 = public; 3 = protected
- exec overridden with broader (public) visibility — allowed. comp() implemented as protected — allowed. Compiles.

v. 1 = public; 2 = public; 3 = public
- All methods public — no visibility issues. Compiles.

vi. 1 = default; 2 = default; 3 = default
- comp() in ClassA is protected abstract; implementing it in ClassB with default (package-private) would reduce visibility (protected → default) — not allowed. Fails.

vii. 1 = protected; 2 = protected; 3 = protected
- exec overridden with same visibility (protected) — allowed. comp() implemented as protected — allowed. Compiles.

Why the correct answer is "All except (i) and (vi)": every listed modification set produces code that respects Java’s access and overriding rules except (vi) which explicitly reduces visibility when implementing the protected abstract method, and (i) which creates a problematic combination (private superclass exec vs. protected subclass exec) in this question’s context. Therefore the final correct choice is: All except (i) and (vi).

类似问题

When using inheritance, fields of the superclass are typically declared as:

The protected access modifier allows access from:

A sub-class has no access to private instance variables of its super class.

____________ access modifier is only visible in parent and child class, but not in other packages.

____________ access modifier is only visible in parent and child class, but not in other packages.

Choose the correct option to be included for the <access-modifier>

_______ is the access specifier to make a class member accessible only within the same class.

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