Given: ``java public interface Moveable<Integer> { public default void walk (Integer distance) { System.out.println("Walking")); } public void run(Integer distance); } `` Which statement is true?

Question

public interface Moveable<Integer> {
 public default void walk (Integer distance) {
 System.out.println("Walking"));
 }
 public void run(Integer distance);
}

Options

• AMoveable can be used as below: Moveable<Integer> animal = n -> System.out.println("Running" + n); animal.run(100); animal.walk(20);
• BMoveable can be used as below: Moveable<Integer> animal = n -> n + 10; animal.run(100); animal.walk(20);
• CMoveable can be used as below: Moveable<Integer> animal = (Integer n) -> System.out.println(n); animal.run(100); Moveable.walk(20);
• DMoveable cannot be used in a lambda expression.

• A
  3% (2)
• B
  5% (3)
• C
  83% (54)
• D
  9% (6)

Explanation

Note on this question: The stated answer of C appears to contain an error that undermines it as the correct choice.

Moveable.walk(20) in option C attempts to call a default method as if it were static on the interface - this does not compile in Java. Only static interface methods can be invoked as Interface.method(); default methods require an instance reference.

Option A is actually the most valid implementation: the lambda n -> System.out.println("Running" + n) correctly implements the single abstract method run, and animal.walk(20) correctly invokes the default method on the instance.

Why the other options fail:

• B is wrong because n -> n + 10 produces a value expression, but run has a void return type - the lambda body is incompatible.
• D is wrong because Moveable has exactly one abstract method (run), making it a functional interface fully eligible for lambda use.
• C is wrong because Moveable.walk(20) is an invalid static-style call on a default (instance) method.

Memory tip: default ≠ static. A default method lives on instances - always call it as instance.method(). Only static interface methods use the Interface.method() syntax. If your exam insists C is correct, push back - Java's compiler will reject Moveable.walk(20).

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