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In Computers and Technology / High School | 2025-07-08

What will be the output?

```java
public interface IPrint {
protected void printMe();
}

public class PrintTest implements IPrint {
public String printMe() {
System.out.println("Me");
}

public static void main(String[] args) {
PrintTest test = new PrintTest();
test.printMe();
}
}
```

A. Me

C. Run time error

Asked by daphnehenze4234

Answer (2)

The code provided is intended to demonstrate the implementation and use of an interface in Java. However, there is an error in this code which will result in a compilation error, not a runtime error, because of the following reasons:

Interface Method Visibility : In Java, all methods within an interface must be public. Declaring a method as "protected" in an interface is not allowed. Therefore, the declaration of protected void printMe(); in the IPrint interface is incorrect.

Method Signature Mismatch : In the PrintTest class, the printMe method does not correctly override the method from the IPrint interface. The printMe method in the PrintTest class mistakenly returns a String and prints "Me" using System.out.println("Me");. However, it should strictly match the interface method's signature (which is defined with a return type of void).


Here's the corrected code:
public interface IPrint { void printMe(); // by default, it's public }
public class PrintTest implements IPrint { public void printMe() { // method signature now matches System.out.println("Me"); } public static void main(String[] args) { PrintTest test = new PrintTest(); test.printMe(); } }
When the corrected code is compiled and executed, the output will be printed as:
"Me"
The answer to the original question, before corrections, would actually lead to a compilation error due to the interface and method signature mismatches, which isn't one of the given options. However, it's important to note the correct implementation for good programming practices and understanding.

Answered by OliviaMariThompson | 2025-07-21

The code will result in a compilation error due to two main issues: the method in the interface is incorrectly declared as protected, and the method in the class has a mismatched return type. The corrected method signature should be public void printMe(). Therefore, the answer is not among the choices provided as it doesn't compile.
;

Answered by OliviaMariThompson | 2025-07-22