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Oracle Java SE 21 Developer Professional Sample Questions (Q10-Q15):

NEW QUESTION # 10
Given:
java
interface SmartPhone {
boolean ring();
}
class Iphone15 implements SmartPhone {
boolean isRinging;
boolean ring() {
isRinging = !isRinging;
return isRinging;
}
}
Choose the right statement.

A. Iphone15 class does not compile
B. SmartPhone interface does not compile
C. Everything compiles
D. An exception is thrown at running Iphone15.ring();

Answer: A

Explanation:
In this code, the SmartPhone interface declares a method ring() with a boolean return type. The Iphone15 class implements the SmartPhone interface and provides an implementation for the ring() method.
However, in the Iphone15 class, the ring() method is declared without the public access modifier. In Java, when a class implements an interface, it must provide implementations for all the interface's methods with the same or a more accessible access level. Since interface methods are implicitly public, the implementing methods in the class must also be public. Failing to do so results in a compilation error.
Therefore, the Iphone15 class does not compile because the ring() method is not declared as public.

NEW QUESTION # 11
Given:
java
package vehicule.parent;
public class Car {
protected String brand = "Peugeot";
}
and
java
package vehicule.child;
import vehicule.parent.Car;
public class MiniVan extends Car {
public static void main(String[] args) {
Car car = new Car();
car.brand = "Peugeot 807";
System.out.println(car.brand);
}
}
What is printed?

A. An exception is thrown at runtime.
B. Compilation fails.
C. Peugeot 807
D. Peugeot

Answer: B

Explanation:
In Java,protected memberscan only be accessedwithin the same packageor bysubclasses, but there is a key restriction:
* A protected member of a superclass is only accessible through inheritance in a subclass but not through an instance of the superclass that is declared outside the package.
Why does compilation fail?
In the MiniVan class, the following line causes acompilation error:
java
Car car = new Car();
car.brand = "Peugeot 807";
* The brand field isprotectedin Car, which means it isnot accessible via an instance of Car outside the vehicule.parent package.
* Even though MiniVan extends Car, itcannotaccess brand using a Car instance (car.brand) because car is declared as an instance of Car, not MiniVan.
* The correct way to access brand inside MiniVan is through inheritance (this.brand or super.brand).
Corrected Code
If we change the MiniVan class like this, it will compile and run successfully:
java
package vehicule.child;
import vehicule.parent.Car;
public class MiniVan extends Car {
public static void main(String[] args) {
MiniVan minivan = new MiniVan(); // Access via inheritance
minivan.brand = "Peugeot 807";
System.out.println(minivan.brand);
}
}
This would output:
nginx
Peugeot 807
Key Rule from Oracle Java Documentation
* Protected membersof a class are accessible withinthe same packageand tosubclasses, butonly through inheritance, not through a superclass instance declared outside the package.
References:
* Java SE 21 & JDK 21 - Controlling Access to Members of a Class
* Java SE 21 & JDK 21 - Inheritance Rules

NEW QUESTION # 12
Given:
java
import java.io.*;
class A implements Serializable {
int number = 1;
}
class B implements Serializable {
int number = 2;
}
public class Test {
public static void main(String[] args) throws Exception {
File file = new File("o.ser");
A a = new A();
var oos = new ObjectOutputStream(new FileOutputStream(file));
oos.writeObject(a);
oos.close();
var ois = new ObjectInputStream(new FileInputStream(file));
B b = (B) ois.readObject();
ois.close();
System.out.println(b.number);
}
}
What is the given program's output?

A. Compilation fails
B. NotSerializableException
C. 0
D. ClassCastException
E. 1

Answer: D

Explanation:
In this program, we have two classes, A and B, both implementing the Serializable interface, and a Test class with the main method.
Program Flow:
* Serialization:
* An instance of class A is created and assigned to the variable a.
* An ObjectOutputStream is created to write to the file "o.ser".
* The object a is serialized and written to the file.
* The ObjectOutputStream is closed.
* Deserialization:
* An ObjectInputStream is created to read from the file "o.ser".
* The program attempts to read an object from the file and cast it to an instance of class B.
* The ObjectInputStream is closed.
Analysis:
* Serialization Process:
* The object a is an instance of class A and is serialized into the file "o.ser".
* Deserialization Process:
* When deserializing, the program reads the object from the file and attempts to cast it to class B.
* However, the object in the file is of type A, not B.
* Since A and B are distinct classes with no inheritance relationship, casting an A instance to B is invalid.
Exception Details:
* Attempting to cast an object of type A to type B results in a ClassCastException.
* The exception message would be similar to:
pgsql
Exception in thread "main" java.lang.ClassCastException: class A cannot be cast to class B Conclusion:
The program compiles successfully but throws a ClassCastException at runtime when it attempts to cast the deserialized object to class B.

NEW QUESTION # 13
What is the output of the following snippet? (Assume the file exists)
java
Path path = Paths.get("C:homejoefoo");
System.out.println(path.getName(0));

A. C:
B. Compilation error
C. home
D. C
E. IllegalArgumentException

Answer: C

Explanation:
In Java's java.nio.file package, the Path class represents a file path in a file system. The Paths.get(String first, String... more) method is used to create a Path instance by converting a path string or URI.
In the provided code snippet, the Path object path is created with the string "C:homejoefoo". This represents an absolute path on a Windows system.
The getName(int index) method of the Path class returns a name element of the path as a Path object. The index is zero-based, where index 0 corresponds to the first element in the path's name sequence. It's important to note that the root component (e.g., "C:" on Windows) is not considered a name element and is not included in this sequence.
Therefore, for the path "C:homejoefoo":
* Root Component:"C:"
* Name Elements:
* Index 0: "home"
* Index 1: "joe"
* Index 2: "foo"
When path.getName(0) is called, it returns the first name element, which is "home". Thus, the output of the System.out.println statement is home.

NEW QUESTION # 14
Which two of the following aren't the correct ways to create a Stream?

A. Stream stream = Stream.of("a");
B. Stream stream = new Stream();
C. Stream stream = Stream.ofNullable("a");
D. Stream stream = Stream.empty();
E. Stream<String> stream = Stream.builder().add("a").build();
F. Stream stream = Stream.generate(() -> "a");
G. Stream stream = Stream.of();

Answer: B,E

Explanation:
In Java, the Stream API provides several methods to create streams. However, not all approaches are valid.

NEW QUESTION # 15
......

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Oracle Java SE 21 Developer Professional Sample Questions (Q10-Q15):

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