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Oracle Java SE 21 Developer Professional Sample Questions (Q41-Q46):

NEW QUESTION # 41
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. ClassCastException
B. Compilation fails
C. NotSerializableException
D. 0
E. 1

Answer: A

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 # 42
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. Peugeot
B. Peugeot 807
C. Compilation fails.
D. An exception is thrown at runtime.

Answer: C

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 # 43
Given:
java
var lyrics = """
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose
""";
for ( int i = 0, int j = 3; i < j; i++ ) {
System.out.println( lyrics.lines()
.toList()
.get( i ) );
}
What is printed?

A. Compilation fails.
B. vbnet
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose
C. An exception is thrown at runtime.
D. Nothing

Answer: A

Explanation:
* Error in for Loop Initialization
* The initialization part of a for loopcannot declare multiple variables with different types in a single statement.
* Error:
java
for (int i = 0, int j = 3; i < j; i++) {
* Fix:Declare variables separately:
java
for (int i = 0, j = 3; i < j; i++) {
* lyrics.lines() in Java 21
* The lines() method of String returns aStream<String>, splitting the string by line breaks.
* Calling .toList() on a streamconverts it to a list.
* Valid Code After Fixing the Loop:
java
var lyrics = """
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose
""";
for (int i = 0, j = 3; i < j; i++) {
System.out.println(lyrics.lines()
toList()
get(i));
}
* Expected Output After Fixing:
vbnet
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - String.lines()
* Java SE 21 - for Statement Rules

NEW QUESTION # 44
Given:
java
public class SpecialAddition extends Addition implements Special {
public static void main(String[] args) {
System.out.println(new SpecialAddition().add());
}
int add() {
return --foo + bar--;
}
}
class Addition {
int foo = 1;
}
interface Special {
int bar = 1;
}
What is printed?

A. Compilation fails.
B. 0
C. 1
D. It throws an exception at runtime.
E. 2

Answer: A

Explanation:
1. Why does the compilation fail?
* The interface Special contains bar as int bar = 1;.
* In Java, all interface fields are implicitly public, static, and final.
* This means that bar is a constant (final variable).
* The method add() contains bar--, which attempts to modify bar.
* Since bar is final, it cannot be modified, causing acompilation error.
2. Correcting the Code
To make the code compile, bar must not be final. One way to fix this:
java
class SpecialImpl implements Special {
int bar = 1;
}
Or modify the add() method:
java
int add() {
return --foo + bar; // No modification of bar
}
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - Interfaces
* Java SE 21 - Final Variables

NEW QUESTION # 45
Given a properties file on the classpath named Person.properties with the content:
ini
name=James
And:
java
public class Person extends ListResourceBundle {
protected Object[][] getContents() {
return new Object[][]{
{"name", "Jeanne"}
};
}
}
And:
java
public class Test {
public static void main(String[] args) {
ResourceBundle bundle = ResourceBundle.getBundle("Person");
String name = bundle.getString("name");
System.out.println(name);
}
}
What is the given program's output?

A. Compilation fails
B. MissingResourceException
C. JeanneJames
D. James
E. Jeanne
F. JamesJeanne

Answer: E

Explanation:
In this scenario, we have a Person class that extends ListResourceBundle and a properties file named Person.
properties. Both define a resource with the key "name" but with different values:
* Person class (ListResourceBundle):Defines the key "name" with the value "Jeanne".
* Person.properties file:Defines the key "name" with the value "James".
When the ResourceBundle.getBundle("Person") method is called, the Java runtime searches for a resource bundle with the base name "Person". The search order is as follows:
* Class-Based Resource Bundle:The runtime first looks for a class named Person (i.e., Person.class).
* Properties File Resource Bundle:If the class is not found, it then looks for a properties file named Person.properties.
In this case, since the Person class is present and accessible, the runtime will load the Person class as the resource bundle. Therefore, the getBundle method returns an instance of the Person class.
Subsequently, when bundle.getString("name") is called, it retrieves the value associated with the key "name" from the Person class, which is "Jeanne".
Thus, the output of the program is:
nginx
Jeanne

NEW QUESTION # 46
......

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