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Oracle Java SE 21 Developer Professional Sample Questions (Q78-Q83):

NEW QUESTION # 78
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. Compilation fails.
B. Peugeot
C. An exception is thrown at runtime.
D. Peugeot 807

Answer: A

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 # 79
Given:
java
Integer frenchRevolution = 1789;
Object o1 = new String("1789");
Object o2 = frenchRevolution;
frenchRevolution = null;
Object o3 = o2.toString();
System.out.println(o1.equals(o3));
What is printed?

A. A ClassCastException is thrown.
B. A NullPointerException is thrown.
C. Compilation fails.
D. false
E. true

Answer: E

Explanation:
* Understanding Variable Assignments
java
Integer frenchRevolution = 1789;
Object o1 = new String("1789");
Object o2 = frenchRevolution;
frenchRevolution = null;
* frenchRevolution is an Integer with value1789.
* o1 is aString with value "1789".
* o2 storesa reference to frenchRevolution, which is an Integer (1789).
* frenchRevolution = null;only nullifies the reference, but o2 still holds the Integer 1789.
* Calling toString() on o2
java
Object o3 = o2.toString();
* o2 refers to an Integer (1789).
* Integer.toString() returns theString representation "1789".
* o3 is assigned "1789" (String).
* Evaluating o1.equals(o3)
java
System.out.println(o1.equals(o3));
* o1.equals(o3) isequivalent to:
java
"1789".equals("1789")
* Since both areequal strings, the output is:
arduino
true
Thus, the correct answer is:true
References:
* Java SE 21 - Integer.toString()
* Java SE 21 - String.equals()

NEW QUESTION # 80
Given:
java
Optional o1 = Optional.empty();
Optional o2 = Optional.of(1);
Optional o3 = Stream.of(o1, o2)
.filter(Optional::isPresent)
.findAny()
.flatMap(o -> o);
System.out.println(o3.orElse(2));
What is the given code fragment's output?

A. An exception is thrown
B. Optional.empty
C. 0
D. Compilation fails
E. 1
F. Optional[1]
G. 2

Answer: E

Explanation:
In this code, two Optional objects are created:
* o1 is an empty Optional.
* o2 is an Optional containing the integer 1.
A stream is created from o1 and o2. The filter method retains only the Optional instances that are present (i.e., non-empty). This results in a stream containing only o2.
The findAny method returns an Optional describing some element of the stream, or an empty Optional if the stream is empty. Since the stream contains o2, findAny returns Optional[Optional[1]].
The flatMap method is then used to flatten this nested Optional. It applies the provided mapping function (o -
> o) to the value, resulting in Optional[1].
Finally, o3.orElse(2) returns the value contained in o3 if it is present; otherwise, it returns 2. Since o3 contains
1, the output is 1.

NEW QUESTION # 81
Given:
java
interface A {
default void ma() {
}
}
interface B extends A {
static void mb() {
}
}
interface C extends B {
void ma();
void mc();
}
interface D extends C {
void md();
}
interface E extends D {
default void ma() {
}
default void mb() {
}
default void mc() {
}
}
Which interface can be the target of a lambda expression?

A. None of the above
B. C
C. B
D. D
E. E
F. A

Answer: A

Explanation:
In Java, a lambda expression can be used where a target type is a functional interface. A functional interface is an interface that contains exactly one abstract method. This concept is also known as a Single Abstract Method (SAM) type.
Analyzing each interface:
* Interface A: Contains a single default method ma(). Since default methods are not abstract, A has no abstract methods.
* Interface B: Extends A and adds a static method mb(). Static methods are also not abstract, so B has no abstract methods.
* Interface C: Extends B and declares two abstract methods: ma() (which overrides the default method from A) and mc(). Therefore, C has two abstract methods.
* Interface D: Extends C and adds another abstract method md(). Thus, D has three abstract methods.
* Interface E: Extends D and provides default implementations for ma(), mb(), and mc(). However, it does not provide an implementation for md(), leaving it as the only abstract method in E.
For an interface to be a functional interface, it must have exactly one abstract method. In this case, E has one abstract method (md()), so it qualifies as a functional interface. However, the question asks which interface can be the target of a lambda expression. Since E is a functional interface, it can be the target of a lambda expression.
Therefore, the correct answer is D (E).

NEW QUESTION # 82
Given:
java
final Stream<String> strings =
Files.readAllLines(Paths.get("orders.csv"));
strings.skip(1)
.limit(2)
.forEach(System.out::println);
And that the orders.csv file contains:
mathematica
OrderID,Customer,Product,Quantity,Price
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99
3,Sebastien Loeb,Monitor,1,199.99
4,Antoine Griezmann,Headset,3,45.00
What is printed?

A. Compilation fails.
B. arduino
2,Teddy Riner,Mouse,1,15.99
3,Sebastien Loeb,Monitor,1,199.99
C. An exception is thrown at runtime.
D. arduino
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99
3,Sebastien Loeb,Monitor,1,199.99
4,Antoine Griezmann,Headset,3,45.00
E. arduino
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99

Answer: A,C

Explanation:
1. Why Does Compilation Fail?
* The error is in this line:
java
final Stream<String> strings = Files.readAllLines(Paths.get("orders.csv"));
* Files.readAllLines(Paths.get("orders.csv")) returns a List<String>,not a Stream<String>.
* A List<String> cannot be assigned to a Stream<String>.
2. Correcting the Code
* The correct way to create a stream from the file:
java
Stream<String> strings = Files.lines(Paths.get("orders.csv"));
* This correctly creates a Stream<String> from the file.
3. Expected Output After Fixing
java
Files.lines(Paths.get("orders.csv"))
skip(1) // Skips the header row
limit(2) // Limits to first two data rows
forEach(System.out::println);
* Output:
arduino
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - Files.readAllLines
* Java SE 21 - Files.lines

NEW QUESTION # 83
......

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