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Oracle Java SE 21 Developer Professional Sample Questions (Q68-Q73):
NEW QUESTION # 68
Which of the following doesnotexist?
Answer: E
Explanation:
1. Understanding Supplier Functional Interfaces
* The Supplier<T> interface is part of java.util.function and provides valueswithout taking any arguments.
* Java also provides primitive specializations of Supplier<T>:
* BooleanSupplier# Returns a boolean. Exists
* DoubleSupplier# Returns a double. Exists
* LongSupplier# Returns a long. Exists
* Supplier<T># Returns a generic T. Exists
2. What about BiSupplier<T, U, R>?
* There is no BiSupplier<T, U, R> in Java.
* In Java, suppliers donot take arguments, so abi-supplierdoes not exist.
* If you need a function thattakes two arguments and returns a value, use BiFunction<T, U, R>.
Thus, the correct answer is:BiSupplier<T, U, R> does not exist.
References:
* Java SE 21 - Supplier<T>
* Java SE 21 - Functional Interfaces
NEW QUESTION # 69
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?
Answer: D
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 # 70
Given:
java
package com.vv;
import java.time.LocalDate;
public class FetchService {
public static void main(String[] args) throws Exception {
FetchService service = new FetchService();
String ack = service.fetch();
LocalDate date = service.fetch();
System.out.println(ack + " the " + date.toString());
}
public String fetch() {
return "ok";
}
public LocalDate fetch() {
return LocalDate.now();
}
}
What will be the output?
Answer: B
Explanation:
In Java, method overloading allows multiple methods with the same name to exist in a class, provided they have different parameter lists (i.e., different number or types of parameters). However, having two methods with the exact same parameter list and only differing in return type is not permitted.
In the provided code, the FetchService class contains two fetch methods:
* public String fetch()
* public LocalDate fetch()
Both methods have identical parameter lists (none) but differ in their return types (String and LocalDate, respectively). This leads to a compilation error because the Java compiler cannot distinguish between the two methods based solely on return type.
The Java Language Specification (JLS) states:
"It is a compile-time error to declare two methods with override-equivalent signatures in a class." In this context, "override-equivalent" means that the methods have the same name and parameter types, regardless of their return types.
Therefore, the code will fail to compile due to the duplicate method signatures, and the correct answer is B:
Compilation fails.
NEW QUESTION # 71
Given:
java
sealed class Vehicle permits Car, Bike {
}
non-sealed class Car extends Vehicle {
}
final class Bike extends Vehicle {
}
public class SealedClassTest {
public static void main(String[] args) {
Class<?> vehicleClass = Vehicle.class;
Class<?> carClass = Car.class;
Class<?> bikeClass = Bike.class;
System.out.print("Is Vehicle sealed? " + vehicleClass.isSealed() +
"; Is Car sealed? " + carClass.isSealed() +
"; Is Bike sealed? " + bikeClass.isSealed());
}
}
What is printed?
Answer: B
Explanation:
* Understanding Sealed Classes in Java
* Asealed classrestricts which other classes can extend it.
* A sealed classmust explicitly declare its permitted subclassesusing the permits keyword.
* Subclasses can be declared as:
* sealed(restricts further extension).
* non-sealed(removes the restriction, allowing unrestricted subclassing).
* final(prevents further subclassing).
* Analyzing the Given Code
* Vehicle is declared as sealed with permits Car, Bike, meaning only Car and Bike can extend it.
* Car is declared as non-sealed, which means itis no longer sealedand can have subclasses.
* Bike is declared as final, meaningit cannot be subclassed.
* Using isSealed() Method
* vehicleClass.isSealed() #truebecause Vehicle is explicitly marked as sealed.
* carClass.isSealed() #falsebecause Car is marked non-sealed.
* bikeClass.isSealed() #falsebecause Bike is final, and a final class isnot considered sealed.
* Final Output
csharp
Is Vehicle sealed? true; Is Car sealed? false; Is Bike sealed? false
Thus, the correct answer is:"Is Vehicle sealed? true; Is Car sealed? false; Is Bike sealed? false" References:
* Java SE 21 - Sealed Classes
* Java SE 21 - isSealed() Method
NEW QUESTION # 72
Given:
java
public static void main(String[] args) {
try {
throw new IOException();
} catch (IOException e) {
throw new RuntimeException();
} finally {
throw new ArithmeticException();
}
}
What is the output?
Answer: A
Explanation:
In this code, the try block throws an IOException. The catch block catches this exception and throws a new RuntimeException. Regardless of exceptions thrown in the try or catch blocks, the finally block is always executed. In this case, the finally block throws an ArithmeticException.
When an exception is thrown in a finally block, it overrides any previous exceptions that were thrown in the try or catch blocks. Therefore, the ArithmeticException thrown in the finally block is the exception that propagates out of the method. As a result, the program terminates with an ArithmeticException.
NEW QUESTION # 73
......
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