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Oracle Java SE 21 Developer Professional Sample Questions (Q37-Q42):

NEW QUESTION # 37
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. A
B. C
C. E
D. D
E. B
F. None of the above

Answer: F

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 # 38
Given:
java
public class Test {
static int count;
synchronized Test() {
count++;
}
public static void main(String[] args) throws InterruptedException {
Runnable task = Test::new;
Thread t1 = new Thread(task);
Thread t2 = new Thread(task);
t1.start();
t2.start();
t1.join();
t2.join();
System.out.println(count);
}
}
What is the given program's output?

A. It's always 1
B. It's either 1 or 2
C. It's always 2
D. It's either 0 or 1
E. Compilation fails

Answer: E

Explanation:
In this code, the Test class has a static integer field count and a constructor that is declared with the synchronized modifier. In Java, the synchronized modifier can be applied to methods to control access to critical sections, but it cannot be applied directly to constructors. Attempting to declare a constructor as synchronized will result in a compilation error.
Compilation Error Details:
The Java Language Specification does not permit the use of the synchronized modifier on constructors.
Therefore, the compiler will produce an error indicating that the synchronized modifier is not allowed in this context.
Correct Usage:
If you need to synchronize the initialization of instances, you can use a synchronized block within the constructor:
java
public class Test {
static int count;
Test() {
synchronized (Test.class) {
count++;
}
}
public static void main(String[] args) throws InterruptedException {
Runnable task = Test::new;
Thread t1 = new Thread(task);
Thread t2 = new Thread(task);
t1.start();
t2.start();
t1.join();
t2.join();
System.out.println(count);
}
}
In this corrected version, the synchronized block within the constructor ensures that the increment operation on count is thread-safe.
Conclusion:
The original program will fail to compile due to the illegal use of the synchronized modifier on the constructor. Therefore, the correct answer is E: Compilation fails.

NEW QUESTION # 39
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?

A. Is Vehicle sealed? true; Is Car sealed? false; Is Bike sealed? false
B. Is Vehicle sealed? false; Is Car sealed? true; Is Bike sealed? true
C. Is Vehicle sealed? false; Is Car sealed? false; Is Bike sealed? false
D. Is Vehicle sealed? true; Is Car sealed? true; Is Bike sealed? true

Answer: A

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 # 40
Which of the following java.io.Console methods doesnotexist?

A. readLine()
B. reader()
C. readPassword()
D. readPassword(String fmt, Object... args)
E. readLine(String fmt, Object... args)
F. read()

Answer: F

Explanation:
* java.io.Console is used for interactive input from the console.
* Existing Methods in java.io.Console
* reader() # Returns a Reader object.
* readLine() # Reads a line of text from the console.
* readLine(String fmt, Object... args) # Reads a formatted line.
* readPassword() # Reads a password, returning a char[].
* readPassword(String fmt, Object... args) # Reads a formatted password.
* read() Does Not Exist
* Consoledoes not have a read() method.
* If character-by-character reading is required, use:
java
Console console = System.console();
Reader reader = console.reader();
int c = reader.read(); // Reads one character
* read() is available inReader, butnot in Console.
Thus, the correct answer is:read() does not exist.
References:
* Java SE 21 - Console API
* Java SE 21 - Reader API

NEW QUESTION # 41
Given:
java
Object input = 42;
String result = switch (input) {
case String s -> "It's a string with value: " + s;
case Double d -> "It's a double with value: " + d;
case Integer i -> "It's an integer with value: " + i;
};
System.out.println(result);
What is printed?

A. It's a string with value: 42
B. null
C. It throws an exception at runtime.
D. Compilation fails.
E. It's an integer with value: 42
F. It's a double with value: 42

Answer: D

Explanation:
* Pattern Matching in switch
* The switch expression introduced inJava 21supportspattern matchingfor different types.
* However,a switch expression must be exhaustive, meaningit must cover all possible cases or provide a default case.
* Why does compilation fail?
* input is an Object, and the switch expression attempts to pattern-match it to String, Double, and Integer.
* If input had been of another type (e.g., Float or Long), there would beno matching case, leading to anon-exhaustive switch.
* Javarequires a default caseto ensure all possible inputs are covered.
* Corrected Code (Adding a default Case)
java
Object input = 42;
String result = switch (input) {
case String s -> "It's a string with value: " + s;
case Double d -> "It's a double with value: " + d;
case Integer i -> "It's an integer with value: " + i;
default -> "Unknown type";
};
System.out.println(result);
* With this change, the codecompiles and runs successfully.
* Output:
vbnet
It's an integer with value: 42
Thus, the correct answer is:Compilation failsdue to a missing default case.
References:
* Java SE 21 - Pattern Matching for switch
* Java SE 21 - switch Expressions

NEW QUESTION # 42
......

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