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The try-with-resources block invokes AutoCloseable.close() after the try body completes but before finally.

Output order: run close finally.

Understanding LongStream.range(1, 1945).boxed().toList();

LongStream.range(1, 1945) generates a stream of numbers from 1 to 1944 .

.boxed() converts the primitive long values to Long objects.

.toList() (introduced in Java 16) creates an immutable list .

Understanding Executors.newVirtualThreadPerTaskExecutor()

Java 21 introduced virtual threads to improve concurrency.

Executors.newVirtualThreadPerTaskExecutor()creates a new virtual thread per submitted task , allowing highly concurrent execution.

Execution Behavior

cannesFestivalfeatureFilms.stream().limit(25) # Limits the stream to the first 25 numbers (1 to 25).

.forEach(film -> executor.submit(() -> System.out.println(film)))

Each film is printed inside a virtual thread.

Virtual threads execute asynchronously, meaning numbers arenot guaranteed to print sequentially .

Output will contain numbers from 1 to 25, but their order is random due to concurrent execution.

Possible Output (Random Order)

python-repl

The order may differ in each run due to concurrent execution.

Thus, the correct answer is:"Numbers from 1 to 25 randomly." References:

Java SE 21 - Virtual Threads

Java SE 21 - Executors.newVirtualThreadPerTaskExecutor()

Why does the compilation fail?

The Java switch statement does not support primitive type pattern matching in switch expressions as of Java 21.

The case pattern case int i -> "integer"; is invalid because pattern matching with primitive types (like int or long) is not yet supported in switch statements .

The error occurs at case int i -> "integer";, leading to a compilation failure .

Correcting the Code

Since myVar is of type Object, autoboxing converts 0 into an Integer .

To make the code compile, we should use Integer instead of int:

java

Object myVar = 0;

String print = switch (myVar) {

case Integer i -> "integer";

case Long l -> "long";

case String s -> "string";
default -> "";

};

System.out.println(print);

Output:

bash

integer

Thus, the correct answer is:Compilation fails.

References:
Java SE 21 - Pattern Matching for switch Java SE 21 - switch Expressions

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 a Stream<String> , splitting the string by line breaks.

Calling .toList() on a stream converts 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

1. private Access Modifier

The private access modifier can only be used for inner classes (nested classes).

Top-level classes cannot be private.

Example of invalid use:

java

private class MyClass {} // Compilation error

Example of valid use (for inner class):

java

class Outer {

private class Inner {}

}

2. protected Access Modifier

Top-level classes cannot be protected.

protectedonly applies to members (fields, methods, and constructors).

Example of invalid use:

java

protected class MyClass {} // Compilation error

Example of valid use (for methods/fields):

java

class Parent {

protected void display() {}

}

3. public Access Modifier

A top-level class can be public , but only one public class per file is allowed .

Example of valid use:

java

public class MyClass {}

Example of invalid use:

java

public class A {}

public class B {} // Compilation error: Only one public class per file

4. final Modifier

final can be used with classes, but not all kinds of classes .

Interfaces cannot be final, because they are meant to be implemented.

Example of invalid use:

java

final interface MyInterface {} // Compilation error

Thus, none of the statements are fully correct , making the correct answer: None

References:
Java SE 21 - Access Modifiers
Java SE 21 - Class Modifiers

In this code, an ArrayDeque named deque is created, and the integers 1 and 2 are added to it using the offer method. The offer method inserts the specified element at the end of the deque.

State of deque after offers:[1, 2]
The peek method retrieves, but does not remove, the head of the deque, returning 1. Therefore, i1 is assigned the value 1.

State of deque after peek:[1, 2]

Value of i1:1

The poll method retrieves and removes the head of the deque, returning 1. Therefore, i2 is assigned the value 1.

State of deque after poll:[2]

Value of i2:1

Another peek operation retrieves the current head of the deque, which is now 2, without removing it.
Therefore, i3 is assigned the value 2.

State of deque after second peek:[2]

Value of i3:2

The System.out.println statement then outputs the values of i1, i2, and i3, resulting in 1 1 2.

In the provided code, a list abc is created containing the strings "a", "b", and "c". The first forEach operation attempts to convert each element to uppercase by assigning x = x.toUpperCase();. However,
this assignment only changes the local variable x within the lambda expression and does not modify the elements in the original list abc. Strings in Java are immutable, meaning their values cannot be changed once created.
Therefore, the original list remains unchanged.

The second forEach operation iterates over the original list and prints each element. Since the list was not modified, the output will be the concatenation of the original elements: abc.

To achieve the output ABC, you would need to collect the transformed elements into a new list, as shown below:

java

List<String> abc = List.of("a", "b", "c");

List<String> upperCaseAbc = abc.stream() map(String::toUpperCase)

collect(Collectors.toList());

upperCaseAbc.forEach(System.out::print);

In this corrected version, the map operation creates a new stream with the uppercase versions of the original elements, which are then collected into a new list upperCaseAbc. The forEach operation then prints ABC.

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

Console does 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 in Reader , but not in Console .

Thus, the correct answer is:read() does not exist.

References:
Java SE 21 - Console API
Java SE 21 - Reader API

The CopyOnWriteArrayList is a thread-safe variant of ArrayList in which all mutative operations (such as add, set, and remove) are implemented by creating a fresh copy of the underlying array. This design allows for safe iteration over the list without requiring external synchronization, as iterators operate over a snapshot of the array at the time the iterator was created. Consequently, modifications made to the list after the creation of an iterator are not reflected in that iterator.

docs.oracle.com

Evaluation of Options:

Option A:Correct. This statement accurately describes the behavior of CopyOnWriteArrayList.

Option B:Incorrect. CopyOnWriteArrayList is thread-safe and is designed to prevent interference among concurrent threads.

Option C:Incorrect. Iterators of CopyOnWriteArrayList do not reflect additions, removals, or changes made to the list after the iterator was created; they operate on a snapshot of the list's state at the time of their
creation.

Option D:Incorrect. CopyOnWriteArrayList allows null elements.

Option E:Incorrect. Element-changing operations on iterators, such as remove, set, and add, are not supported in CopyOnWriteArrayList and will throw UnsupportedOperationException.

In this code, the class Main contains both an instance variable instanceVar and a static variable staticVar.
The sequence of initialization and execution is as follows:

Static Variable Initialization:

staticVar is declared and initialized to 666.

Static Block Execution:

The static block executes, updating staticVar to 42.

Instance Variable Initialization:

When a new instance of Main is created, instanceVar is initialized to the current value of staticVar, which is
42.

main Method Execution:

The main method creates a new instance of Main and prints the values of instanceVar and staticVar.

Therefore, the output of the program is 42 42.

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.

In Java, a lambda expression can have two forms for its body:

Single Expression:A concise form where the body consists of a single expression. The result of this expression is implicitly returned.

Example:

java

(a, b) -> a + b

In this example, (a, b) are the parameters, and a + b is the single expression that adds them together.

Statement Block:A more detailed form where the body consists of a block of statements enclosed in braces {}. Within this block, you can have multiple statements, and if a return value is expected, you must
explicitly use the return statement.

Example:

java

(a, b) -> {

int sum = a + b;

System.out.println("Sum is: " + sum);

return sum;

}
In this example, the lambda body is a statement block that performs multiple actions: it calculates the sum, prints it, and then returns the sum.

Given the options:

A. Two statements:While a lambda body can contain multiple statements, they must be enclosed within a
statement block {}. Simply having two statements without braces is not valid syntax for a lambda
expression.

B. An expression and a statement:Similar to option A, if a lambda body contains more than one element
(be it expressions or statements), they need to be enclosed in a statement block.

C. A statement block:This is correct. A lambda expression can have a body that is a statement block,
allowing multiple statements enclosed in braces.

D. None of the above:This is incorrect since option C is valid.

E. Two expressions:As with options A and B, multiple expressions must be enclosed in a statement block
to form a valid lambda body.

Therefore, the correct answer is C: A statement block.

In this code, a Map named map is created using Map.of with the following key-value pairs:

"b": 1

"a": 3

"c": 2

The Map.of method returns an immutable map containing these mappings.

Next, a TreeMap named treeMap is instantiated by passing the map to its constructor:

java

TreeMap<String, Integer> treeMap = new TreeMap<>(map);

The TreeMap constructor with a Map parameter creates a new tree map containing the same mappings as the given map, ordered according to the natural ordering of its keys. In Java, the natural ordering for String keys is lexicographical order.

Therefore, the TreeMap will store the entries in the following order:

"a": 3

"b": 1

"c": 2

When System.out.println(treeMap); is executed, it outputs the TreeMap in its natural order, resulting in:

r

{a=3, b=1, c=2}

Thus, the correct answer is option F: {a=3, b=1, c=2}.

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 a compilation 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

In this code, the Calculable interface defines a single abstract method calculate that takes an int parameter and returns a long. The main method contains three lambda expressions assigned to variables c1, c2, and c3 of type Calculable.

Line 1:Calculable c1 = i -> i + 1;

This lambda expression takes an integer i and returns the result of i + 1. Since the expression i + 1 results in an int, and Java allows implicit widening conversion from int to long, this line compiles successfully.

Line 2:Calculable c2 = i -> Long.valueOf(i);

Here, the lambda expression takes an integer i and returns the result of Long.valueOf(i). The Long.valueOf (int i) method returns a Long object. However, Java allows unboxing of the Long object to a long primitive type when necessary. Therefore, this line compiles successfully.

Line 3:Calculable c3 = i -> { throw new ArithmeticException(); };

This lambda expression takes an integer i and throws an ArithmeticException. Since the method calculate has a return type of long, and throwing an exception is a valid way to exit the method without returning a value, this line compiles successfully.

Since all three lines adhere to the method signature defined in the Calculable interface and there are no type mismatches or syntax errors, the program compiles successfully.