New 2025 1z0-830 Dumps for Java SE Certified Exam Questions & Answer [Q41-Q63]

New 2025 1z0-830 Dumps for Java SE Certified Exam Questions and Answer

Realistic Verified 1z0-830 exam dumps Q&As - 1z0-830 Free Update

NEW QUESTION # 41
Given:
java
public class BoomBoom implements AutoCloseable {
public static void main(String[] args) {
try (BoomBoom boomBoom = new BoomBoom()) {
System.out.print("bim ");
throw new Exception();
} catch (Exception e) {
System.out.print("boom ");
}
}
@Override
public void close() throws Exception {
System.out.print("bam ");
throw new RuntimeException();
}
}
What is printed?

A. Compilation fails.
B. bim boom
C. bim bam followed by an exception
D. bim boom bam
E. bim bam boom

Answer: E

Explanation:
* Understanding Try-With-Resources (AutoCloseable)
* BoomBoom implements AutoCloseable, meaning its close() method isautomatically calledat the end of the try block.
* Step-by-Step Execution
* Step 1: Enter Try Block
java
try (BoomBoom boomBoom = new BoomBoom()) {
System.out.print("bim ");
throw new Exception();
}
* "bim " is printed.
* Anexception (Exception) is thrown, butbefore it is handled, the close() method is executed.
* Step 2: close() is Called
java
@Override
public void close() throws Exception {
System.out.print("bam ");
throw new RuntimeException();
}
* "bam " is printed.
* A new RuntimeException is thrown, but it doesnot override the existing Exception yet.
* Step 3: Exception Handling
java
} catch (Exception e) {
System.out.print("boom ");
}
* The catch (Exception e)catches the original Exception from the try block.
* "boom " is printed.
* Final Output
nginx
bim bam boom
* Theoriginal Exception is caught, not the RuntimeException from close().
* TheRuntimeException from close() is ignoredbecause thecatch block is already handling Exception.
Thus, the correct answer is:bim bam boom
References:
* Java SE 21 - Try-With-Resources
* Java SE 21 - AutoCloseable Interface

NEW QUESTION # 42
Given:
java
List<String> frenchAuthors = new ArrayList<>();
frenchAuthors.add("Victor Hugo");
frenchAuthors.add("Gustave Flaubert");
Which compiles?

A. Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>>(); java authorsMap4.put("FR", frenchAuthors);
B. Map<String, List<String>> authorsMap5 = new HashMap<String, List<String>>(); java authorsMap5.put("FR", frenchAuthors);
C. Map<String, ? extends List<String>> authorsMap2 = new HashMap<String, ArrayList<String>> (); java authorsMap2.put("FR", frenchAuthors);
D. Map<String, ArrayList<String>> authorsMap1 = new HashMap<>();
java
authorsMap1.put("FR", frenchAuthors);
E. var authorsMap3 = new HashMap<>();
java
authorsMap3.put("FR", frenchAuthors);

Answer: A,B,E

Explanation:
* Option A (Map<String, ArrayList<String>> authorsMap1 = new HashMap<>();)
* #Compilation Fails
* frenchAuthors is declared as List<String>,notArrayList<String>.
* The correct way to declare a Map that allows storing List<String> is to use List<String> as the generic type,notArrayList<String>.
* Fix:
java
Map<String, List<String>> authorsMap1 = new HashMap<>();
authorsMap1.put("FR", frenchAuthors);
* Reason:The type ArrayList<String> is more specific than List<String>, and this would cause a type mismatcherror.
* Option B (Map<String, ? extends List<String>> authorsMap2 = new HashMap<String, ArrayList<String>>();)
* #Compilation Fails
* ? extends List<String>makes the map read-onlyfor adding new elements.
* The line authorsMap2.put("FR", frenchAuthors); causes acompilation errorbecause wildcard (?
extends List<String>) prevents modifying the map.
* Fix:Remove the wildcard:
java
Map<String, List<String>> authorsMap2 = new HashMap<>();
authorsMap2.put("FR", frenchAuthors);
* Option C (var authorsMap3 = new HashMap<>();)
* Compiles Successfully
* The var keyword allows the compiler to infer the type.
* However,the inferred type is HashMap<Object, Object>, which may cause issues when retrieving values.
* Option D (Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>
>();)
* Compiles Successfully
* Valid declaration:HashMap<K, V> can be assigned to Map<K, V>.
* Using new HashMap<String, ArrayList<String>>() with Map<String, List<String>> isallowed due to polymorphism.
* Correct syntax:
java
Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>>(); authorsMap4.put("FR", frenchAuthors);
* Option E (Map<String, List<String>> authorsMap5 = new HashMap<String, List<String>>();)
* Compiles Successfully
* HashMap<String, List<String>> isa valid instantiation.
* Correct usage:
java
Map<String, List<String>> authorsMap5 = new HashMap<>();
authorsMap5.put("FR", frenchAuthors);
Thus, the correct answers are:C, D, E
References:
* Java SE 21 - Generics and Type Inference
* Java SE 21 - var Keyword

NEW QUESTION # 43
Given:
java
interface Calculable {
long calculate(int i);
}
public class Test {
public static void main(String[] args) {
Calculable c1 = i -> i + 1; // Line 1
Calculable c2 = i -> Long.valueOf(i); // Line 2
Calculable c3 = i -> { throw new ArithmeticException(); }; // Line 3
}
}
Which lines fail to compile?

A. The program successfully compiles
B. Line 2 only
C. Line 3 only
D. Line 2 and line 3
E. Line 1 and line 2
F. Line 1 only
G. Line 1 and line 3

Answer: A

Explanation:
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.

NEW QUESTION # 44
Which of the following suggestions compile?(Choose two.)

A. java
public sealed class Figure
permits Circle, Rectangle {}
final sealed class Circle extends Figure {
float radius;
}
non-sealed class Rectangle extends Figure {
float length, width;
}
B. java
sealed class Figure permits Rectangle {}
public class Rectangle extends Figure {
float length, width;
}
C. java
sealed class Figure permits Rectangle {}
final class Rectangle extends Figure {
float length, width;
}
D. java
public sealed class Figure
permits Circle, Rectangle {}
final class Circle extends Figure {
float radius;
}
non-sealed class Rectangle extends Figure {
float length, width;
}

Answer: C,D

Explanation:
Option A (sealed class Figure permits Rectangle {} and final class Rectangle extends Figure {}) - Valid
* Why it compiles?
* Figure issealed, meaning itmust explicitly declareits subclasses.
* Rectangle ispermittedto extend Figure and isdeclared final, meaning itcannot be extended further.
* This followsvalid sealed class rules.
Option B (sealed class Figure permits Rectangle {} and public class Rectangle extends Figure {}) -# Invalid
* Why it fails?
* Rectangle extends Figure, but it doesnot specify if it is sealed, final, or non-sealed.
* Fix:The correct declaration must be one of the following:
java
final class Rectangle extends Figure {} // OR
sealed class Rectangle permits OtherClass {} // OR
non-sealed class Rectangle extends Figure {}
Option C (final sealed class Circle extends Figure {}) -#Invalid
* Why it fails?
* A class cannot be both final and sealedat the same time.
* sealed meansit must have permitted subclasses, but final meansit cannot be extended.
* Fix:Change final sealed to just final:
java
final class Circle extends Figure {}
Option D (public sealed class Figure permits Circle, Rectangle {} with final class Circle and non-sealed class Rectangle) - Valid
* Why it compiles?
* Figure issealed, meaning it mustdeclare its permitted subclasses(Circle and Rectangle).
* Circle is declaredfinal, so itcannot have subclasses.
* Rectangle is declarednon-sealed, meaningit can be subclassedfreely.
* This correctly followsJava's sealed class rules.
Thus, the correct answers are:A, D
References:
* Java SE 21 - Sealed Classes
* Java SE 21 - Class Modifiers

NEW QUESTION # 45
What is the output of the following snippet? (Assume the file exists)
java
Path path = Paths.get("C:\\home\\joe\\foo");
System.out.println(path.getName(0));

A. Compilation error
B. C
C. C:
D. IllegalArgumentException
E. home

Answer: E

Explanation:
In Java's java.nio.file package, the Path class represents a file path in a file system. The Paths.get(String first, String... more) method is used to create a Path instance by converting a path string or URI.
In the provided code snippet, the Path object path is created with the string "C:\\home\\joe\\foo". This represents an absolute path on a Windows system.
The getName(int index) method of the Path class returns a name element of the path as a Path object. The index is zero-based, where index 0 corresponds to the first element in the path's name sequence. It's important to note that the root component (e.g., "C:\" on Windows) is not considered a name element and is not included in this sequence.
Therefore, for the path "C:\\home\\joe\\foo":
* Root Component:"C:\"
* Name Elements:
* Index 0: "home"
* Index 1: "joe"
* Index 2: "foo"
When path.getName(0) is called, it returns the first name element, which is "home". Thus, the output of the System.out.println statement is home.

NEW QUESTION # 46
Given:
java
Period p = Period.between(
LocalDate.of(2023, Month.MAY, 4),
LocalDate.of(2024, Month.MAY, 4));
System.out.println(p);
Duration d = Duration.between(
LocalDate.of(2023, Month.MAY, 4),
LocalDate.of(2024, Month.MAY, 4));
System.out.println(d);
What is the output?

A. P1Y
UnsupportedTemporalTypeException
B. PT8784H
P1Y
C. P1Y
PT8784H
D. UnsupportedTemporalTypeException

Answer: A

Explanation:
In this code, two LocalDate instances are created representing May 4, 2023, and May 4, 2024. The Period.
between() method is used to calculate the period between these two dates, and the Duration.between() method is used to calculate the duration between them.
Period Calculation:
The Period.between() method calculates the amount of time between two LocalDate objects in terms of years, months, and days. In this case, the period between May 4, 2023, and May 4, 2024, is exactly one year.
Therefore, p is P1Y, which stands for a period of one year. Printing p will output P1Y.
Duration Calculation:
The Duration.between() method is intended to calculate the duration between two temporal objects that have time components, such as LocalDateTime or Instant. However, LocalDate represents a date without a time component. Attempting to use Duration.between() with LocalDate instances will result in an UnsupportedTemporalTypeException because Duration requires time-based units, which LocalDate does not support.
Exception Details:
The UnsupportedTemporalTypeException is thrown when an unsupported unit is used. In this case, Duration.
between() internally attempts to access time-based fields (like seconds), which are not supported by LocalDate. This behavior is documented in the Java Bug System underJDK-8170275.
Correct Usage:
To calculate the duration between two dates, including time components, you should use LocalDateTime or Instant. For example:
java
LocalDateTime start = LocalDateTime.of(2023, Month.MAY, 4, 0, 0);
LocalDateTime end = LocalDateTime.of(2024, Month.MAY, 4, 0, 0);
Duration d = Duration.between(start, end);
System.out.println(d); // Outputs: PT8784H
This will correctly calculate the duration as PT8784H, representing 8,784 hours (which is 366 days, accounting for a leap year).
Conclusion:
The output of the given code will be:
pgsql
P1Y
Exception in thread "main" java.time.temporal.UnsupportedTemporalTypeException: Unsupported unit:
Seconds
Therefore, the correct answer is D:
nginx
P1Y
UnsupportedTemporalTypeException

NEW QUESTION # 47
Given:
java
var sList = new CopyOnWriteArrayList<Customer>();
Which of the following statements is correct?

A. The CopyOnWriteArrayList class does not allow null elements.
B. Element-changing operations on iterators of CopyOnWriteArrayList, such as remove, set, and add, are supported and do not throw UnsupportedOperationException.
C. The CopyOnWriteArrayList class is not thread-safe and does not prevent interference amongconcurrent threads.
D. The CopyOnWriteArrayList class's iterator reflects all additions, removals, or changes to the list since the iterator was created.
E. The CopyOnWriteArrayList class is a thread-safe variant of ArrayList where all mutative operations are implemented by making a fresh copy of the underlying array.

Answer: E

Explanation:
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.

NEW QUESTION # 48
Given:
java
var counter = 0;
do {
System.out.print(counter + " ");
} while (++counter < 3);
What is printed?

A. 1 2 3 4
B. 1 2 3
C. 0 1 2 3
D. Compilation fails.
E. An exception is thrown.
F. 0 1 2

Answer: F

Explanation:
* Understanding do-while Execution
* A do-while loopexecutes at least oncebefore checking the condition.
* ++counter < 3 increments counterbeforeevaluating the condition.
* Step-by-Step Execution
* Iteration 1:counter = 0, print "0", then ++counter becomes 1, condition 1 < 3 istrue.
* Iteration 2:counter = 1, print "1", then ++counter becomes 2, condition 2 < 3 istrue.
* Iteration 3:counter = 2, print "2", then ++counter becomes 3, condition 3 < 3 isfalse, so loop exits.
* Final Output
0 1 2
Thus, the correct answer is:0 1 2
References:
* Java SE 21 - Control Flow Statements
* Java SE 21 - do-while Loop

NEW QUESTION # 49
Given:
java
double amount = 42_000.00;
NumberFormat format = NumberFormat.getCompactNumberInstance(Locale.FRANCE, NumberFormat.Style.
SHORT);
System.out.println(format.format(amount));
What is the output?

A. 42 k
B. 42000E
C. 0
D. 42 000,00 €

Answer: A

Explanation:
In this code, a double variable amount is initialized to 42,000.00. The NumberFormat.
getCompactNumberInstance(Locale.FRANCE, NumberFormat.Style.SHORT) method is used to obtain a compact number formatter for the French locale with the short style. The format method is then called to format the amount.
The compact number formatting is designed to represent numbers in a shorter form, based on the patterns provided for a given locale. In the French locale, the short style represents thousands with a lowercase 'k'.
Therefore, 42,000 is formatted as 42 k.
* Option Evaluations:
* A. 42000E: This format is not standard in the French locale for compact number formatting.
* B. 42 000,00 €: This represents the number as a currency with two decimal places, which is not the compact form.
* C. 42000: This is the plain number without any formatting, which does not match the compact number format.
* D. 42 k: This is the correct compact representation of 42,000 in the French locale with the short style.
Thus, option D (42 k) is the correct output.

NEW QUESTION # 50
Which of the following statements are correct?

A. You can use 'private' access modifier with all kinds of classes
B. None
C. You can use 'final' modifier with all kinds of classes
D. You can use 'public' access modifier with all kinds of classes
E. You can use 'protected' access modifier with all kinds of classes

Answer: B

Explanation:
1. private Access Modifier
* The private access modifiercan only be used for inner classes(nested classes).
* Top-level classes cannot be private.
* Example ofinvaliduse:
java
private class MyClass {} // Compilation error
* Example ofvaliduse (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 ofinvaliduse:
java
protected class MyClass {} // Compilation error
* Example ofvaliduse (for methods/fields):
java
class Parent {
protected void display() {}
}
3. public Access Modifier
* Atop-level class can be public, butonly one public class per file is allowed.
* Example ofvaliduse:
java
public class MyClass {}
* Example ofinvaliduse:
java
public class A {}
public class B {} // Compilation error: Only one public class per file
4. final Modifier
* finalcan be used with classes, but not all kinds of classes.
* Interfaces cannot be final, because they are meant to be implemented.
* Example ofinvaliduse:
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

NEW QUESTION # 51
What do the following print?
java
import java.time.Duration;
public class DividedDuration {
public static void main(String[] args) {
var day = Duration.ofDays(2);
System.out.print(day.dividedBy(8));
}
}

A. PT6H
B. PT0D
C. It throws an exception
D. PT0H
E. Compilation fails

Answer: A

Explanation:
In this code, a Duration object day is created representing a duration of 2 days using the Duration.ofDays(2) method. The dividedBy(long divisor) method is then called on this Duration object with the argument 8.
The dividedBy(long divisor) method returns a copy of the original Duration divided by the specified value. In this case, dividing 2 days by 8 results in a duration of 0.25 days. In the ISO-8601 duration format used by Java's Duration class, this is represented as PT6H, which stands for a period of 6 hours.
Therefore, the output of the System.out.print statement is PT6H.

NEW QUESTION # 52
Which two of the following aren't the correct ways to create a Stream?

A. Stream stream = new Stream();
B. Stream stream = Stream.ofNullable("a");
C. Stream stream = Stream.empty();
D. Stream<String> stream = Stream.builder().add("a").build();
E. Stream stream = Stream.generate(() -> "a");
F. Stream stream = Stream.of();

Answer: A,D

NEW QUESTION # 53
Given:
java
public class OuterClass {
String outerField = "Outer field";
class InnerClass {
void accessMembers() {
System.out.println(outerField);
}
}
public static void main(String[] args) {
System.out.println("Inner class:");
System.out.println("------------");
OuterClass outerObject = new OuterClass();
InnerClass innerObject = new InnerClass(); // n1
innerObject.accessMembers(); // n2
}
}
What is printed?

A. markdown
Inner class:
------------
Outer field
B. Compilation fails at line n2.
C. An exception is thrown at runtime.
D. Compilation fails at line n1.
E. Nothing

Answer: D

Explanation:
* Understanding Inner Classes in Java
* Aninner class (non-static nested class)requires an instance of the outer classbefore it can be instantiated.
* Incorrect instantiationof the inner class at n1:
java
InnerClass innerObject = new InnerClass(); // Compilation error
* Since InnerClass is anon-staticinner class, itmust be created from an instance of OuterClass.
* Correct Way to Instantiate the Inner Class
java
OuterClass outerObject = new OuterClass();
OuterClass.InnerClass innerObject = outerObject.new InnerClass(); // Correct
* Thiscorrectly associatesthe inner class with an instance of OuterClass.
* Why Does Compilation Fail?
* The error occurs atline n1because InnerClass is beinginstantiated incorrectly.
Thus, the correct answer is:Compilation fails at line n1.
References:
* Java SE 21 - Nested and Inner Classes
* Java SE 21 - Accessing Outer Class Members

NEW QUESTION # 54
Given:
java
int post = 5;
int pre = 5;
int postResult = post++ + 10;
int preResult = ++pre + 10;
System.out.println("postResult: " + postResult +
", preResult: " + preResult +
", Final value of post: " + post +
", Final value of pre: " + pre);
What is printed?

A. postResult: 16, preResult: 16, Final value of post: 6, Final value of pre: 6
B. postResult: 15, preResult: 16, Final value of post: 5, Final value of pre: 6
C. postResult: 15, preResult: 16, Final value of post: 6, Final value of pre: 6
D. postResult: 16, preResult: 15, Final value of post: 6, Final value of pre: 5

Answer: C

Explanation:
* Understanding post++ (Post-increment)
* post++uses the value first, then increments it.
* postResult = post++ + 10;
* post starts as 5.
* post++ returns 5, then post is incremented to 6.
* postResult = 5 + 10 = 15.
* Final value of post after this line is 6.
* Understanding ++pre (Pre-increment)
* ++preincrements the value first, then uses it.
* preResult = ++pre + 10;
* pre starts as 5.
* ++pre increments pre to 6, then returns 6.
* preResult = 6 + 10 = 16.
* Final value of pre after this line is 6.
Thus, the final output is:
yaml
postResult: 15, preResult: 16, Final value of post: 6, Final value of pre: 6 References:
* Java SE 21 - Operators and Expressions
* Java SE 21 - Arithmetic Operators

NEW QUESTION # 55
Given:
java
DoubleSummaryStatistics stats1 = new DoubleSummaryStatistics();
stats1.accept(4.5);
stats1.accept(6.0);
DoubleSummaryStatistics stats2 = new DoubleSummaryStatistics();
stats2.accept(3.0);
stats2.accept(8.5);
stats1.combine(stats2);
System.out.println("Sum: " + stats1.getSum() + ", Max: " + stats1.getMax() + ", Avg: " + stats1.getAverage()); What is printed?

A. Compilation fails.
B. An exception is thrown at runtime.
C. Sum: 22.0, Max: 8.5, Avg: 5.5
D. Sum: 22.0, Max: 8.5, Avg: 5.0

Answer: C

Explanation:
The DoubleSummaryStatistics class in Java is part of the java.util package and is used to collect and summarize statistics for a stream of double values. Let's analyze how the methods work:
* Initialization and Data Insertion
* stats1.accept(4.5); # Adds 4.5 to stats1.
* stats1.accept(6.0); # Adds 6.0 to stats1.
* stats2.accept(3.0); # Adds 3.0 to stats2.
* stats2.accept(8.5); # Adds 8.5 to stats2.
* Combining stats1 and stats2
* stats1.combine(stats2); merges stats2 into stats1, resulting in one statistics summary containing all values {4.5, 6.0, 3.0, 8.5}.
* Calculating Output Values
* Sum= 4.5 + 6.0 + 3.0 + 8.5 = 22.0
* Max= 8.5
* Average= (22.0) / 4 = 5.5
Thus, the output is:
yaml
Sum: 22.0, Max: 8.5, Avg: 5.5
References:
* Java SE 21 & JDK 21 - DoubleSummaryStatistics
* Java SE 21 - Streams and Statistical Operations

NEW QUESTION # 56
Given:
java
var frenchCities = new TreeSet<String>();
frenchCities.add("Paris");
frenchCities.add("Marseille");
frenchCities.add("Lyon");
frenchCities.add("Lille");
frenchCities.add("Toulouse");
System.out.println(frenchCities.headSet("Marseille"));
What will be printed?

A. [Lyon, Lille, Toulouse]
B. [Paris, Toulouse]
C. [Paris]
D. [Lille, Lyon]
E. Compilation fails

Answer: D

Explanation:
In this code, a TreeSet named frenchCities is created and populated with the following cities: "Paris",
"Marseille", "Lyon", "Lille", and "Toulouse". The TreeSet class in Java stores elements in a sorted order according to their natural ordering, which, for strings, is lexicographical order.
Sorted Order of Elements:
When the elements are added to the TreeSet, they are stored in the following order:
* "Lille"
* "Lyon"
* "Marseille"
* "Paris"
* "Toulouse"
headSet Method:
The headSet(E toElement) method of the TreeSet class returns a view of the portion of this set whose elements are strictly less than toElement. In this case, frenchCities.headSet("Marseille") will return a subset of frenchCities containing all elements that are lexicographically less than "Marseille".
Elements Less Than "Marseille":
From the sorted order, the elements that are less than "Marseille" are:
* "Lille"
* "Lyon"
Therefore, the output of the System.out.println statement will be [Lille, Lyon].
Option Evaluations:
* A. [Paris]: Incorrect. "Paris" is lexicographically greater than "Marseille".
* B. [Paris, Toulouse]: Incorrect. Both "Paris" and "Toulouse" are lexicographically greater than
"Marseille".
* C. [Lille, Lyon]: Correct. These are the elements less than "Marseille".
* D. Compilation fails: Incorrect. The code compiles successfully.
* E. [Lyon, Lille, Toulouse]: Incorrect. "Toulouse" is lexicographically greater than "Marseille".

NEW QUESTION # 57
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 an integer with value: 42
B. It throws an exception at runtime.
C. Compilation fails.
D. It's a double with value: 42
E. It's a string with value: 42
F. null

Answer: C

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 # 58
Which of the following statements oflocal variables declared with varareinvalid?(Choose 4)

A. var a = 1;(Valid: var correctly infers int)
B. var e;
C. var h = (g = 7);
D. var f = { 6 };
E. var b = 2, c = 3.0;
F. var d[] = new int[4];

Answer: B,D,E,F

Explanation:
1. Valid Use Cases of var
* var is alocal variable type inferencefeature.
* The compilerinfers the type from the assigned value.
* Example of valid use:
java
var a = 10; // Type inferred as int
var str = "Hello"; // Type inferred as String
2. Analyzing the Given Statements
Statement
Valid/Invalid
Reason
var a = 1;
Valid
Type inferred as int.
var b = 2, c = 3.0;
#Invalid
var doesnot allow multiple declarationsin one statement.
var d[] = new int[4];
#Invalid
Array brackets []are not allowedwith var.
var e;
#Invalid
varrequires an initializer(cannot be declared without assignment).
var f = { 6 };
#Invalid
{ 6 } is anarray initializer, which must have an explicit type.
var h = (g = 7);
Valid
g is assigned 7, and h gets its value.
Thus, the correct answers are:B, C, D, E
References:
* Java SE 21 - Local Variable Type Inference (var)
* Java SE 21 - var Restrictions

NEW QUESTION # 59
Which of the following isn't a valid option of the jdeps command?

A. --print-module-deps
B. --list-deps
C. --generate-open-module
D. --generate-module-info
E. --check-deps
F. --list-reduced-deps

Answer: E

Explanation:
The jdeps tool is a Java class dependency analyzer that can be used to understand the static dependencies of applications and libraries. It provides several command-line options to customize its behavior.
Valid jdeps Options:
* --generate-open-module: Generates a module declaration (module-info.java) with open directives for the given JAR files or classes.
* --list-deps: Lists the immediate dependencies of the specified classes or JAR files.
* --generate-module-info: Generates a module declaration (module-info.java) for the given JAR files or classes.
* --print-module-deps: Prints the module dependencies of the specified modules or JAR files.
* --list-reduced-deps: Lists the reduced dependencies, showing only the packages that are directly depended upon.
Invalid Option:
* --check-deps: There is no --check-deps option in the jdeps tool.
Conclusion:
Option A (--check-deps) is not a valid option of the jdeps command.

NEW QUESTION # 60
Given:
java
public class Test {
public static void main(String[] args) throws IOException {
Path p1 = Path.of("f1.txt");
Path p2 = Path.of("f2.txt");
Files.move(p1, p2);
Files.delete(p1);
}
}
In which case does the given program throw an exception?

A. Both files f1.txt and f2.txt exist
B. Neither files f1.txt nor f2.txt exist
C. File f1.txt exists while file f2.txt doesn't
D. File f2.txt exists while file f1.txt doesn't
E. An exception is always thrown

Answer: E

Explanation:
In this program, the following operations are performed:
* Paths Initialization:
* Path p1 is set to "f1.txt".
* Path p2 is set to "f2.txt".
* File Move Operation:
* Files.move(p1, p2); attempts to move (or rename) f1.txt to f2.txt.
* File Delete Operation:
* Files.delete(p1); attempts to delete f1.txt.
Analysis:
* If f1.txt Does Not Exist:
* The Files.move(p1, p2); operation will throw a NoSuchFileException because the source file f1.
txt is missing.
* If f1.txt Exists and f2.txt Does Not Exist:
* The Files.move(p1, p2); operation will successfully rename f1.txt to f2.txt.
* Subsequently, the Files.delete(p1); operation will throw a NoSuchFileException because p1 (now f1.txt) no longer exists after the move.
* If Both f1.txt and f2.txt Exist:
* The Files.move(p1, p2); operation will throw a FileAlreadyExistsException because the target file f2.txt already exists.
* If f2.txt Exists While f1.txt Does Not:
* Similar to the first scenario, the Files.move(p1, p2); operation will throw a NoSuchFileException due to the absence of f1.txt.
In all possible scenarios, an exception is thrown during the execution of the program.

NEW QUESTION # 61
Given:
java
ExecutorService service = Executors.newFixedThreadPool(2);
Runnable task = () -> System.out.println("Task is complete");
service.submit(task);
service.shutdown();
service.submit(task);
What happens when executing the given code fragment?

A. It prints "Task is complete" twice and throws an exception.
B. It exits normally without printing anything to the console.
C. It prints "Task is complete" once, then exits normally.
D. It prints "Task is complete" twice, then exits normally.
E. It prints "Task is complete" once and throws an exception.

Answer: E

Explanation:
In this code, an ExecutorService is created with a fixed thread pool of size 2 using Executors.
newFixedThreadPool(2). A Runnable task is defined to print "Task is complete" to the console.
The sequence of operations is as follows:
* service.submit(task);
This submits the task to the executor service for execution. Since the thread pool has a size of 2 and no other tasks are running, this task will be executed promptly, printing "Task is complete" to the console.
* service.shutdown();
This initiates an orderly shutdown of the executor service. In this state, the service stops accepting new tasks

NEW QUESTION # 62
Given:
java
Object myVar = 0;
String print = switch (myVar) {
case int i -> "integer";
case long l -> "long";
case String s -> "string";
default -> "";
};
System.out.println(print);
What is printed?

A. string
B. It throws an exception at runtime.
C. nothing
D. Compilation fails.
E. long
F. integer

Answer: D

Explanation:
* Why does the compilation fail?
* TheJava switch statement does not support primitive type pattern matchingin switch expressions as of Java 21.
* The case pattern case int i -> "integer"; isinvalidbecausepattern 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 acompilation 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

NEW QUESTION # 63
......

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