Get Real 1z0-830 Exam Dumps [Aug-2025] Practice Tests [Q35-Q58]

Get Real 1z0-830 Exam Dumps [Aug-2025] Practice Tests

Last 1z0-830 practice test reviews: Practice Test Oracle dumps

NEW QUESTION # 35
Which of the following statements are correct?

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

Answer: A

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 # 36
Given:
java
public class SpecialAddition extends Addition implements Special {
public static void main(String[] args) {
System.out.println(new SpecialAddition().add());
}
int add() {
return --foo + bar--;
}
}
class Addition {
int foo = 1;
}
interface Special {
int bar = 1;
}
What is printed?

• A. 0
• B. 1
• C. It throws an exception at runtime.
• D. 2
• E. Compilation fails.

Answer: E

Explanation:
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 acompilation 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

NEW QUESTION # 37
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. 0
• B. Optional[1]
• C. 1
• D. Compilation fails
• E. Optional.empty
• F. An exception is thrown
• G. 2

Answer: G

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 # 38
Consider the following methods to load an implementation of MyService using ServiceLoader. Which of the methods are correct? (Choose all that apply)

• A. MyService service = ServiceLoader.getService(MyService.class);
• B. MyService service = ServiceLoader.load(MyService.class).iterator().next();
• C. MyService service = ServiceLoader.services(MyService.class).getFirstInstance();
• D. MyService service = ServiceLoader.load(MyService.class).findFirst().get();

Answer: B,D

Explanation:
The ServiceLoader class in Java is used to load service providers implementing a given service interface. The following methods are evaluated for their correctness in loading an implementation of MyService:
* A. MyService service = ServiceLoader.load(MyService.class).iterator().next(); This method uses the ServiceLoader.load(MyService.class) to create a ServiceLoader instance for MyService.
Calling iterator().next() retrieves the next available service provider. If no providers are available, a NoSuchElementException will be thrown. This approach is correct but requires handling the potential exception if no providers are found.
* B. MyService service = ServiceLoader.load(MyService.class).findFirst().get(); This method utilizes the findFirst() method introduced in Java 9, which returns an Optional describing the first available service provider. Calling get() on the Optional retrieves the service provider if present; otherwise, a NoSuchElementException is thrown. This approach is correct and provides a more concise way to obtain the first service provider.
* C. MyService service = ServiceLoader.getService(MyService.class);
The ServiceLoader class does not have a method named getService. Therefore, this method is incorrect and will result in a compilation error.
* D. MyService service = ServiceLoader.services(MyService.class).getFirstInstance(); The ServiceLoader class does not have a method named services or getFirstInstance. Therefore, this method is incorrect and will result in a compilation error.
In summary, options A and B are correct methods to load an implementation of MyService using ServiceLoader.

NEW QUESTION # 39
Given:
java
var now = LocalDate.now();
var format1 = new DateTimeFormatter(ISO_WEEK_DATE);
var format2 = DateTimeFormatter.ISO_WEEK_DATE;
var format3 = new DateFormat(WEEK_OF_YEAR_FIELD);
var format4 = DateFormat.getDateInstance(WEEK_OF_YEAR_FIELD);
System.out.println(now.format(REPLACE_HERE));
Which variable prints 2025-W01-2 (present-day is 12/31/2024)?

• A. format3
• B. format4
• C. format1
• D. format2

Answer: D

Explanation:
In this code, now is assigned the current date using LocalDate.now(). The goal is to format this date to the ISO week date format, which represents dates in the YYYY-'W'WW-E pattern, where:
* YYYY: Week-based year
* 'W': Literal 'W' character
* WW: Week number
* E: Day of the week
Given that the present day is December 31, 2024, this date falls in the first week of the week-based year 2025.
Therefore, the ISO week date representation would be 2025-W01-2, where '2' denotes Tuesday.
Among the provided formatters:
* format1: This line attempts to create a DateTimeFormatter using a constructor, which is incorrect because DateTimeFormatter does not have a public constructor that accepts a pattern directly. This would result in a compilation error.
* format2: This is correctly assigned the predefined DateTimeFormatter.ISO_WEEK_DATE, which formats dates in the ISO week date format.
* format3: This line attempts to create a DateFormat instance using a field, which is incorrect because DateFormat does not have such a constructor. This would result in a compilation error.
* format4: This line attempts to get a DateFormat instance using an integer field, which is incorrect because DateFormat.getDateInstance() does not accept such parameters. This would result in a compilation error.
Therefore, the only correct and applicable formatter is format2. Using format2 in the now.format() method will produce the desired output: 2025-W01-2.

NEW QUESTION # 40
What does the following code print?
java
import java.util.stream.Stream;
public class StreamReduce {
public static void main(String[] args) {
Stream<String> stream = Stream.of("J", "a", "v", "a");
System.out.print(stream.reduce(String::concat));
}
}

• A. Optional[Java]
• B. null
• C. Java
• D. Compilation fails

Answer: A

Explanation:
In this code, a Stream of String elements is created containing the characters "J", "a", "v", and "a". The reduce method is then used with String::concat as the accumulator function.
The reduce method with a single BinaryOperator parameter performs a reduction on the elements of the stream, using an associative accumulation function, and returns an Optional describing the reduced value, if any. In this case, it concatenates the strings in the stream.
Since the stream contains elements, the reduction operation concatenates them to form the string "Java". The result is wrapped in an Optional, resulting in Optional[Java]. The print statement outputs this Optional object, displaying Optional[Java].

NEW QUESTION # 41
How would you create a ConcurrentHashMap configured to allow a maximum of 10 concurrent writer threads and an initial capacity of 42?
Which of the following options meets this requirement?

• A. var concurrentHashMap = new ConcurrentHashMap(42, 0.88f, 10);
• B. var concurrentHashMap = new ConcurrentHashMap(42, 10);
• C. var concurrentHashMap = new ConcurrentHashMap(42);
• D. None of the suggestions.
• E. var concurrentHashMap = new ConcurrentHashMap();

Answer: A

Explanation:
In Java, the ConcurrentHashMap class provides several constructors that allow for the customization of its initial capacity, load factor, and concurrency level. To configure a ConcurrentHashMap with an initial capacity of 42 and a concurrency level of 10, you can use the following constructor:
java
public ConcurrentHashMap(int initialCapacity, float loadFactor, int concurrencyLevel) Parameters:
* initialCapacity: The initial capacity of the hash table. This is the number of buckets that the hash table will have when it is created. In this case, it is set to 42.
* loadFactor: A measure of how full the hash table is allowed to get before it is resized. The default value is 0.75, but in this case, it is set to 0.88.
* concurrencyLevel: The estimated number of concurrently updating threads. This is used as a hint for internal sizing. In this case, it is set to 10.
Therefore, to create a ConcurrentHashMap with an initial capacity of 42, a load factor of 0.88, and a concurrency level of 10, you can use the following code:
java
var concurrentHashMap = new ConcurrentHashMap<>(42, 0.88f, 10);
Option Evaluations:
* A. var concurrentHashMap = new ConcurrentHashMap(42);: This constructor sets the initial capacity to 42 but uses the default load factor (0.75) and concurrency level (16). It does not meet the requirement of setting the concurrency level to 10.
* B. None of the suggestions.: This is incorrect because option E provides the correct configuration.
* C. var concurrentHashMap = new ConcurrentHashMap();: This uses the default constructor, which sets the initial capacity to 16, the load factor to 0.75, and the concurrency level to 16. It does not meet the specified requirements.
* D. var concurrentHashMap = new ConcurrentHashMap(42, 10);: This constructor sets the initial capacity to 42 and the load factor to 10, which is incorrect because the load factor should be a float value between 0 and 1.
* E. var concurrentHashMap = new ConcurrentHashMap(42, 0.88f, 10);: This correctly sets the initial capacity to 42, the load factor to 0.88, and the concurrency level to 10, meeting all the specified requirements.
Therefore, the correct answer is option E.

NEW QUESTION # 42
Given:
java
void verifyNotNull(Object input) {
boolean enabled = false;
assert enabled = true;
assert enabled;
System.out.println(input.toString());
assert input != null;
}
When does the given method throw a NullPointerException?

• A. Only if assertions are enabled and the input argument isn't null
• B. A NullPointerException is never thrown
• C. Only if assertions are enabled and the input argument is null
• D. Only if assertions are disabled and the input argument isn't null
• E. Only if assertions are disabled and the input argument is null

Answer: E

Explanation:
In the verifyNotNull method, the following operations are performed:
* Assertion to Enable Assertions:
java
boolean enabled = false;
assert enabled = true;
assert enabled;
* The variable enabled is initially set to false.
* The first assertion assert enabled = true; assigns true to enabled if assertions are enabled. If assertions are disabled, this assignment does not occur.
* The second assertion assert enabled; checks if enabled is true. If assertions are enabled and the previous assignment occurred, this assertion passes. If assertions are disabled, this assertion is ignored.
* Dereferencing the input Object:
java
System.out.println(input.toString());
* This line attempts to call the toString() method on the input object. If input is null, this will throw a NullPointerException.
* Assertion to Check input for null:
java
assert input != null;
* This assertion checks that input is not null. If input is null and assertions are enabled, this assertion will fail, throwing an AssertionError. If assertions are disabled, this assertion is ignored.
Analysis:
* If Assertions Are Enabled:
* The enabled variable is set to true by the first assertion, and the second assertion passes.
* If input is null, calling input.toString() will throw a NullPointerException before the final assertion is reached.
* If input is not null, input.toString() executes without issue, and the final assertion assert input != null; passes.
* If Assertions Are Disabled:
* The enabled variable remains false, but the assertions are ignored, so this has no effect.
* If input is null, calling input.toString() will throw a NullPointerException.
* If input is not null, input.toString() executes without issue.
Conclusion:
A NullPointerException is thrown if input is null, regardless of whether assertions are enabled or disabled.
Therefore, the correct answer is:
C: Only if assertions are disabled and the input argument is null

NEW QUESTION # 43
Given:
java
List<Long> cannesFestivalfeatureFilms = LongStream.range(1, 1945)
.boxed()
.toList();
try (var executor = Executors.newVirtualThreadPerTaskExecutor()) {
cannesFestivalfeatureFilms.stream()
.limit(25)
.forEach(film -> executor.submit(() -> {
System.out.println(film);
}));
}
What is printed?

• A. Numbers from 1 to 25 randomly
• B. An exception is thrown at runtime
• C. Compilation fails
• D. Numbers from 1 to 25 sequentially
• E. Numbers from 1 to 1945 randomly

Answer: A

Explanation:
* Understanding LongStream.range(1, 1945).boxed().toList();
* LongStream.range(1, 1945) generates a stream of numbersfrom 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 introducedvirtual threadsto 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 thefirst 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
3
1
5
2
4
7
25
* The ordermay differ in each rundue 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()

NEW QUESTION # 44
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, ArrayList<String>> authorsMap1 = new HashMap<>();
  java
  authorsMap1.put("FR", frenchAuthors);
• C. Map<String, List<String>> authorsMap5 = new HashMap<String, List<String>>(); java authorsMap5.put("FR", frenchAuthors);
• D. Map<String, ? extends List<String>> authorsMap2 = new HashMap<String, ArrayList<String>> (); java authorsMap2.put("FR", frenchAuthors);
• E. var authorsMap3 = new HashMap<>();
  java
  authorsMap3.put("FR", frenchAuthors);

Answer: A,C,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 # 45
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?

• A. Compilation fails
• B. ok the 2024-07-10
• C. ok the 2024-07-10T07:17:45.523939600
• D. An exception is thrown

Answer: A

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 # 46
Which of the following suggestions compile?(Choose two.)

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

Answer: A,C

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 # 47
Given:
java
StringBuilder result = Stream.of("a", "b")
.collect(
() -> new StringBuilder("c"),
StringBuilder::append,
(a, b) -> b.append(a)
);
System.out.println(result);
What is the output of the given code fragment?

• A. cacb
• B. cba
• C. cbca
• D. bac
• E. bca
• F. acb
• G. abc

Answer: B

Explanation:
In this code, a Stream containing the elements "a" and "b" is processed using the collect method. The collect method is a terminal operation that performs a mutable reduction on the elements of the stream using a Collector. In this case, custom implementations for the supplier, accumulator, and combiner are provided.
Components of the collect Method:
* Supplier:
* () -> new StringBuilder("c")
* This supplier creates a new StringBuilder initialized with the string "c".
* Accumulator:
* StringBuilder::append
* This accumulator appends each element of the stream to the StringBuilder.
* Combiner:
* (a, b) -> b.append(a)
* This combiner is used in parallel stream operations to merge two StringBuilder instances. It appends the contents of a to b.
Execution Flow:
* Stream Elements:"a", "b"
* Initial StringBuilder:"c"
* Accumulation:
* The first element "a" is appended to "c", resulting in "ca".
* The second element "b" is appended to "ca", resulting in "cab".
* Combiner:
* In this sequential stream, the combiner is not utilized. The combiner is primarily used in parallel streams to merge partial results.
Final Result:
The StringBuilder contains "cab". Therefore, the output of the program is:
nginx
cab

NEW QUESTION # 48
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. Peugeot
• B. Peugeot 807
• C. An exception is thrown at runtime.
• D. Compilation fails.

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 # 49
Given:
java
var counter = 0;
do {
System.out.print(counter + " ");
} while (++counter < 3);
What is printed?

• A. 1 2 3 4
• B. An exception is thrown.
• C. 0 1 2 3
• D. Compilation fails.
• E. 1 2 3
• 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 # 50
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. It throws an exception at runtime.
• B. integer
• C. string
• D. long
• E. nothing
• F. Compilation fails.

Answer: F

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 # 51
Which of the following doesnotexist?

• A. Supplier<T>
• B. BooleanSupplier
• C. BiSupplier<T, U, R>
• D. DoubleSupplier
• E. LongSupplier
• F. They all exist.

Answer: C

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

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

Answer: C,E

NEW QUESTION # 53
Given:
java
System.out.print(Boolean.logicalAnd(1 == 1, 2 < 1));
System.out.print(Boolean.logicalOr(1 == 1, 2 < 1));
System.out.print(Boolean.logicalXor(1 == 1, 2 < 1));
What is printed?

• A. truetruetrue
• B. falsetruetrue
• C. Compilation fails
• D. truefalsetrue
• E. truetruefalse

Answer: D

Explanation:
In this code, three static methods from the Boolean class are used: logicalAnd, logicalOr, and logicalXor.
Each method takes two boolean arguments and returns a boolean result based on the respective logical operation.
Evaluation of Each Statement:
* Boolean.logicalAnd(1 == 1, 2 < 1)
* Operands:
* 1 == 1 evaluates to true.
* 2 < 1 evaluates to false.
* Operation:
* Boolean.logicalAnd(true, false) performs a logical AND operation.
* The result is false because both operands must be true for the AND operation to return true.
* Output:
* System.out.print(false); prints false.
* Boolean.logicalOr(1 == 1, 2 < 1)
* Operands:
* 1 == 1 evaluates to true.
* 2 < 1 evaluates to false.
* Operation:
* Boolean.logicalOr(true, false) performs a logical OR operation.
* The result is true because at least one operand is true.
* Output:
* System.out.print(true); prints true.
* Boolean.logicalXor(1 == 1, 2 < 1)
* Operands:
* 1 == 1 evaluates to true.
* 2 < 1 evaluates to false.
* Operation:
* Boolean.logicalXor(true, false) performs a logical XOR (exclusive OR) operation.
* The result is true because exactly one operand is true.
* Output:
* System.out.print(true); prints true.
Combined Output:
Combining the outputs from each statement, the final printed result is:
nginx
falsetruetrue

NEW QUESTION # 54
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. 0
• C. 42 000,00 €
• D. 42000E

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 # 55
Which of the following can be the body of a lambda expression?

• A. A statement block
• B. Two statements
• C. An expression and a statement
• D. Two expressions
• E. None of the above

Answer: A

Explanation:
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.

NEW QUESTION # 56
Given:
java
String textBlock = """
j \
a \t
v \s
a \
""";
System.out.println(textBlock.length());
What is the output?

• A. 0
• B. 1
• C. 2
• D. 3

Answer: C

Explanation:
In this code, a text block is defined using the """ syntax introduced in Java 13. Text blocks allow for multiline string literals, preserving the format as written in the code.
Text Block Analysis:
The text block is defined as:
java
String textBlock = """
j \
a \t
contentReference[oaicite:0]{index=0}

NEW QUESTION # 57
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 prints "Task is complete" twice, then exits normally.
• C. It prints "Task is complete" once, then exits normally.
• D. It exits normally without printing anything to the console.
• 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 # 58
......

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