1z0-830 Oracle Java SE 21 Developer Professional exact Exam Questions

Java SE 21 Developer Professional

Last Update 4 hours ago Total Questions : 84

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Question # 11

Which of the following statements is correct about a final class?

The final keyword in its declaration must go right before the class keyword.

It must contain at least a final method.

It cannot be extended by any other class.

It cannot implement any interface.

It cannot extend another class.

Question # 12

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?

42000E

42 000,00 €

42000

42 k

Question # 13

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

Stream stream = Stream.of("a");

Stream stream = Stream.ofNullable("a");

Stream stream = Stream.generate(() - > "a");

Stream stream = Stream.of();

Stream stream = new Stream();

Stream < String > stream = Stream.builder().add("a").build();

Stream stream = Stream.empty();

Question # 14

Given:

java

public static void main(String[] args) {

try {

throw new IOException();

} catch (IOException e) {

throw new RuntimeException();

} finally {

throw new ArithmeticException();

}

What is the output?

Compilation fails

IOException

RuntimeException

ArithmeticException

Question # 15

Given:

java

import java.io.*;

class A implements Serializable {

int number = 1;

}

class B implements Serializable {

int number = 2;

}

public class Test {

public static void main(String[] args) throws Exception {

File file = new File("o.ser");

A a = new A();

var oos = new ObjectOutputStream(new FileOutputStream(file));

oos.writeObject(a);

oos.close();

var ois = new ObjectInputStream(new FileInputStream(file));

B b = (B) ois.readObject();

ois.close();

System.out.println(b.number);

}

What is the given program's output?

Compilation fails

ClassCastException

NotSerializableException

Question # 16

Given:

java

Optional < String > optionalName = Optional.ofNullable(null);

String bread = optionalName.orElse("Baguette");

System.out.print("bread:" + bread);

String dish = optionalName.orElseGet(() - > "Frog legs");

System.out.print(", dish:" + dish);

try {

String cheese = optionalName.orElseThrow(() - > new Exception());

System.out.println(", cheese:" + cheese);

} catch (Exception exc) {

System.out.println(", no cheese.");

}

What is printed?

bread:Baguette, dish:Frog legs, cheese.

bread:Baguette, dish:Frog legs, no cheese.

bread:bread, dish:dish, cheese.

Compilation fails.

Question # 17

Given:

java

List < String > frenchAuthors = new ArrayList < > ();

frenchAuthors.add("Victor Hugo");

frenchAuthors.add("Gustave Flaubert");

Which compiles?

Map < String, ArrayList < String > > authorsMap1 = new HashMap < > ();

java

authorsMap1.put("FR", frenchAuthors);

B.
Map < String, ? extends List < String > > authorsMap2 = new HashMap < String, ArrayList < String > > ();
java
authorsMap2.put("FR", frenchAuthors);

C.
var authorsMap3 = new HashMap < > ();
java
authorsMap3.put("FR", frenchAuthors);

D.
Map < String, List < String > > authorsMap4 = new HashMap < String, ArrayList < String > > ();
java
authorsMap4.put("FR", frenchAuthors);

E.
Map < String, List < String > > authorsMap5 = new HashMap < String, List < String > > ();
java
authorsMap5.put("FR", frenchAuthors);

Answer:

C, D, E

Explanation:

Option A (Map < String, ArrayList < String > > authorsMap1 = new HashMap < > ();)
❌ Compilation Fails
frenchAuthors is declared as List < String > , not ArrayList < String > .
The correct way to declare a Map that allows storing List < String > is to use List < String > as the generic type, not ArrayList < 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 mismatch error.
Option B (Map < String, ? extends List < String > > authorsMap2 = new HashMap < String, ArrayList < String > > ();)
❌ Compilation Fails
? extends List < String > makes the map read-only for adding new elements.
The line authorsMap2.put("FR", frenchAuthors); causes a compilation error because 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 > > is allowed 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 > > is a 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

Question # 18

Which of the following methods of java.util.function.Predicate are default methods ?

A.
and(Predicate < ? super T > other)

B.
isEqual(Object targetRef)

C.
negate()

D.
not(Predicate < ? super T > target)

E.
or(Predicate < ? super T > other)

F.
test(T t)

Answer:

A, C, E

Explanation:

Understanding java.util.function.Predicate < T >
The Predicate < T > interface represents a function that takes an input and returns a boolean (true or false).
It is often used for filtering operations in functional programming and streams.
Analyzing the Methods:
and(Predicate < ? super T > other) → Default method
Combines two predicates using logical AND ( & & ).
java
Predicate < String > startsWithA = s - > s.startsWith("A");
Predicate < String > hasLength3 = s - > s.length() == 3;
Predicate < String > combined = startsWithA.and(hasLength3);
❌ isEqual(Object targetRef) → Static method
Not a default method , because it does not operate on an instance .
java
Predicate < String > isEqualToHello = Predicate.isEqual("Hello");
negate() → Default method
Negates a predicate (! operator).
java
Predicate < String > notEmpty = s - > !s.isEmpty();
Predicate < String > isEmpty = notEmpty.negate();
❌ not(Predicate < ? super T > target) → Static method (introduced in Java 11)
Not a default method , since it is static.
or(Predicate < ? super T > other) → Default method
Combines two predicates using logical OR (||).
❌ test(T t) → Abstract method
Not a default method , because every predicate must implement this method .
Thus, the correct answers are: and(Predicate < ? super T > other), negate(), or(Predicate < ? super T > other)
References:
Java SE 21 - Predicate Interface
Java SE 21 - Functional Interfaces

Question # 19

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 sequentially

B.
Numbers from 1 to 25 randomly

C.
Numbers from 1 to 1945 randomly

D.
An exception is thrown at runtime

E.
Compilation fails

Answer:

B

Explanation:

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 are not 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 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()

Question # 20

Given:
java
var array1 = new String[]{ "foo", "bar", "buz" };
var array2[] = { "foo", "bar", "buz" };
var array3 = new String[3] { "foo", "bar", "buz" };
var array4 = { "foo", "bar", "buz" };
String array5[] = new String[] { "foo", "bar", "buz" };
Which arrays compile? (Select 2)

A.
array1

B.
array2

C.
array3

D.
array4

E.
array5

Answer:

A, E

Explanation:

In Java, array initialization can be performed in several ways, but certain syntaxes are invalid and will cause compilation errors. Let's analyze each declaration:
var array1 = new String[]{ "foo", "bar", "buz" };
This is a valid declaration. The var keyword allows the compiler to infer the type from the initializer. Here, new String[]{ "foo", "bar", "buz" } creates an anonymous array of String with three elements. The compiler infers array1 as String[] . This syntax is correct and compiles successfully.
var array2[] = { "foo", "bar", "buz" };
This declaration is invalid. While var can be used for type inference, appending [] after var is not allowed. The correct syntax would be either String[] array2 = { "foo", "bar", "buz" }; or var array2 = new String[]{ "foo", "bar", "buz" };. Therefore, this line will cause a compilation error.
var array3 = new String[3] { "foo", "bar", "buz" };
This declaration is invalid. In Java, when specifying the size of the array (new String[3]), you cannot simultaneously provide an initializer. The correct approach is either to provide the size without an initializer (new String[3] ) or to provide the initializer without specifying the size (new String[]{ "foo", "bar", "buz" }). Therefore, this line will cause a compilation error.
var array4 = { "foo", "bar", "buz" };
This declaration is invalid. The array initializer { "foo", "bar", "buz" } can only be used in an array declaration when the type is explicitly provided. Since var relies on type inference and there's no explicit type provided here, this will cause a compilation error. The correct syntax would be String[] array4 = { "foo", "bar", "buz" };.
String array5[] = new String[] { "foo", "bar", "buz" };
This is a valid declaration. Here, String array5[] declares array5 as an array of String. The initializer new String[] { "foo", "bar", "buz" } creates an array with three elements. This syntax is correct and compiles successfully.
Therefore, the declarations that compile successfully are array1 and array5.
References:
Java SE 21 & JDK 21 - Local Variable Type Inference
Java SE 21 & JDK 21 - Arrays

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