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The study material is available in three easy-to-access formats. The first one is PDF format which is printable and portable. You can access it anywhere with your smart devices like smartphones, tablets, and laptops. In addition, you can even print PDF questions in order to study anywhere and pass Java SE 21 Developer Professional (1z1-830) certification exam.

Oracle Java SE 21 Developer Professional Sample Questions (Q52-Q57):

NEW QUESTION # 52
Given:
java
public class Versailles {
int mirrorsCount;
int gardensHectares;
void Versailles() { // n1
this.mirrorsCount = 17;
this.gardensHectares = 800;
System.out.println("Hall of Mirrors has " + mirrorsCount + " mirrors."); System.out.println("The gardens cover " + gardensHectares + " hectares.");
}
public static void main(String[] args) {
var castle = new Versailles(); // n2
}
}
What is printed?

A. Compilation fails at line n1.
B. An exception is thrown at runtime.
C. Nothing
D. nginx
Hall of Mirrors has 17 mirrors.
The gardens cover 800 hectares.
E. Compilation fails at line n2.

Answer: A

Explanation:
* Understanding Constructors vs. Methods in Java
* In Java, aconstructormustnot have a return type.
* The followingis NOT a constructorbut aregular method:
java
void Versailles() { // This is NOT a constructor!
* Correct way to define a constructor:
java
public Versailles() { // Constructor must not have a return type
* Since there isno constructor explicitly defined,Java provides a default no-argument constructor, which does nothing.
* Why Does Compilation Fail?
* void Versailles() is interpreted as amethod,not a constructor.
* This means the default constructor (which does nothing) is called.
* Since the method Versailles() is never called, the object fields remain uninitialized.
* If the constructor were correctly defined, the output would be:
nginx
Hall of Mirrors has 17 mirrors.
The gardens cover 800 hectares.
* How to Fix It
java
public Versailles() { // Corrected constructor
this.mirrorsCount = 17;
this.gardensHectares = 800;
System.out.println("Hall of Mirrors has " + mirrorsCount + " mirrors."); System.out.println("The gardens cover " + gardensHectares + " hectares.");
}
Thus, the correct answer is:Compilation fails at line n1.
References:
* Java SE 21 - Constructors
* Java SE 21 - Methods vs. Constructors

NEW QUESTION # 53
Given:
java
List<String> abc = List.of("a", "b", "c");
abc.stream()
.forEach(x -> {
x = x.toUpperCase();
});
abc.stream()
.forEach(System.out::print);
What is the output?

A. Compilation fails.
B. abc
C. ABC
D. An exception is thrown.

Answer: B

Explanation:
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.

NEW QUESTION # 54
Which methods compile?

A. ```java public List<? super IOException> getListSuper() { return new ArrayList<Exception>(); } csharp
B. ```java
public List<? super IOException> getListSuper() {
return new ArrayList<FileNotFoundException>();
}
C. ```java public List<? extends IOException> getListExtends() { return new ArrayList<Exception>(); } csharp
D. ```java
public List<? extends IOException> getListExtends() {
return new ArrayList<FileNotFoundException>();
}

Answer: A,D

Explanation:
In Java generics, wildcards are used to relax the type constraints of generic types. The extends wildcard (<?
extends Type>) denotes an upper bounded wildcard, allowing any type that is a subclass of Type. Conversely, the super wildcard (<? super Type>) denotes a lower bounded wildcard, allowing any type that is a superclass of Type.
Option A:
java
public List<? super IOException> getListSuper() {
return new ArrayList<Exception>();
}
Here, List<? super IOException> represents a list that can hold IOException objects and objects of its supertypes. Since Exception is a superclass of IOException, ArrayList<Exception> is compatible with List<?
super IOException>. Therefore, this method compiles successfully.
Option B:
java
public List<? extends IOException> getListExtends() {
return new ArrayList<FileNotFoundException>();
}
In this case, List<? extends IOException> represents a list that can hold objects of IOException and its subclasses. Since FileNotFoundException is a subclass of IOException, ArrayList<FileNotFoundException> is compatible with List<? extends IOException>. Thus, this method compiles successfully.
Option C:
java
public List<? extends IOException> getListExtends() {
return new ArrayList<Exception>();
}
Here, List<? extends IOException> expects a list of IOException or its subclasses. However, Exception is a superclass of IOException, not a subclass. Therefore, ArrayList<Exception> is not compatible with List<?
extends IOException>, and this method will not compile.
Option D:
java
public List<? super IOException> getListSuper() {
return new ArrayList<FileNotFoundException>();
}
In this scenario, List<? super IOException> expects a list that can hold IOException objects and objects of its supertypes. Since FileNotFoundException is a subclass of IOException, ArrayList<FileNotFoundException> is not compatible with List<? super IOException>, and this method will not compile.
Therefore, the methods in options A and B compile successfully, while those in options C and D do not.

NEW QUESTION # 55
A module com.eiffeltower.shop with the related sources in the src directory.
That module requires com.eiffeltower.membership, available in a JAR located in the lib directory.
What is the command to compile the module com.eiffeltower.shop?

A. css
CopyEdit
javac --module-source-path src -p lib/com.eiffel.membership.jar -d out -m com.eiffeltower.shop
B. css
CopyEdit
javac -path src -p lib/com.eiffel.membership.jar -d out -m com.eiffeltower.shop
C. bash
CopyEdit
javac -source src -p lib/com.eiffel.membership.jar -d out -m com.eiffeltower.shop
D. css
CopyEdit
javac --module-source-path src -p lib/com.eiffel.membership.jar -s out -m com.eiffeltower.shop

Answer: A

Explanation:
Comprehensive and Detailed In-Depth Explanation:
Understanding Java Module Compilation (javac)
Java modules are compiled using the javac command with specific options to specify:
* Where the source files are located (--module-source-path)
* Where required dependencies (external modules) are located (-p / --module-path)
* Where the compiled output should be placed (-d)
Breaking Down the Correct Compilation Command
css
CopyEdit
javac --module-source-path src -p lib/com.eiffel.membership.jar -d out -m com.eiffeltower.shop
* --module-source-path src # Specifies the directory where module sources are located.
* -p lib/com.eiffel.membership.jar # Specifies the module path (JAR dependency in lib).
* -d out # Specifies the output directory for compiled .class files.
* -m com.eiffeltower.shop # Specifies the module to compile (com.eiffeltower.shop).

NEW QUESTION # 56
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 is null
B. Only if assertions are disabled and the input argument isn't null
C. A NullPointerException is never thrown
D. Only if assertions are enabled 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 # 57
......

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2025 Valid Dumps 1z1-830 Files | High-quality Java SE 21 Developer Professional 100% Free Valid Exam Camp

2025 Accurate 100% Free 1z1-830 – 100% Free Valid Dumps Files | Java SE 21 Developer Professional Valid Exam Camp

Oracle Java SE 21 Developer Professional Sample Questions (Q52-Q57):

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