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Oracle Java SE 21 Developer Professional Sample Questions (Q29-Q34):

NEW QUESTION # 29
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. Compilation fails at line n2.
C. An exception is thrown at runtime.
D. nginx
Hall of Mirrors has 17 mirrors.
The gardens cover 800 hectares.
E. Nothing

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 # 30
Given:
java
var lyrics = """
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose
""";
for ( int i = 0, int j = 3; i < j; i++ ) {
System.out.println( lyrics.lines()
.toList()
.get( i ) );
}
What is printed?

A. Compilation fails.
B. vbnet
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose
C. An exception is thrown at runtime.
D. Nothing

Answer: A

Explanation:
* Error in for Loop Initialization
* The initialization part of a for loopcannot declare multiple variables with different types in a single statement.
* Error:
java
for (int i = 0, int j = 3; i < j; i++) {
* Fix:Declare variables separately:
java
for (int i = 0, j = 3; i < j; i++) {
* lyrics.lines() in Java 21
* The lines() method of String returns aStream<String>, splitting the string by line breaks.
* Calling .toList() on a streamconverts it to a list.
* Valid Code After Fixing the Loop:
java
var lyrics = """
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose
""";
for (int i = 0, j = 3; i < j; i++) {
System.out.println(lyrics.lines()
toList()
get(i));
}
* Expected Output After Fixing:
vbnet
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - String.lines()
* Java SE 21 - for Statement Rules

NEW QUESTION # 31
Given:
java
record WithInstanceField(String foo, int bar) {
double fuz;
}
record WithStaticField(String foo, int bar) {
static double wiz;
}
record ExtendingClass(String foo) extends Exception {}
record ImplementingInterface(String foo) implements Cloneable {}
Which records compile? (Select 2)

A. WithInstanceField
B. ImplementingInterface
C. ExtendingClass
D. WithStaticField

Answer: B,D

Explanation:
In Java, records are a special kind of class designed to act as transparent carriers for immutabledata. They automatically provide implementations for equals(), hashCode(), and toString(), and their fields are final and private by default.
* Option A: ExtendingClass
* Analysis: Records in Java implicitly extend java.lang.Record and cannot extend any other class because Java does not support multiple inheritance. Attempting to extend another class, such as Exception, will result in a compilation error.
* Conclusion: Does not compile.
* Option B: WithInstanceField
* Analysis: Records do not allow the declaration of instance fields outside of their components.
The declaration of double fuz; is not permitted and will cause a compilation error.
* Conclusion: Does not compile.
* Option C: ImplementingInterface
* Analysis: Records can implement interfaces. In this case, ImplementingInterface implements Cloneable, which is valid.
* Conclusion: Compiles successfully.

NEW QUESTION # 32
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?

A. bread:Baguette, dish:Frog legs, no cheese.
B. Compilation fails.
C. bread:bread, dish:dish, cheese.
D. bread:Baguette, dish:Frog legs, cheese.

Answer: A

Explanation:
Understanding Optional.ofNullable(null)
* Optional.ofNullable(null); creates an empty Optional (i.e., it contains no value).
* Optional.of(null); would throw a NullPointerException, but ofNullable(null); safely creates an empty Optional.
Execution of orElse, orElseGet, and orElseThrow
* orElse("Baguette")
* Since optionalName is empty, "Baguette" is returned.
* bread = "Baguette"
* Output:"bread:Baguette"
* orElseGet(() -> "Frog legs")
* Since optionalName is empty, "Frog legs" is returned from the lambda expression.
* dish = "Frog legs"
* Output:", dish:Frog legs"
* orElseThrow(() -> new Exception())
* Since optionalName is empty, an exception is thrown.
* The catch block catches this exception and prints ", no cheese.".
Thus, the final output is:
makefile
bread:Baguette, dish:Frog legs, no cheese.
References:
* Java SE 21 & JDK 21 - Optional
* Java SE 21 - Functional Interfaces

NEW QUESTION # 33
Given:
java
Map<String, Integer> map = Map.of("b", 1, "a", 3, "c", 2);
TreeMap<String, Integer> treeMap = new TreeMap<>(map);
System.out.println(treeMap);
What is the output of the given code fragment?

A. Compilation fails
B. {c=1, b=2, a=3}
C. {b=1, c=2, a=3}
D. {a=3, b=1, c=2}
E. {c=2, a=3, b=1}
F. {b=1, a=3, c=2}
G. {a=1, b=2, c=3}

Answer: D

Explanation:
In this code, a Map named map is created using Map.of with the following key-value pairs:
* "b": 1
* "a": 3
* "c": 2
The Map.of method returns an immutable map containing these mappings.
Next, a TreeMap named treeMap is instantiated by passing the map to its constructor:
java
TreeMap<String, Integer> treeMap = new TreeMap<>(map);
The TreeMap constructor with a Map parameter creates a new tree map containing the same mappings as the given map, ordered according to the natural ordering of its keys. In Java, the natural ordering for String keys is lexicographical order.
Therefore, the TreeMap will store the entries in the following order:
* "a": 3
* "b": 1
* "c": 2
When System.out.println(treeMap); is executed, it outputs the TreeMap in its natural order, resulting in:
r
{a=3, b=1, c=2}
Thus, the correct answer is option F: {a=3, b=1, c=2}.

NEW QUESTION # 34
......

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