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Oracle Java SE 21 Developer Professional Sample Questions (Q79-Q84):

NEW QUESTION # 79
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, ArrayList<String>> authorsMap1 = new HashMap<>();
java
authorsMap1.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,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 # 80
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. long
B. nothing
C. integer
D. Compilation fails.
E. It throws an exception at runtime.
F. string

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 # 81
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 a string with value: 42
B. Compilation fails.
C. It throws an exception at runtime.
D. null
E. It's an integer with value: 42
F. It's a double with value: 42

Answer: B

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 # 82
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. Nothing
D. An exception is thrown at runtime.
E. nginx
Hall of Mirrors has 17 mirrors.
The gardens cover 800 hectares.

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 # 83
Given:
java
public class ExceptionPropagation {
public static void main(String[] args) {
try {
thrower();
System.out.print("Dom Perignon, ");
} catch (Exception e) {
System.out.print("Chablis, ");
} finally {
System.out.print("Saint-Emilion");
}
}
static int thrower() {
try {
int i = 0;
return i / i;
} catch (NumberFormatException e) {
System.out.print("Rose");
return -1;
} finally {
System.out.print("Beaujolais Nouveau, ");
}
}
}
What is printed?

A. Saint-Emilion
B. Rose
C. Beaujolais Nouveau, Chablis, Dom Perignon, Saint-Emilion
D. Beaujolais Nouveau, Chablis, Saint-Emilion

Answer: D

Explanation:
* Analyzing the thrower() Method Execution
java
int i = 0;
return i / i;
* i / i evaluates to 0 / 0, whichthrows ArithmeticException (/ by zero).
* Since catch (NumberFormatException e) doesnot matchArithmeticException, it is skipped.
* The finally block always executes, printing:
nginx
Beaujolais Nouveau,
* The exceptionpropagates backto main().
* Handling the Exception in main()
java
try {
thrower();
System.out.print("Dom Perignon, ");
} catch (Exception e) {
System.out.print("Chablis, ");
} finally {
System.out.print("Saint-Emilion");
}
* Since thrower() throws ArithmeticException, it is caught by catch (Exception e).
* "Chablis, "is printed.
* Thefinally block always executes, printing "Saint-Emilion".
* Final Output
nginx
Beaujolais Nouveau, Chablis, Saint-Emilion
Thus, the correct answer is:Beaujolais Nouveau, Chablis, Saint-Emilion
References:
* Java SE 21 - Exception Handling
* Java SE 21 - finally Block Execution

NEW QUESTION # 84
......

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