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Oracle Java SE 21 Developer Professional Sample Questions (Q61-Q66):

NEW QUESTION # 61
Given:
java
List<String> frenchAuthors = new ArrayList<>();
frenchAuthors.add("Victor Hugo");
frenchAuthors.add("Gustave Flaubert");
Which compiles?

A. Map<String, ArrayList<String>> authorsMap1 = new HashMap<>();
java
authorsMap1.put("FR", frenchAuthors);
B. var authorsMap3 = new HashMap<>();
java
authorsMap3.put("FR", frenchAuthors);
C. Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>>(); java authorsMap4.put("FR", frenchAuthors);
D. Map<String, ? extends List<String>> authorsMap2 = new HashMap<String, ArrayList<String>> (); java authorsMap2.put("FR", frenchAuthors);
E. Map<String, List<String>> authorsMap5 = new HashMap<String, List<String>>(); java authorsMap5.put("FR", frenchAuthors);

Answer: B,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 # 62
Which of the following statements is correct about a final class?

A. It must contain at least a final method.
B. It cannot implement any interface.
C. The final keyword in its declaration must go right before the class keyword.
D. It cannot be extended by any other class.
E. It cannot extend another class.

Answer: D

Explanation:
In Java, the final keyword can be applied to classes, methods, and variables to impose certain restrictions.
Final Classes:
* Definition:A class declared with the final keyword is known as a final class.
* Purpose:Declaring a class as final prevents it from being subclassed. This is useful when you want to ensure that the class's implementation remains unchanged and cannot be extended or modified through inheritance.
Option Evaluations:
* A. The final keyword in its declaration must go right before the class keyword.
* This is correct. The syntax for declaring a final class is:
java
public final class ClassName {
// class body
}
* However, this statement is about syntax rather than the core characteristic of a final class.
* B. It must contain at least a final method.
* Incorrect. A final class can have zero or more methods, and none of them are required to be declared as final. The final keyword at the class level prevents inheritance, regardless of the methods' finality.
* C. It cannot be extended by any other class.
* Correct. The primary characteristic of a final class is that it cannot be subclassed. Attempting to do so will result in a compilation error.
* D. It cannot implement any interface.
* Incorrect. A final class can implement interfaces. Declaring a class as final restricts inheritance but does not prevent the class from implementing interfaces.
* E. It cannot extend another class.
* Incorrect. A final class can extend another class. The final keyword prevents the class from being subclassed but does not prevent it from being a subclass itself.
Therefore, the correct statement about a final class is option C: "It cannot be extended by any other class."

NEW QUESTION # 63
Given:
java
StringBuffer us = new StringBuffer("US");
StringBuffer uk = new StringBuffer("UK");
Stream<StringBuffer> stream = Stream.of(us, uk);
String output = stream.collect(Collectors.joining("-", "=", ""));
System.out.println(output);
What is the given code fragment's output?

A. -US=UK
B. An exception is thrown.
C. Compilation fails.
D. US-UK
E. US=UK
F. =US-UK

Answer: F

Explanation:
In this code, two StringBuffer objects, us and uk, are created with the values "US" and "UK", respectively. A stream is then created from these objects using Stream.of(us, uk).
The collect method is used with Collectors.joining("-", "=", ""). The joining collector concatenates the elements of the stream into a single String with the following parameters:
* Delimiter ("-"):Inserted between each element.
* Prefix ("="):Inserted at the beginning of the result.
* Suffix (""):Inserted at the end of the result.
Therefore, the elements "US" and "UK" are concatenated with "-" between them, resulting in "US-UK". The prefix "=" is added at the beginning, resulting in the final output =US-UK.

NEW QUESTION # 64
Given:
java
Deque<Integer> deque = new ArrayDeque<>();
deque.offer(1);
deque.offer(2);
var i1 = deque.peek();
var i2 = deque.poll();
var i3 = deque.peek();
System.out.println(i1 + " " + i2 + " " + i3);
What is the output of the given code fragment?

A. 2 2 2
B. 1 1 2
C. An exception is thrown.
D. 1 2 1
E. 1 1 1
F. 1 2 2
G. 2 2 1
H. 2 1 2
I. 2 1 1

Answer: F

Explanation:
In this code, an ArrayDeque named deque is created, and the integers 1 and 2 are added to it using the offer method. The offer method inserts the specified element at the end of the deque.
* State of deque after offers:[1, 2]
The peek method retrieves, but does not remove, the head of the deque, returning 1. Therefore, i1 is assigned the value 1.
* State of deque after peek:[1, 2]
* Value of i1:1
The poll method retrieves and removes the head of the deque, returning 1. Therefore, i2 is assigned the value
1.
* State of deque after poll:[2]
* Value of i2:1
Another peek operation retrieves the current head of the deque, which is now 2, without removing it.
Therefore, i3 is assigned the value 2.
* State of deque after second peek:[2]
* Value of i3:2
The System.out.println statement then outputs the values of i1, i2, and i3, resulting in 1 1 2.

NEW QUESTION # 65
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 throws an exception at runtime.
B. null
C. It's a string with value: 42
D. It's an integer with value: 42
E. It's a double with value: 42
F. Compilation fails.

Answer: F

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 # 66
......

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