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Oracle Java SE 21 Developer Professional Sample Questions (Q68-Q73):

NEW QUESTION # 68
Given:
java
List<String> frenchAuthors = new ArrayList<>();
frenchAuthors.add("Victor Hugo");
frenchAuthors.add("Gustave Flaubert");
Which compiles?

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

Answer: A,B,D

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 # 69
Given:
java
DoubleStream doubleStream = DoubleStream.of(3.3, 4, 5.25, 6.66);
Predicate<Double> doublePredicate = d -> d < 5;
System.out.println(doubleStream.anyMatch(doublePredicate));
What is printed?

A. Compilation fails
B. An exception is thrown at runtime
C. true
D. false
E. 3.3

Answer: A

Explanation:
In this code, there is a type mismatch between the DoubleStream and the Predicate<Double>.
* DoubleStream: A sequence of primitive double values.
* Predicate<Double>: A functional interface that operates on objects of type Double (the wrapper class), not on primitive double values.
The DoubleStream class provides a method anyMatch(DoublePredicate predicate), where DoublePredicate is a functional interface that operates on primitive double values. However, in the code, a Predicate<Double> is used instead of a DoublePredicate. This mismatch leads to a compilation error because anyMatch cannot accept a Predicate<Double> when working with a DoubleStream.
To correct this, the predicate should be defined as a DoublePredicate to match the primitive double type:
java
DoubleStream doubleStream = DoubleStream.of(3.3, 4, 5.25, 6.66);
DoublePredicate doublePredicate = d -> d < 5;
System.out.println(doubleStream.anyMatch(doublePredicate));
With this correction, the code will compile and print true because there are elements in the stream (e.g., 3.3 and 4.0) that are less than 5.

NEW QUESTION # 70
Which of the following statements is correct about a final class?

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

Answer: C

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 # 71
Given:
java
var sList = new CopyOnWriteArrayList<Customer>();
Which of the following statements is correct?

A. The CopyOnWriteArrayList class does not allow null elements.
B. The CopyOnWriteArrayList class is not thread-safe and does not prevent interference amongconcurrent threads.
C. The CopyOnWriteArrayList class is a thread-safe variant of ArrayList where all mutative operations are implemented by making a fresh copy of the underlying array.
D. Element-changing operations on iterators of CopyOnWriteArrayList, such as remove, set, and add, are supported and do not throw UnsupportedOperationException.
E. The CopyOnWriteArrayList class's iterator reflects all additions, removals, or changes to the list since the iterator was created.

Answer: C

Explanation:
The CopyOnWriteArrayList is a thread-safe variant of ArrayList in which all mutative operations (such as add, set, and remove) are implemented by creating a fresh copy of the underlying array. This design allows for safe iteration over the list without requiring external synchronization, as iterators operate over a snapshot of the array at the time the iterator was created. Consequently, modifications made to the list after the creation of an iterator are not reflected in that iterator.
docs.oracle.com
Evaluation of Options:
* Option A:Correct. This statement accurately describes the behavior of CopyOnWriteArrayList.
* Option B:Incorrect. CopyOnWriteArrayList is thread-safe and is designed to prevent interference among concurrent threads.
* Option C:Incorrect. Iterators of CopyOnWriteArrayList do not reflect additions, removals, or changes made to the list after the iterator was created; they operate on a snapshot of the list's state at the time of their creation.
* Option D:Incorrect. CopyOnWriteArrayList allows null elements.
* Option E:Incorrect. Element-changing operations on iterators, such as remove, set, and add, are not supported in CopyOnWriteArrayList and will throw UnsupportedOperationException.

NEW QUESTION # 72
Which of the following suggestions compile?(Choose two.)

A. java
sealed class Figure permits Rectangle {}
final class Rectangle extends Figure {
float length, width;
}
B. java
sealed class Figure permits Rectangle {}
public class Rectangle extends Figure {
float length, width;
}
C. java
public sealed class Figure
permits Circle, Rectangle {}
final class Circle extends Figure {
float radius;
}
non-sealed class Rectangle extends Figure {
float length, width;
}
D. java
public sealed class Figure
permits Circle, Rectangle {}
final sealed class Circle extends Figure {
float radius;
}
non-sealed class Rectangle extends Figure {
float length, width;
}

Answer: A,C

Explanation:
Option A (sealed class Figure permits Rectangle {} and final class Rectangle extends Figure {}) - Valid
* Why it compiles?
* Figure issealed, meaning itmust explicitly declareits subclasses.
* Rectangle ispermittedto extend Figure and isdeclared final, meaning itcannot be extended further.
* This followsvalid sealed class rules.
Option B (sealed class Figure permits Rectangle {} and public class Rectangle extends Figure {}) -# Invalid
* Why it fails?
* Rectangle extends Figure, but it doesnot specify if it is sealed, final, or non-sealed.
* Fix:The correct declaration must be one of the following:
java
final class Rectangle extends Figure {} // OR
sealed class Rectangle permits OtherClass {} // OR
non-sealed class Rectangle extends Figure {}
Option C (final sealed class Circle extends Figure {}) -#Invalid
* Why it fails?
* A class cannot be both final and sealedat the same time.
* sealed meansit must have permitted subclasses, but final meansit cannot be extended.
* Fix:Change final sealed to just final:
java
final class Circle extends Figure {}
Option D (public sealed class Figure permits Circle, Rectangle {} with final class Circle and non-sealed class Rectangle) - Valid
* Why it compiles?
* Figure issealed, meaning it mustdeclare its permitted subclasses(Circle and Rectangle).
* Circle is declaredfinal, so itcannot have subclasses.
* Rectangle is declarednon-sealed, meaningit can be subclassedfreely.
* This correctly followsJava's sealed class rules.
Thus, the correct answers are:A, D
References:
* Java SE 21 - Sealed Classes
* Java SE 21 - Class Modifiers

NEW QUESTION # 73
......

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Oracle Java SE 21 Developer Professional Sample Questions (Q68-Q73):

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