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Quiz 2025 1z0-830: Java SE 21 Developer Professional–Trustable Accurate Answers

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Oracle Java SE 21 Developer Professional Sample Questions (Q77-Q82):

NEW QUESTION # 77
Which of the following statements is correct about a final class?

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

Answer: A

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 # 78
Which of the following methods of java.util.function.Predicate aredefault methods?

A. test(T t)
B. isEqual(Object targetRef)
C. or(Predicate<? super T> other)
D. and(Predicate<? super T> other)
E. negate()
F. not(Predicate<? super T> target)

Answer: C,D,E

Explanation:
* Understanding java.util.function.Predicate<T>
* The Predicate<T> interface represents a function thattakes an input and returns a boolean(true or false).
* It is often used for filtering operations in functional programming and streams.
* Analyzing the Methods:
* and(Predicate<? super T> other)#Default method
* Combines two predicates usinglogical AND(&&).
java
Predicate<String> startsWithA = s -> s.startsWith("A");
Predicate<String> hasLength3 = s -> s.length() == 3;
Predicate<String> combined = startsWithA.and(hasLength3);
* #isEqual(Object targetRef)#Static method
* Not a default method, because it doesnot operate on an instance.
java
Predicate<String> isEqualToHello = Predicate.isEqual("Hello");
* negate()#Default method
* Negates a predicate (! operator).
java
Predicate<String> notEmpty = s -> !s.isEmpty();
Predicate<String> isEmpty = notEmpty.negate();
* #not(Predicate<? super T> target)#Static method (introduced in Java 11)
* Not a default method, since it is static.
* or(Predicate<? super T> other)#Default method
* Combines two predicates usinglogical OR(||).
* #test(T t)#Abstract method
* Not a default method, because every predicatemust implement this method.
Thus, the correct answers are:and(Predicate<? super T> other), negate(), or(Predicate<? super T> other) References:
* Java SE 21 - Predicate Interface
* Java SE 21 - Functional Interfaces

NEW QUESTION # 79
Given:
java
public class SpecialAddition extends Addition implements Special {
public static void main(String[] args) {
System.out.println(new SpecialAddition().add());
}
int add() {
return --foo + bar--;
}
}
class Addition {
int foo = 1;
}
interface Special {
int bar = 1;
}
What is printed?

A. 0
B. It throws an exception at runtime.
C. 1
D. Compilation fails.
E. 2

Answer: D

Explanation:
1. Why does the compilation fail?
* The interface Special contains bar as int bar = 1;.
* In Java, all interface fields are implicitly public, static, and final.
* This means that bar is a constant (final variable).
* The method add() contains bar--, which attempts to modify bar.
* Since bar is final, it cannot be modified, causing acompilation error.
2. Correcting the Code
To make the code compile, bar must not be final. One way to fix this:
java
class SpecialImpl implements Special {
int bar = 1;
}
Or modify the add() method:
java
int add() {
return --foo + bar; // No modification of bar
}
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - Interfaces
* Java SE 21 - Final Variables

NEW QUESTION # 80
Given:
java
final Stream<String> strings =
Files.readAllLines(Paths.get("orders.csv"));
strings.skip(1)
.limit(2)
.forEach(System.out::println);
And that the orders.csv file contains:
mathematica
OrderID,Customer,Product,Quantity,Price
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99
3,Sebastien Loeb,Monitor,1,199.99
4,Antoine Griezmann,Headset,3,45.00
What is printed?

A. arduino
2,Teddy Riner,Mouse,1,15.99
3,Sebastien Loeb,Monitor,1,199.99
B. Compilation fails.
C. An exception is thrown at runtime.
D. arduino
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99
3,Sebastien Loeb,Monitor,1,199.99
4,Antoine Griezmann,Headset,3,45.00
E. arduino
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99

Answer: B,C

Explanation:
1. Why Does Compilation Fail?
* The error is in this line:
java
final Stream<String> strings = Files.readAllLines(Paths.get("orders.csv"));
* Files.readAllLines(Paths.get("orders.csv")) returns a List<String>,not a Stream<String>.
* A List<String> cannot be assigned to a Stream<String>.
2. Correcting the Code
* The correct way to create a stream from the file:
java
Stream<String> strings = Files.lines(Paths.get("orders.csv"));
* This correctly creates a Stream<String> from the file.
3. Expected Output After Fixing
java
Files.lines(Paths.get("orders.csv"))
skip(1) // Skips the header row
limit(2) // Limits to first two data rows
forEach(System.out::println);
* Output:
arduino
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - Files.readAllLines
* Java SE 21 - Files.lines

NEW QUESTION # 81
Given:
java
var array1 = new String[]{ "foo", "bar", "buz" };
var array2[] = { "foo", "bar", "buz" };
var array3 = new String[3] { "foo", "bar", "buz" };
var array4 = { "foo", "bar", "buz" };
String array5[] = new String[]{ "foo", "bar", "buz" };
Which arrays compile? (Select 2)

A. array4
B. array1
C. array3
D. array2
E. array5

Answer: B,E

Explanation:
In Java, array initialization can be performed in several ways, but certain syntaxes are invalid and will cause compilation errors. Let's analyze each declaration:
* var array1 = new String[]{ "foo", "bar", "buz" };
This is a valid declaration. The var keyword allows the compiler to infer the type from the initializer. Here, new String[]{ "foo", "bar", "buz" } creates an anonymous array of String with three elements. The compiler infers array1 as String[]. This syntax is correct and compiles successfully.
* var array2[] = { "foo", "bar", "buz" };
This declaration is invalid. While var can be used for type inference, appending [] after var is not allowed.
The correct syntax would be either String[] array2 = { "foo", "bar", "buz" }; or var array2 = new String[]{
"foo", "bar", "buz" };. Therefore, this line will cause a compilation error.
* var array3 = new String[3] { "foo", "bar", "buz" };
This declaration is invalid. In Java, when specifying the size of the array (new String[3]), you cannot simultaneously provide an initializer. The correct approach is either to provide the size without an initializer (new String[3]) or to provide the initializer without specifying the size (new String[]{ "foo", "bar", "buz" }).
Therefore, this line will cause a compilation error.
* var array4 = { "foo", "bar", "buz" };
This declaration is invalid. The array initializer { "foo", "bar", "buz" } can only be used in an array declaration when the type is explicitly provided. Since var relies on type inference and there's no explicit type provided here, this will cause a compilation error. The correct syntax would be String[] array4 = { "foo",
"bar", "buz" };.
* String array5[] = new String[]{ "foo", "bar", "buz" };
This is a valid declaration. Here, String array5[] declares array5 as an array of String. The initializer new String[]{ "foo", "bar", "buz" } creates an array with three elements. This syntax is correct and compiles successfully.
Therefore, the declarations that compile successfully are array1 and array5.
References:
* Java SE 21 & JDK 21 - Local Variable Type Inference
* Java SE 21 & JDK 21 - Arrays

NEW QUESTION # 82
......

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