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Oracle Java SE 21 Developer Professional Sample Questions (Q21-Q26):
NEW QUESTION # 21
Which of the following doesnotexist?
- A. BooleanSupplier
- B. LongSupplier
- C. BiSupplier<T, U, R>
- D. Supplier<T>
- E. They all exist.
- F. DoubleSupplier
Answer: C
Explanation:
1. Understanding Supplier Functional Interfaces
* The Supplier<T> interface is part of java.util.function and provides valueswithout taking any arguments.
* Java also provides primitive specializations of Supplier<T>:
* BooleanSupplier# Returns a boolean. Exists
* DoubleSupplier# Returns a double. Exists
* LongSupplier# Returns a long. Exists
* Supplier<T># Returns a generic T. Exists
2. What about BiSupplier<T, U, R>?
* There is no BiSupplier<T, U, R> in Java.
* In Java, suppliers donot take arguments, so abi-supplierdoes not exist.
* If you need a function thattakes two arguments and returns a value, use BiFunction<T, U, R>.
Thus, the correct answer is:BiSupplier<T, U, R> does not exist.
References:
* Java SE 21 - Supplier<T>
* Java SE 21 - Functional Interfaces
NEW QUESTION # 22
Which of the following methods of java.util.function.Predicate aredefault methods?
- A. not(Predicate<? super T> target)
- B. or(Predicate<? super T> other)
- C. isEqual(Object targetRef)
- D. and(Predicate<? super T> other)
- E. negate()
- F. test(T t)
Answer: B,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 # 23
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. An exception is thrown.
- B. 1 2 2
- C. 2 1 1
- D. 2 2 2
- E. 1 1 1
- F. 2 2 1
- G. 1 1 2
- H. 1 2 1
- I. 2 1 2
Answer: B
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 # 24
Which of the following suggestions compile?(Choose two.)
- A. java
sealed class Figure permits Rectangle {}
public class Rectangle extends Figure {
float length, width;
}
- B. java
sealed class Figure permits Rectangle {}
final class Rectangle extends Figure {
float length, width;
}
- C. 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;
}
- D. java
public sealed class Figure
permits Circle, Rectangle {}
final class Circle extends Figure {
float radius;
}
non-sealed class Rectangle extends Figure {
float length, width;
}
Answer: B,D
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 # 25
Given:
java
interface Calculable {
long calculate(int i);
}
public class Test {
public static void main(String[] args) {
Calculable c1 = i -> i + 1; // Line 1
Calculable c2 = i -> Long.valueOf(i); // Line 2
Calculable c3 = i -> { throw new ArithmeticException(); }; // Line 3
}
}
Which lines fail to compile?
- A. Line 2 only
- B. Line 3 only
- C. Line 1 only
- D. Line 2 and line 3
- E. Line 1 and line 2
- F. The program successfully compiles
- G. Line 1 and line 3
Answer: F
Explanation:
In this code, the Calculable interface defines a single abstract method calculate that takes an int parameter and returns a long. The main method contains three lambda expressions assigned to variables c1, c2, and c3 of type Calculable.
* Line 1:Calculable c1 = i -> i + 1;
This lambda expression takes an integer i and returns the result of i + 1. Since the expression i + 1 results in an int, and Java allows implicit widening conversion from int to long, this line compiles successfully.
* Line 2:Calculable c2 = i -> Long.valueOf(i);
Here, the lambda expression takes an integer i and returns the result of Long.valueOf(i). The Long.valueOf (int i) method returns a Long object. However, Java allows unboxing of the Long object to a long primitive type when necessary. Therefore, this line compiles successfully.
* Line 3:Calculable c3 = i -> { throw new ArithmeticException(); };
This lambda expression takes an integer i and throws an ArithmeticException. Since the method calculate has a return type of long, and throwing an exception is a valid way to exit the method without returning a value, this line compiles successfully.
Since all three lines adhere to the method signature defined in the Calculable interface and there are no type mismatches or syntax errors, the program compiles successfully.
NEW QUESTION # 26
......
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