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

NEW QUESTION # 77
Given:
java
var deque = new ArrayDeque<>();
deque.add(1);
deque.add(2);
deque.add(3);
deque.add(4);
deque.add(5);
System.out.print(deque.peek() + " ");
System.out.print(deque.poll() + " ");
System.out.print(deque.pop() + " ");
System.out.print(deque.element() + " ");
What is printed?

A. 1 5 5 1
B. 5 5 2 3
C. 1 1 2 2
D. 1 1 1 1
E. 1 1 2 3

Answer: E

Explanation:
* Understanding ArrayDeque Behavior
* ArrayDeque<E>is a double-ended queue (deque), working as aFIFO (queue) and LIFO (stack).
* Thedefault behaviorisqueue-like (FIFO)unless explicitly used as a stack.
* Step-by-Step Execution
java
var deque = new ArrayDeque<>();
deque.add(1);
deque.add(2);
deque.add(3);
deque.add(4);
deque.add(5);
* Deque after additions# [1, 2, 3, 4, 5]
* Operations Breakdown
* deque.peek()# Returns thehead(first element)without removal.
makefile
Output: 1
* deque.poll()# Removes and returns thehead.
go
Output: 1, Deque after poll # `[2, 3, 4, 5]`
* deque.pop()#Same as removeFirst(); removes and returns thehead.
perl
Output: 2, Deque after pop # `[3, 4, 5]`
* deque.element()# Returns thehead(same as peek(), but throws an exception if empty).
makefile
Output: 3
* Final Output
1 1 2 3
Thus, the correct answer is:1 1 2 3
References:
* Java SE 21 - ArrayDeque
* Java SE 21 - Queue Operations

NEW QUESTION # 78
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. Compilation fails.
D. 1
E. 2

Answer: C

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 # 79
Given:
java
CopyOnWriteArrayList<String> list = new CopyOnWriteArrayList<>();
list.add("A");
list.add("B");
list.add("C");
// Writing in one thread
new Thread(() -> {
list.add("D");
System.out.println("Element added: D");
}).start();
// Reading in another thread
new Thread(() -> {
for (String element : list) {
System.out.println("Read element: " + element);
}
}).start();
What is printed?

A. It prints all elements, but changes made during iteration may not be visible.
B. It throws an exception.
C. Compilation fails.
D. It prints all elements, including changes made during iteration.

Answer: A

Explanation:
* Understanding CopyOnWriteArrayList
* CopyOnWriteArrayList is a thread-safe variant of ArrayList whereall mutative operations (add, set, remove, etc.) create a new copy of the underlying array.
* This meansiterations will not reflect modifications made after the iterator was created.
* Instead of modifying the existing array, a new copy is created for modifications, ensuring that readers always see a consistent snapshot.
* Thread Execution Behavior
* Thread 1 (Writer Thread)adds "D" to the list.
* Thread 2 (Reader Thread)iterates over the list.
* The reader thread gets a snapshot of the listbefore"D" is added.
* The output may look like:
mathematica
Read element: A
Read element: B
Read element: C
Element added: D
* "D" may not appear in the output of the reader threadbecause the iteration occurs on a snapshot before the modification.
* Why doesn't it print all elements including changes?
* Since CopyOnWriteArrayList doesnot allow changes to be visible during iteration, the reader threadwill not see "D"if it started iterating before "D" was added.
Thus, the correct answer is:"It prints all elements, but changes made during iteration may not be visible." References:
* Java SE 21 - CopyOnWriteArrayList

NEW QUESTION # 80
Which of the following suggestions compile?(Choose two.)

A. 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;
}
B. java
sealed class Figure permits Rectangle {}
final class Rectangle extends Figure {
float length, width;
}
C. java
sealed class Figure permits Rectangle {}
public 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 # 81
Given:
java
System.out.print(Boolean.logicalAnd(1 == 1, 2 < 1));
System.out.print(Boolean.logicalOr(1 == 1, 2 < 1));
System.out.print(Boolean.logicalXor(1 == 1, 2 < 1));
What is printed?

A. truefalsetrue
B. truetruetrue
C. falsetruetrue
D. truetruefalse
E. Compilation fails

Answer: A

Explanation:
In this code, three static methods from the Boolean class are used: logicalAnd, logicalOr, and logicalXor.
Each method takes two boolean arguments and returns a boolean result based on the respective logical operation.
Evaluation of Each Statement:
* Boolean.logicalAnd(1 == 1, 2 < 1)
* Operands:
* 1 == 1 evaluates to true.
* 2 < 1 evaluates to false.
* Operation:
* Boolean.logicalAnd(true, false) performs a logical AND operation.
* The result is false because both operands must be true for the AND operation to return true.
* Output:
* System.out.print(false); prints false.
* Boolean.logicalOr(1 == 1, 2 < 1)
* Operands:
* 1 == 1 evaluates to true.
* 2 < 1 evaluates to false.
* Operation:
* Boolean.logicalOr(true, false) performs a logical OR operation.
* The result is true because at least one operand is true.
* Output:
* System.out.print(true); prints true.
* Boolean.logicalXor(1 == 1, 2 < 1)
* Operands:
* 1 == 1 evaluates to true.
* 2 < 1 evaluates to false.
* Operation:
* Boolean.logicalXor(true, false) performs a logical XOR (exclusive OR) operation.
* The result is true because exactly one operand is true.
* Output:
* System.out.print(true); prints true.
Combined Output:
Combining the outputs from each statement, the final printed result is:
nginx
falsetruetrue

NEW QUESTION # 82
......

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