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Oracle Java SE 21 Developer Professional Sample Questions (Q38-Q43):

NEW QUESTION # 38
Given:
java
var _ = 3;
var $ = 7;
System.out.println(_ + $);
What is printed?

A. _$
B. It throws an exception.
C. Compilation fails.
D. 0

Answer: C

Explanation:
* The var keyword and identifier rules:
* The var keyword is used for local variable type inference introduced inJava 10.
* However,Java does not allow _ (underscore) as an identifiersinceJava 9.
* If we try to use _ as a variable name, the compiler will throw an error:
pgsql
error: as of release 9, '_' is a keyword, and may not be used as an identifier
* The $ symbol as an identifier:
* The $ characteris a valid identifierin Java.
* However, since _ is not allowed, the codefails to compile before even reaching $.
Thus,the correct answer is "Compilation fails."
References:
* Java SE 21 - var Local Variable Type Inference
* Java SE 9 - Restrictions on _ Identifier

NEW QUESTION # 39
Given:
java
var hauteCouture = new String[]{ "Chanel", "Dior", "Louis Vuitton" };
var i = 0;
do {
System.out.print(hauteCouture[i] + " ");
} while (i++ > 0);
What is printed?

A. Compilation fails.
B. Chanel Dior Louis Vuitton
C. Chanel
D. An ArrayIndexOutOfBoundsException is thrown at runtime.

Answer: C

Explanation:
* Understanding the do-while Loop
* The do-while loopexecutes at least oncebefore checking the condition.
* The condition i++ > 0 increments iafterchecking.
* Step-by-Step Execution
* Iteration 1:
* i = 0
* Prints: "Chanel"
* i++ updates i to 1
* Condition 1 > 0is true, so the loop exits.
* Why Doesn't the Loop Continue?
* Since i starts at 0, the conditioni++ > 0 is false after the first iteration.
* The loopexits immediately after printing "Chanel".
* Final Output
nginx
Chanel
Thus, the correct answer is:Chanel
References:
* Java SE 21 - do-while Loop
* Java SE 21 - Post-Increment Behavior

NEW QUESTION # 40
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 1 1 1
B. 1 1 2 2
C. 1 5 5 1
D. 5 5 2 3
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 # 41
Which of the followingisn'ta correct way to write a string to a file?

A. java
Path path = Paths.get("file.txt");
byte[] strBytes = "Hello".getBytes();
Files.write(path, strBytes);
B. java
try (FileOutputStream outputStream = new FileOutputStream("file.txt")) { byte[] strBytes = "Hello".getBytes(); outputStream.write(strBytes);
}
C. java
try (PrintWriter printWriter = new PrintWriter("file.txt")) {
printWriter.printf("Hello %s", "James");
}
D. java
try (BufferedWriter writer = new BufferedWriter("file.txt")) {
writer.write("Hello");
}
E. java
try (FileWriter writer = new FileWriter("file.txt")) {
writer.write("Hello");
}
F. None of the suggestions

Answer: D

Explanation:
(BufferedWriter writer = new BufferedWriter("file.txt") is incorrect.)
Theincorrect statementisoption Bbecause BufferedWriterdoes nothave a constructor that accepts a String (file name) directly. The correct way to use BufferedWriter is to wrap it around a FileWriter, like this:
java
try (BufferedWriter writer = new BufferedWriter(new FileWriter("file.txt"))) { writer.write("Hello");
}
Evaluation of Other Options:
Option A (Files.write)# Correct
* Uses Files.write() to write bytes to a file.
* Efficient and concise method for writing small text files.
Option C (FileOutputStream)# Correct
* Uses a FileOutputStream to write raw bytes to a file.
* Works for both text and binary data.
Option D (PrintWriter)# Correct
* Uses PrintWriter for formatted text output.
Option F (FileWriter)# Correct
* Uses FileWriter to write text data.
Option E (None of the suggestions)# Incorrect becauseoption Bis incorrect.

NEW QUESTION # 42
Given:
java
List<String> l1 = new ArrayList<>(List.of("a", "b"));
List<String> l2 = new ArrayList<>(Collections.singletonList("c"));
Collections.copy(l1, l2);
l2.set(0, "d");
System.out.println(l1);
What is the output of the given code fragment?

A. [c, b]
B. [d, b]
C. [a, b]
D. An UnsupportedOperationException is thrown
E. [d]
F. An IndexOutOfBoundsException is thrown

Answer: A

Explanation:
In this code, two lists l1 and l2 are created and initialized as follows:
* l1 Initialization:
* Created using List.of("a", "b"), which returns an immutable list containing the elements "a" and
"b".
* Wrapped with new ArrayList<>(...) to create a mutable ArrayList containing the same elements.
* l2 Initialization:
* Created using Collections.singletonList("c"), which returns an immutable list containing the single element "c".
* Wrapped with new ArrayList<>(...) to create a mutable ArrayList containing the same element.
State of Lists Before Collections.copy:
* l1: ["a", "b"]
* l2: ["c"]
Collections.copy(l1, l2):
The Collections.copy method copies elements from the source list (l2) into the destination list (l1). The destination list must have at least as many elements as the source list; otherwise, an IndexOutOfBoundsException is thrown.
In this case, l1 has two elements, and l2 has one element, so the copy operation is valid. After copying, the first element of l1 is replaced with the first element of l2:
* l1 after copy: ["c", "b"]
l2.set(0, "d"):
This line sets the first element of l2 to "d".
* l2 after set: ["d"]
Final State of Lists:
* l1: ["c", "b"]
* l2: ["d"]
The System.out.println(l1); statement outputs the current state of l1, which is ["c", "b"]. Therefore, the correct answer is C: [c, b].

NEW QUESTION # 43
......

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