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

NEW QUESTION # 77
Given:
java
var counter = 0;
do {
System.out.print(counter + " ");
} while (++counter < 3);
What is printed?

A. 1 2 3
B. 0 1 2 3
C. 0 1 2
D. 1 2 3 4
E. Compilation fails.
F. An exception is thrown.

Answer: C

Explanation:
* Understanding do-while Execution
* A do-while loopexecutes at least oncebefore checking the condition.
* ++counter < 3 increments counterbeforeevaluating the condition.
* Step-by-Step Execution
* Iteration 1:counter = 0, print "0", then ++counter becomes 1, condition 1 < 3 istrue.
* Iteration 2:counter = 1, print "1", then ++counter becomes 2, condition 2 < 3 istrue.
* Iteration 3:counter = 2, print "2", then ++counter becomes 3, condition 3 < 3 isfalse, so loop exits.
* Final Output
0 1 2
Thus, the correct answer is:0 1 2
References:
* Java SE 21 - Control Flow Statements
* Java SE 21 - do-while Loop

NEW QUESTION # 78
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. true
B. false
C. An exception is thrown at runtime
D. Compilation fails
E. 3.3

Answer: D

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 # 79
Which of the followingisn'ta correct way to write a string to a file?

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

Answer: A

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 # 80
Given:
java
interface A {
default void ma() {
}
}
interface B extends A {
static void mb() {
}
}
interface C extends B {
void ma();
void mc();
}
interface D extends C {
void md();
}
interface E extends D {
default void ma() {
}
default void mb() {
}
default void mc() {
}
}
Which interface can be the target of a lambda expression?

A. E
B. B
C. C
D. None of the above
E. D
F. A

Answer: D

Explanation:
In Java, a lambda expression can be used where a target type is a functional interface. A functional interface is an interface that contains exactly one abstract method. This concept is also known as a Single Abstract Method (SAM) type.
Analyzing each interface:
* Interface A: Contains a single default method ma(). Since default methods are not abstract, A has no abstract methods.
* Interface B: Extends A and adds a static method mb(). Static methods are also not abstract, so B has no abstract methods.
* Interface C: Extends B and declares two abstract methods: ma() (which overrides the default method from A) and mc(). Therefore, C has two abstract methods.
* Interface D: Extends C and adds another abstract method md(). Thus, D has three abstract methods.
* Interface E: Extends D and provides default implementations for ma(), mb(), and mc(). However, it does not provide an implementation for md(), leaving it as the only abstract method in E.
For an interface to be a functional interface, it must have exactly one abstract method. In this case, E has one abstract method (md()), so it qualifies as a functional interface. However, the question asks which interface can be the target of a lambda expression. Since E is a functional interface, it can be the target of a lambda expression.
Therefore, the correct answer is D (E).

NEW QUESTION # 81
Given:
java
ExecutorService service = Executors.newFixedThreadPool(2);
Runnable task = () -> System.out.println("Task is complete");
service.submit(task);
service.shutdown();
service.submit(task);
What happens when executing the given code fragment?

A. It prints "Task is complete" twice, then exits normally.
B. It prints "Task is complete" once, then exits normally.
C. It prints "Task is complete" once and throws an exception.
D. It prints "Task is complete" twice and throws an exception.
E. It exits normally without printing anything to the console.

Answer: C

Explanation:
In this code, an ExecutorService is created with a fixed thread pool of size 2 using Executors.
newFixedThreadPool(2). A Runnable task is defined to print "Task is complete" to the console.
The sequence of operations is as follows:
* service.submit(task);
This submits the task to the executor service for execution. Since the thread pool has a size of 2 and no other tasks are running, this task will be executed promptly, printing "Task is complete" to the console.
* service.shutdown();
This initiates an orderly shutdown of the executor service. In this state, the service stops accepting new tasks

NEW QUESTION # 82
......

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