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Oracle Java SE 21 Developer Professional Sample Questions (Q43-Q48):

NEW QUESTION # 43
How would you create a ConcurrentHashMap configured to allow a maximum of 10 concurrent writer threads and an initial capacity of 42?
Which of the following options meets this requirement?

A. None of the suggestions.
B. var concurrentHashMap = new ConcurrentHashMap(42, 10);
C. var concurrentHashMap = new ConcurrentHashMap(42, 0.88f, 10);
D. var concurrentHashMap = new ConcurrentHashMap(42);
E. var concurrentHashMap = new ConcurrentHashMap();

Answer: C

Explanation:
In Java, the ConcurrentHashMap class provides several constructors that allow for the customization of its initial capacity, load factor, and concurrency level. To configure a ConcurrentHashMap with an initial capacity of 42 and a concurrency level of 10, you can use the following constructor:
java
public ConcurrentHashMap(int initialCapacity, float loadFactor, int concurrencyLevel) Parameters:
* initialCapacity: The initial capacity of the hash table. This is the number of buckets that the hash table will have when it is created. In this case, it is set to 42.
* loadFactor: A measure of how full the hash table is allowed to get before it is resized. The default value is 0.75, but in this case, it is set to 0.88.
* concurrencyLevel: The estimated number of concurrently updating threads. This is used as a hint for internal sizing. In this case, it is set to 10.
Therefore, to create a ConcurrentHashMap with an initial capacity of 42, a load factor of 0.88, and a concurrency level of 10, you can use the following code:
java
var concurrentHashMap = new ConcurrentHashMap<>(42, 0.88f, 10);
Option Evaluations:
* A. var concurrentHashMap = new ConcurrentHashMap(42);: This constructor sets the initial capacity to 42 but uses the default load factor (0.75) and concurrency level (16). It does not meet the requirement of setting the concurrency level to 10.
* B. None of the suggestions.: This is incorrect because option E provides the correct configuration.
* C. var concurrentHashMap = new ConcurrentHashMap();: This uses the default constructor, which sets the initial capacity to 16, the load factor to 0.75, and the concurrency level to 16. It does not meet the specified requirements.
* D. var concurrentHashMap = new ConcurrentHashMap(42, 10);: This constructor sets the initial capacity to 42 and the load factor to 10, which is incorrect because the load factor should be a float value between 0 and 1.
* E. var concurrentHashMap = new ConcurrentHashMap(42, 0.88f, 10);: This correctly sets the initial capacity to 42, the load factor to 0.88, and the concurrency level to 10, meeting all the specified requirements.
Therefore, the correct answer is option E.

NEW QUESTION # 44
What do the following print?
java
public class DefaultAndStaticMethods {
public static void main(String[] args) {
WithStaticMethod.print();
}
}
interface WithDefaultMethod {
default void print() {
System.out.print("default");
}
}
interface WithStaticMethod extends WithDefaultMethod {
static void print() {
System.out.print("static");
}
}

A. nothing
B. Compilation fails
C. default
D. static

Answer: D

Explanation:
In this code, we have two interfaces and a class with a main method:
* WithDefaultMethod Interface:
* Declares a default method print() that outputs "default".
* WithStaticMethod Interface:
* Extends WithDefaultMethod.
* Declares a static method print() that outputs "static".
* DefaultAndStaticMethods Class:
* Contains the main method, which calls WithStaticMethod.print().
Key Points:
* Static Methods in Interfaces:
* Static methods in interfaces are not inherited by implementing or extending classes or interfaces.
They belong solely to the interface in which they are declared.
* Default Methods in Interfaces:
* Default methods can be inherited by implementing classes, but they cannot be overridden by static methods in subinterfaces.
Execution Flow:
* The main method calls WithStaticMethod.print().
* This invokes the static method print() defined in the WithStaticMethod interface, which outputs "static".
Therefore, the program compiles successfully and prints static.

NEW QUESTION # 45
Given:
java
List<Integer> integers = List.of(0, 1, 2);
integers.stream()
.peek(System.out::print)
.limit(2)
.forEach(i -> {});
What is the output of the given code fragment?

A. 01
B. An exception is thrown
C. Nothing
D. Compilation fails
E. 012

Answer: A

Explanation:
In this code, a list of integers integers is created containing the elements 0, 1, and 2. A stream is then created from this list, and the following operations are performed in sequence:
* peek(System.out::print):
* The peek method is an intermediate operation that allows performing an action on each element as it is encountered in the stream. In this case, System.out::print is used to print each element.
However, since peek is intermediate, the printing occurs only when a terminal operation is executed.
* limit(2):
* The limit method is another intermediate operation that truncates the stream to contain no more than the specified number of elements. Here, it limits the stream to the first 2 elements.
* forEach(i -> {}):
* The forEach method is a terminal operation that performs the given action on each element of the stream. In this case, the action is an empty lambda expression (i -> {}), which does nothing for each element.
The sequence of operations can be visualized as follows:
* Original Stream Elements: 0, 1, 2
* After peek(System.out::print): Elements are printed as they are encountered.
* After limit(2): Stream is truncated to 0, 1.
* After forEach(i -> {}): No additional action; serves to trigger the processing.
Therefore, the output of the code is 01, corresponding to the first two elements of the list being printed due to the peek operation.

NEW QUESTION # 46
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. 3.3
B. Compilation fails
C. false
D. true
E. An exception is thrown at runtime

Answer: B

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 # 47
Given:
java
Runnable task1 = () -> System.out.println("Executing Task-1");
Callable<String> task2 = () -> {
System.out.println("Executing Task-2");
return "Task-2 Finish.";
};
ExecutorService execService = Executors.newCachedThreadPool();
// INSERT CODE HERE
execService.awaitTermination(3, TimeUnit.SECONDS);
execService.shutdownNow();
Which of the following statements, inserted in the code above, printsboth:
"Executing Task-2" and "Executing Task-1"?

A. execService.run(task1);
B. execService.call(task2);
C. execService.execute(task1);
D. execService.run(task2);
E. execService.submit(task2);
F. execService.execute(task2);
G. execService.submit(task1);
H. execService.call(task1);

Answer: E,G

Explanation:
* Understanding ExecutorService Methods
* execute(Runnable command)
* Runs the task but only supports Runnable (not Callable).
* #execService.execute(task2); fails because task2 is Callable<String>.
* submit(Runnable task)
* Submits a Runnable task for execution.
* execService.submit(task1); executes "Executing Task-1".
* submit(Callable<T> task)
* Submits a Callable<T> task for execution.
* execService.submit(task2); executes "Executing Task-2".
* call() Does Not Exist in ExecutorService
* #execService.call(task1); and execService.call(task2); are invalid.
* run() Does Not Exist in ExecutorService
* #execService.run(task1); and execService.run(task2); are invalid.
* Correct Code to Print Both Messages:
java
execService.submit(task1);
execService.submit(task2);
Thus, the correct answer is:execService.submit(task1); execService.submit(task2); References:
* Java SE 21 - ExecutorService
* Java SE 21 - Callable and Runnable

NEW QUESTION # 48
......

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Oracle 1z0-830 Exam Dumps & 1z0-830 Free Exam

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Oracle Java SE 21 Developer Professional Sample Questions (Q43-Q48):

Free PDF 2025 1z0-830: Perfect Java SE 21 Developer Professional Exam Dumps

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