INTRODUCING DUMPSMATERIALS: YOUR PATH TO 1Z1-830 SUCCESS

Introducing DumpsMaterials: Your Path to 1z1-830 Success

Introducing DumpsMaterials: Your Path to 1z1-830 Success

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Oracle Java SE 21 Developer Professional Sample Questions (Q14-Q19):

NEW QUESTION # 14
Which of the following methods of java.util.function.Predicate aredefault methods?

  • A. negate()
  • B. test(T t)
  • C. and(Predicate<? super T> other)
  • D. isEqual(Object targetRef)
  • E. or(Predicate<? super T> other)
  • F. not(Predicate<? super T> target)

Answer: A,C,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 # 15
Given:
java
public class Test {
static int count;
synchronized Test() {
count++;
}
public static void main(String[] args) throws InterruptedException {
Runnable task = Test::new;
Thread t1 = new Thread(task);
Thread t2 = new Thread(task);
t1.start();
t2.start();
t1.join();
t2.join();
System.out.println(count);
}
}
What is the given program's output?

  • A. It's always 1
  • B. Compilation fails
  • C. It's always 2
  • D. It's either 0 or 1
  • E. It's either 1 or 2

Answer: B

Explanation:
In this code, the Test class has a static integer field count and a constructor that is declared with the synchronized modifier. In Java, the synchronized modifier can be applied to methods to control access to critical sections, but it cannot be applied directly to constructors. Attempting to declare a constructor as synchronized will result in a compilation error.
Compilation Error Details:
The Java Language Specification does not permit the use of the synchronized modifier on constructors.
Therefore, the compiler will produce an error indicating that the synchronized modifier is not allowed in this context.
Correct Usage:
If you need to synchronize the initialization of instances, you can use a synchronized block within the constructor:
java
public class Test {
static int count;
Test() {
synchronized (Test.class) {
count++;
}
}
public static void main(String[] args) throws InterruptedException {
Runnable task = Test::new;
Thread t1 = new Thread(task);
Thread t2 = new Thread(task);
t1.start();
t2.start();
t1.join();
t2.join();
System.out.println(count);
}
}
In this corrected version, the synchronized block within the constructor ensures that the increment operation on count is thread-safe.
Conclusion:
The original program will fail to compile due to the illegal use of the synchronized modifier on the constructor. Therefore, the correct answer is E: Compilation fails.


NEW QUESTION # 16
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.submit(task1);
  • B. execService.execute(task1);
  • C. execService.call(task1);
  • D. execService.execute(task2);
  • E. execService.run(task1);
  • F. execService.call(task2);
  • G. execService.run(task2);
  • H. execService.submit(task2);

Answer: A,H

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 # 17
Given:
java
double amount = 42_000.00;
NumberFormat format = NumberFormat.getCompactNumberInstance(Locale.FRANCE, NumberFormat.Style.
SHORT);
System.out.println(format.format(amount));
What is the output?

  • A. 42000E
  • B. 42 k
  • C. 42 000,00 €
  • D. 0

Answer: B

Explanation:
In this code, a double variable amount is initialized to 42,000.00. The NumberFormat.
getCompactNumberInstance(Locale.FRANCE, NumberFormat.Style.SHORT) method is used to obtain a compact number formatter for the French locale with the short style. The format method is then called to format the amount.
The compact number formatting is designed to represent numbers in a shorter form, based on the patterns provided for a given locale. In the French locale, the short style represents thousands with a lowercase 'k'.
Therefore, 42,000 is formatted as 42 k.
* Option Evaluations:
* A. 42000E: This format is not standard in the French locale for compact number formatting.
* B. 42 000,00 €: This represents the number as a currency with two decimal places, which is not the compact form.
* C. 42000: This is the plain number without any formatting, which does not match the compact number format.
* D. 42 k: This is the correct compact representation of 42,000 in the French locale with the short style.
Thus, option D (42 k) is the correct output.


NEW QUESTION # 18
Given:
java
var array1 = new String[]{ "foo", "bar", "buz" };
var array2[] = { "foo", "bar", "buz" };
var array3 = new String[3] { "foo", "bar", "buz" };
var array4 = { "foo", "bar", "buz" };
String array5[] = new String[]{ "foo", "bar", "buz" };
Which arrays compile? (Select 2)

  • A. array2
  • B. array3
  • C. array1
  • D. array4
  • E. array5

Answer: C,E

Explanation:
In Java, array initialization can be performed in several ways, but certain syntaxes are invalid and will cause compilation errors. Let's analyze each declaration:
* var array1 = new String[]{ "foo", "bar", "buz" };
This is a valid declaration. The var keyword allows the compiler to infer the type from the initializer. Here, new String[]{ "foo", "bar", "buz" } creates an anonymous array of String with three elements. The compiler infers array1 as String[]. This syntax is correct and compiles successfully.
* var array2[] = { "foo", "bar", "buz" };
This declaration is invalid. While var can be used for type inference, appending [] after var is not allowed.
The correct syntax would be either String[] array2 = { "foo", "bar", "buz" }; or var array2 = new String[]{
"foo", "bar", "buz" };. Therefore, this line will cause a compilation error.
* var array3 = new String[3] { "foo", "bar", "buz" };
This declaration is invalid. In Java, when specifying the size of the array (new String[3]), you cannot simultaneously provide an initializer. The correct approach is either to provide the size without an initializer (new String[3]) or to provide the initializer without specifying the size (new String[]{ "foo", "bar", "buz" }).
Therefore, this line will cause a compilation error.
* var array4 = { "foo", "bar", "buz" };
This declaration is invalid. The array initializer { "foo", "bar", "buz" } can only be used in an array declaration when the type is explicitly provided. Since var relies on type inference and there's no explicit type provided here, this will cause a compilation error. The correct syntax would be String[] array4 = { "foo",
"bar", "buz" };.
* String array5[] = new String[]{ "foo", "bar", "buz" };
This is a valid declaration. Here, String array5[] declares array5 as an array of String. The initializer new String[]{ "foo", "bar", "buz" } creates an array with three elements. This syntax is correct and compiles successfully.
Therefore, the declarations that compile successfully are array1 and array5.
References:
* Java SE 21 & JDK 21 - Local Variable Type Inference
* Java SE 21 & JDK 21 - Arrays


NEW QUESTION # 19
......

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