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NEW QUESTION # 16
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?
Answer: B
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 # 17
Given:
java
List<Long> cannesFestivalfeatureFilms = LongStream.range(1, 1945)
.boxed()
.toList();
try (var executor = Executors.newVirtualThreadPerTaskExecutor()) {
cannesFestivalfeatureFilms.stream()
.limit(25)
.forEach(film -> executor.submit(() -> {
System.out.println(film);
}));
}
What is printed?
Answer: A
Explanation:
* Understanding LongStream.range(1, 1945).boxed().toList();
* LongStream.range(1, 1945) generates a stream of numbersfrom 1 to 1944.
* .boxed() converts the primitive long values to Long objects.
* .toList() (introduced in Java 16)creates an immutable list.
* Understanding Executors.newVirtualThreadPerTaskExecutor()
* Java 21 introducedvirtual threadsto improve concurrency.
* Executors.newVirtualThreadPerTaskExecutor()creates a new virtual thread per submitted task
, allowing highly concurrent execution.
* Execution Behavior
* cannesFestivalfeatureFilms.stream().limit(25) # Limits the stream to thefirst 25 numbers(1 to
25).
* .forEach(film -> executor.submit(() -> System.out.println(film)))
* Each film is printed inside a virtual thread.
* Virtual threads execute asynchronously, meaning numbers arenot guaranteed to print sequentially.
* Output will contain numbers from 1 to 25, but their order is random due to concurrent execution.
* Possible Output (Random Order)
python-repl
3
1
5
2
4
7
25
* The ordermay differ in each rundue to concurrent execution.
Thus, the correct answer is:"Numbers from 1 to 25 randomly."
References:
* Java SE 21 - Virtual Threads
* Java SE 21 - Executors.newVirtualThreadPerTaskExecutor()
NEW QUESTION # 18
Given:
java
Optional o1 = Optional.empty();
Optional o2 = Optional.of(1);
Optional o3 = Stream.of(o1, o2)
.filter(Optional::isPresent)
.findAny()
.flatMap(o -> o);
System.out.println(o3.orElse(2));
What is the given code fragment's output?
Answer: A
Explanation:
In this code, two Optional objects are created:
* o1 is an empty Optional.
* o2 is an Optional containing the integer 1.
A stream is created from o1 and o2. The filter method retains only the Optional instances that are present (i.e., non-empty). This results in a stream containing only o2.
The findAny method returns an Optional describing some element of the stream, or an empty Optional if the stream is empty. Since the stream contains o2, findAny returns Optional[Optional[1]].
The flatMap method is then used to flatten this nested Optional. It applies the provided mapping function (o -
> o) to the value, resulting in Optional[1].
Finally, o3.orElse(2) returns the value contained in o3 if it is present; otherwise, it returns 2. Since o3 contains
1, the output is 1.
NEW QUESTION # 19
Given:
java
Map<String, Integer> map = Map.of("b", 1, "a", 3, "c", 2);
TreeMap<String, Integer> treeMap = new TreeMap<>(map);
System.out.println(treeMap);
What is the output of the given code fragment?
Answer: A
Explanation:
In this code, a Map named map is created using Map.of with the following key-value pairs:
* "b": 1
* "a": 3
* "c": 2
The Map.of method returns an immutable map containing these mappings.
Next, a TreeMap named treeMap is instantiated by passing the map to its constructor:
java
TreeMap<String, Integer> treeMap = new TreeMap<>(map);
The TreeMap constructor with a Map parameter creates a new tree map containing the same mappings as the given map, ordered according to the natural ordering of its keys. In Java, the natural ordering for String keys is lexicographical order.
Therefore, the TreeMap will store the entries in the following order:
* "a": 3
* "b": 1
* "c": 2
When System.out.println(treeMap); is executed, it outputs the TreeMap in its natural order, resulting in:
r
{a=3, b=1, c=2}
Thus, the correct answer is option F: {a=3, b=1, c=2}.
NEW QUESTION # 20
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?
Answer: C
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 # 21
......
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