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Oracle Java SE 21 Developer Professional Sample Questions (Q66-Q71):
NEW QUESTION # 66
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" once, then exits normally.
- B. It prints "Task is complete" once and throws an exception.
- C. It exits normally without printing anything to the console.
- D. It prints "Task is complete" twice, then exits normally.
- E. It prints "Task is complete" twice and throws an exception.
Answer: B
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 # 67
Given:
java
var counter = 0;
do {
System.out.print(counter + " ");
} while (++counter < 3);
What is printed?
- A. 0 1 2 3
- B. An exception is thrown.
- C. 0 1 2
- D. 1 2 3
- E. 1 2 3 4
- F. Compilation fails.
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 # 68
Given:
java
package com.vv;
import java.time.LocalDate;
public class FetchService {
public static void main(String[] args) throws Exception {
FetchService service = new FetchService();
String ack = service.fetch();
LocalDate date = service.fetch();
System.out.println(ack + " the " + date.toString());
}
public String fetch() {
return "ok";
}
public LocalDate fetch() {
return LocalDate.now();
}
}
What will be the output?
- A. An exception is thrown
- B. Compilation fails
- C. ok the 2024-07-10T07:17:45.523939600
- D. ok the 2024-07-10
Answer: B
Explanation:
In Java, method overloading allows multiple methods with the same name to exist in a class, provided they have different parameter lists (i.e., different number or types of parameters). However, having two methods with the exact same parameter list and only differing in return type is not permitted.
In the provided code, the FetchService class contains two fetch methods:
* public String fetch()
* public LocalDate fetch()
Both methods have identical parameter lists (none) but differ in their return types (String and LocalDate, respectively). This leads to a compilation error because the Java compiler cannot distinguish between the two methods based solely on return type.
The Java Language Specification (JLS) states:
"It is a compile-time error to declare two methods with override-equivalent signatures in a class." In this context, "override-equivalent" means that the methods have the same name and parameter types, regardless of their return types.
Therefore, the code will fail to compile due to the duplicate method signatures, and the correct answer is B:
Compilation fails.
NEW QUESTION # 69
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. Compilation fails
- B. 3.3
- C. An exception is thrown at runtime
- D. true
- E. false
Answer: A
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 # 70
Consider the following methods to load an implementation of MyService using ServiceLoader. Which of the methods are correct? (Choose all that apply)
- A. MyService service = ServiceLoader.services(MyService.class).getFirstInstance();
- B. MyService service = ServiceLoader.load(MyService.class).iterator().next();
- C. MyService service = ServiceLoader.load(MyService.class).findFirst().get();
- D. MyService service = ServiceLoader.getService(MyService.class);
Answer: B,C
Explanation:
The ServiceLoader class in Java is used to load service providers implementing a given service interface. The following methods are evaluated for their correctness in loading an implementation of MyService:
* A. MyService service = ServiceLoader.load(MyService.class).iterator().next(); This method uses the ServiceLoader.load(MyService.class) to create a ServiceLoader instance for MyService.
Calling iterator().next() retrieves the next available service provider. If no providers are available, a NoSuchElementException will be thrown. This approach is correct but requires handling the potential exception if no providers are found.
* B. MyService service = ServiceLoader.load(MyService.class).findFirst().get(); This method utilizes the findFirst() method introduced in Java 9, which returns an Optional describing the first available service provider. Calling get() on the Optional retrieves the service provider if present; otherwise, a NoSuchElementException is thrown. This approach is correct and provides a more concise way to obtain the first service provider.
* C. MyService service = ServiceLoader.getService(MyService.class);
The ServiceLoader class does not have a method named getService. Therefore, this method is incorrect and will result in a compilation error.
* D. MyService service = ServiceLoader.services(MyService.class).getFirstInstance(); The ServiceLoader class does not have a method named services or getFirstInstance. Therefore, this method is incorrect and will result in a compilation error.
In summary, options A and B are correct methods to load an implementation of MyService using ServiceLoader.
NEW QUESTION # 71
......
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