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Oracle Java SE 21 Developer Professional Sample Questions (Q28-Q33):

NEW QUESTION # 28
Which three of the following are correct about the Java module system?

A. If a package is defined in both a named module and the unnamed module, then the package in the unnamed module is ignored.
B. Code in an explicitly named module can access types in the unnamed module.
C. We must add a module descriptor to make an application developed using a Java version prior to SE9 run on Java 11.
D. The unnamed module exports all of its packages.
E. If a request is made to load a type whose package is not defined in any known module, then the module system will attempt to load it from the classpath.
F. The unnamed module can only access packages defined in the unnamed module.

Answer: A,D,E

Explanation:
The Java Platform Module System (JPMS), introduced in Java 9, modularizes the Java platform and applications. Understanding the behavior of named and unnamed modules is crucial.
* B. The unnamed module exports all of its packages.
Correct. The unnamed module, which includes all code on the classpath, exports all of its packages. This means that any code can access the public types in these packages. However, the unnamed module cannot be explicitly required by named modules.
* C. If a package is defined in both a named module and the unnamed module, then the package in the unnamed module is ignored.
Correct. In cases where a package is present in both a named module and the unnamed module, the version in the named module takes precedence. The package in the unnamed module is ignored to maintain module integrity and avoid conflicts.
* F. If a request is made to load a type whose package is not defined in any known module, then the module system will attempt to load it from the classpath.
Correct. When the module system cannot find a requested type in any known module, it defaults to searching the classpath (i.e., the unnamed module) to locate the type.
Incorrect Options:
* A. Code in an explicitly named module can access types in the unnamed module.
Incorrect. Named modules cannot access types in the unnamed module. The unnamed module can read from named modules, but the reverse is not allowed to ensure strong encapsulation.
* D. We must add a module descriptor to make an application developed using a Java version prior to SE9 run on Java 11.
Incorrect. Adding a module descriptor (module-info.java) is not mandatory for applications developed before Java 9 to run on Java 11. Such applications can run in the unnamed module without modification.
* E. The unnamed module can only access packages defined in the unnamed module.
Incorrect. The unnamed module can access all packages exported by all named modules, in addition to its own packages.

NEW QUESTION # 29
Given:
java
sealed class Vehicle permits Car, Bike {
}
non-sealed class Car extends Vehicle {
}
final class Bike extends Vehicle {
}
public class SealedClassTest {
public static void main(String[] args) {
Class<?> vehicleClass = Vehicle.class;
Class<?> carClass = Car.class;
Class<?> bikeClass = Bike.class;
System.out.print("Is Vehicle sealed? " + vehicleClass.isSealed() +
"; Is Car sealed? " + carClass.isSealed() +
"; Is Bike sealed? " + bikeClass.isSealed());
}
}
What is printed?

A. Is Vehicle sealed? false; Is Car sealed? true; Is Bike sealed? true
B. Is Vehicle sealed? true; Is Car sealed? false; Is Bike sealed? false
C. Is Vehicle sealed? true; Is Car sealed? true; Is Bike sealed? true
D. Is Vehicle sealed? false; Is Car sealed? false; Is Bike sealed? false

Answer: B

Explanation:
* Understanding Sealed Classes in Java
* Asealed classrestricts which other classes can extend it.
* A sealed classmust explicitly declare its permitted subclassesusing the permits keyword.
* Subclasses can be declared as:
* sealed(restricts further extension).
* non-sealed(removes the restriction, allowing unrestricted subclassing).
* final(prevents further subclassing).
* Analyzing the Given Code
* Vehicle is declared as sealed with permits Car, Bike, meaning only Car and Bike can extend it.
* Car is declared as non-sealed, which means itis no longer sealedand can have subclasses.
* Bike is declared as final, meaningit cannot be subclassed.
* Using isSealed() Method
* vehicleClass.isSealed() #truebecause Vehicle is explicitly marked as sealed.
* carClass.isSealed() #falsebecause Car is marked non-sealed.
* bikeClass.isSealed() #falsebecause Bike is final, and a final class isnot considered sealed.
* Final Output
csharp
Is Vehicle sealed? true; Is Car sealed? false; Is Bike sealed? false
Thus, the correct answer is:"Is Vehicle sealed? true; Is Car sealed? false; Is Bike sealed? false" References:
* Java SE 21 - Sealed Classes
* Java SE 21 - isSealed() Method

NEW QUESTION # 30
Given:
java
try (FileOutputStream fos = new FileOutputStream("t.tmp");
ObjectOutputStream oos = new ObjectOutputStream(fos)) {
fos.write("Today");
fos.writeObject("Today");
oos.write("Today");
oos.writeObject("Today");
} catch (Exception ex) {
// handle exception
}
Which statement compiles?

A. oos.writeObject("Today");
B. fos.writeObject("Today");
C. fos.write("Today");
D. oos.write("Today");

Answer: A

Explanation:
In Java, FileOutputStream and ObjectOutputStream are used for writing data to files, but they have different purposes and methods. Let's analyze each statement:
* fos.write("Today");
The FileOutputStream class is designed to write raw byte streams to files. The write method in FileOutputStream expects a parameter of type int or byte[]. Since "Today" is a String, passing it directly to fos.
write("Today"); will cause a compilation error because there is no write method in FileOutputStream that accepts a String parameter.
* fos.writeObject("Today");
The FileOutputStream class does not have a method named writeObject. The writeObject method is specific to ObjectOutputStream. Therefore, attempting to call fos.writeObject("Today"); will result in a compilation error.
* oos.write("Today");
The ObjectOutputStream class is used to write objects to an output stream. However, it does not have a write method that accepts a String parameter. The available write methods in ObjectOutputStream are for writing primitive data types and objects. Therefore, oos.write("Today"); will cause a compilation error.
* oos.writeObject("Today");
The ObjectOutputStream class provides the writeObject method, which is used to serialize objects and write them to the output stream. Since String implements the Serializable interface, "Today" can be serialized.
Therefore, oos.writeObject("Today"); is valid and compiles successfully.
In summary, the only statement that compiles without errors is oos.writeObject("Today");.
References:
* Java SE 21 & JDK 21 - ObjectOutputStream
* Java SE 21 & JDK 21 - FileOutputStream

NEW QUESTION # 31
Which of the following statements oflocal variables declared with varareinvalid?(Choose 4)

A. var d[] = new int[4];
B. var f = { 6 };
C. var b = 2, c = 3.0;
D. var a = 1;(Valid: var correctly infers int)
E. var e;
F. var h = (g = 7);

Answer: A,B,C,E

Explanation:
1. Valid Use Cases of var
* var is alocal variable type inferencefeature.
* The compilerinfers the type from the assigned value.
* Example of valid use:
java
var a = 10; // Type inferred as int
var str = "Hello"; // Type inferred as String
2. Analyzing the Given Statements
Statement
Valid/Invalid
Reason
var a = 1;
Valid
Type inferred as int.
var b = 2, c = 3.0;
#Invalid
var doesnot allow multiple declarationsin one statement.
var d[] = new int[4];
#Invalid
Array brackets []are not allowedwith var.
var e;
#Invalid
varrequires an initializer(cannot be declared without assignment).
var f = { 6 };
#Invalid
{ 6 } is anarray initializer, which must have an explicit type.
var h = (g = 7);
Valid
g is assigned 7, and h gets its value.
Thus, the correct answers are:B, C, D, E
References:
* Java SE 21 - Local Variable Type Inference (var)
* Java SE 21 - var Restrictions

NEW QUESTION # 32
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 2
B. It's either 0 or 1
C. Compilation fails
D. It's always 1
E. It's either 1 or 2

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 # 33
......

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