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Oracle Java SE 21 Developer Professional Sample Questions (Q21-Q26):

NEW QUESTION # 21
Given a properties file on the classpath named Person.properties with the content:
ini
name=James
And:
java
public class Person extends ListResourceBundle {
protected Object[][] getContents() {
return new Object[][]{
{"name", "Jeanne"}
};
}
}
And:
java
public class Test {
public static void main(String[] args) {
ResourceBundle bundle = ResourceBundle.getBundle("Person");
String name = bundle.getString("name");
System.out.println(name);
}
}
What is the given program's output?

A. JamesJeanne
B. James
C. MissingResourceException
D. Jeanne
E. Compilation fails
F. JeanneJames

Answer: D

Explanation:
In this scenario, we have a Person class that extends ListResourceBundle and a properties file named Person.
properties. Both define a resource with the key "name" but with different values:
* Person class (ListResourceBundle):Defines the key "name" with the value "Jeanne".
* Person.properties file:Defines the key "name" with the value "James".
When the ResourceBundle.getBundle("Person") method is called, the Java runtime searches for a resource bundle with the base name "Person". The search order is as follows:
* Class-Based Resource Bundle:The runtime first looks for a class named Person (i.e., Person.class).
* Properties File Resource Bundle:If the class is not found, it then looks for a properties file named Person.properties.
In this case, since the Person class is present and accessible, the runtime will load the Person class as the resource bundle. Therefore, the getBundle method returns an instance of the Person class.
Subsequently, when bundle.getString("name") is called, it retrieves the value associated with the key "name" from the Person class, which is "Jeanne".
Thus, the output of the program is:
nginx
Jeanne

NEW QUESTION # 22
Given:
java
public class Test {
public static void main(String[] args) throws IOException {
Path p1 = Path.of("f1.txt");
Path p2 = Path.of("f2.txt");
Files.move(p1, p2);
Files.delete(p1);
}
}
In which case does the given program throw an exception?

A. Neither files f1.txt nor f2.txt exist
B. An exception is always thrown
C. File f1.txt exists while file f2.txt doesn't
D. File f2.txt exists while file f1.txt doesn't
E. Both files f1.txt and f2.txt exist

Answer: B

Explanation:
In this program, the following operations are performed:
* Paths Initialization:
* Path p1 is set to "f1.txt".
* Path p2 is set to "f2.txt".
* File Move Operation:
* Files.move(p1, p2); attempts to move (or rename) f1.txt to f2.txt.
* File Delete Operation:
* Files.delete(p1); attempts to delete f1.txt.
Analysis:
* If f1.txt Does Not Exist:
* The Files.move(p1, p2); operation will throw a NoSuchFileException because the source file f1.
txt is missing.
* If f1.txt Exists and f2.txt Does Not Exist:
* The Files.move(p1, p2); operation will successfully rename f1.txt to f2.txt.
* Subsequently, the Files.delete(p1); operation will throw a NoSuchFileException because p1 (now f1.txt) no longer exists after the move.
* If Both f1.txt and f2.txt Exist:
* The Files.move(p1, p2); operation will throw a FileAlreadyExistsException because the target file f2.txt already exists.
* If f2.txt Exists While f1.txt Does Not:
* Similar to the first scenario, the Files.move(p1, p2); operation will throw a NoSuchFileException due to the absence of f1.txt.
In all possible scenarios, an exception is thrown during the execution of the program.

NEW QUESTION # 23
Given:
java
interface Calculable {
long calculate(int i);
}
public class Test {
public static void main(String[] args) {
Calculable c1 = i -> i + 1; // Line 1
Calculable c2 = i -> Long.valueOf(i); // Line 2
Calculable c3 = i -> { throw new ArithmeticException(); }; // Line 3
}
}
Which lines fail to compile?

A. Line 1 and line 3
B. Line 2 and line 3
C. The program successfully compiles
D. Line 1 only
E. Line 1 and line 2
F. Line 3 only
G. Line 2 only

Answer: C

Explanation:
In this code, the Calculable interface defines a single abstract method calculate that takes an int parameter and returns a long. The main method contains three lambda expressions assigned to variables c1, c2, and c3 of type Calculable.
* Line 1:Calculable c1 = i -> i + 1;
This lambda expression takes an integer i and returns the result of i + 1. Since the expression i + 1 results in an int, and Java allows implicit widening conversion from int to long, this line compiles successfully.
* Line 2:Calculable c2 = i -> Long.valueOf(i);
Here, the lambda expression takes an integer i and returns the result of Long.valueOf(i). The Long.valueOf (int i) method returns a Long object. However, Java allows unboxing of the Long object to a long primitive type when necessary. Therefore, this line compiles successfully.
* Line 3:Calculable c3 = i -> { throw new ArithmeticException(); };
This lambda expression takes an integer i and throws an ArithmeticException. Since the method calculate has a return type of long, and throwing an exception is a valid way to exit the method without returning a value, this line compiles successfully.
Since all three lines adhere to the method signature defined in the Calculable interface and there are no type mismatches or syntax errors, the program compiles successfully.

NEW QUESTION # 24
A module com.eiffeltower.shop with the related sources in the src directory.
That module requires com.eiffeltower.membership, available in a JAR located in the lib directory.
What is the command to compile the module com.eiffeltower.shop?

A. bash
CopyEdit
javac -source src -p lib/com.eiffel.membership.jar -d out -m com.eiffeltower.shop
B. css
CopyEdit
javac --module-source-path src -p lib/com.eiffel.membership.jar -d out -m com.eiffeltower.shop
C. css
CopyEdit
javac -path src -p lib/com.eiffel.membership.jar -d out -m com.eiffeltower.shop
D. css
CopyEdit
javac --module-source-path src -p lib/com.eiffel.membership.jar -s out -m com.eiffeltower.shop

Answer: B

Explanation:
Comprehensive and Detailed In-Depth Explanation:
Understanding Java Module Compilation (javac)
Java modules are compiled using the javac command with specific options to specify:
* Where the source files are located (--module-source-path)
* Where required dependencies (external modules) are located (-p / --module-path)
* Where the compiled output should be placed (-d)
Breaking Down the Correct Compilation Command
css
CopyEdit
javac --module-source-path src -p lib/com.eiffel.membership.jar -d out -m com.eiffeltower.shop
* --module-source-path src # Specifies the directory where module sources are located.
* -p lib/com.eiffel.membership.jar # Specifies the module path (JAR dependency in lib).
* -d out # Specifies the output directory for compiled .class files.
* -m com.eiffeltower.shop # Specifies the module to compile (com.eiffeltower.shop).

NEW QUESTION # 25
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. ok the 2024-07-10
B. An exception is thrown
C. Compilation fails
D. ok the 2024-07-10T07:17:45.523939600

Answer: C

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 # 26
......

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Pass Guaranteed Quiz Oracle - 1z1-830 - Trustable Java SE 21 Developer Professional Certification Cost

Free PDF 2025 Oracle 1z1-830: Java SE 21 Developer Professional Useful Certification Cost

Oracle Java SE 21 Developer Professional Sample Questions (Q21-Q26):

1z1-830 Certification Cost - Quiz 2025 Oracle Java SE 21 Developer Professional Realistic Interactive Practice Exam

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