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Oracle Java SE 21 Developer Professional Sample Questions (Q74-Q79):

NEW QUESTION # 74
Which two of the following aren't the correct ways to create a Stream?

A. Stream stream = new Stream();
B. Stream stream = Stream.generate(() -> "a");
C. Stream stream = Stream.empty();
D. Stream stream = Stream.of("a");
E. Stream<String> stream = Stream.builder().add("a").build();
F. Stream stream = Stream.ofNullable("a");
G. Stream stream = Stream.of();

Answer: A,E

Explanation:
In Java, the Stream API provides several methods to create streams. However, not all approaches are valid.

NEW QUESTION # 75
Given:
java
final Stream<String> strings =
Files.readAllLines(Paths.get("orders.csv"));
strings.skip(1)
.limit(2)
.forEach(System.out::println);
And that the orders.csv file contains:
mathematica
OrderID,Customer,Product,Quantity,Price
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99
3,Sebastien Loeb,Monitor,1,199.99
4,Antoine Griezmann,Headset,3,45.00
What is printed?

A. arduino
2,Teddy Riner,Mouse,1,15.99
3,Sebastien Loeb,Monitor,1,199.99
B. Compilation fails.
C. An exception is thrown at runtime.
D. arduino
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99
E. arduino
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99
3,Sebastien Loeb,Monitor,1,199.99
4,Antoine Griezmann,Headset,3,45.00

Answer: B,C

Explanation:
1. Why Does Compilation Fail?
* The error is in this line:
java
final Stream<String> strings = Files.readAllLines(Paths.get("orders.csv"));
* Files.readAllLines(Paths.get("orders.csv")) returns a List<String>,not a Stream<String>.
* A List<String> cannot be assigned to a Stream<String>.
2. Correcting the Code
* The correct way to create a stream from the file:
java
Stream<String> strings = Files.lines(Paths.get("orders.csv"));
* This correctly creates a Stream<String> from the file.
3. Expected Output After Fixing
java
Files.lines(Paths.get("orders.csv"))
skip(1) // Skips the header row
limit(2) // Limits to first two data rows
forEach(System.out::println);
* Output:
arduino
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - Files.readAllLines
* Java SE 21 - Files.lines

NEW QUESTION # 76
Which of the following statements are correct?

A. You can use 'public' access modifier with all kinds of classes
B. You can use 'private' access modifier with all kinds of classes
C. You can use 'protected' access modifier with all kinds of classes
D. None
E. You can use 'final' modifier with all kinds of classes

Answer: D

Explanation:
1. private Access Modifier
* The private access modifiercan only be used for inner classes(nested classes).
* Top-level classes cannot be private.
* Example ofinvaliduse:
java
private class MyClass {} // Compilation error
* Example ofvaliduse (for inner class):
java
class Outer {
private class Inner {}
}
2. protected Access Modifier
* Top-level classes cannot be protected.
* protectedonly applies to members (fields, methods, and constructors).
* Example ofinvaliduse:
java
protected class MyClass {} // Compilation error
* Example ofvaliduse (for methods/fields):
java
class Parent {
protected void display() {}
}
3. public Access Modifier
* Atop-level class can be public, butonly one public class per file is allowed.
* Example ofvaliduse:
java
public class MyClass {}
* Example ofinvaliduse:
java
public class A {}
public class B {} // Compilation error: Only one public class per file
4. final Modifier
* finalcan be used with classes, but not all kinds of classes.
* Interfaces cannot be final, because they are meant to be implemented.
* Example ofinvaliduse:
java
final interface MyInterface {} // Compilation error
Thus,none of the statements are fully correct, making the correct answer:None References:
* Java SE 21 - Access Modifiers
* Java SE 21 - Class Modifiers

NEW QUESTION # 77
Given:
java
import java.io.*;
class A implements Serializable {
int number = 1;
}
class B implements Serializable {
int number = 2;
}
public class Test {
public static void main(String[] args) throws Exception {
File file = new File("o.ser");
A a = new A();
var oos = new ObjectOutputStream(new FileOutputStream(file));
oos.writeObject(a);
oos.close();
var ois = new ObjectInputStream(new FileInputStream(file));
B b = (B) ois.readObject();
ois.close();
System.out.println(b.number);
}
}
What is the given program's output?

A. NotSerializableException
B. 0
C. 1
D. ClassCastException
E. Compilation fails

Answer: D

Explanation:
In this program, we have two classes, A and B, both implementing the Serializable interface, and a Test class with the main method.
Program Flow:
* Serialization:
* An instance of class A is created and assigned to the variable a.
* An ObjectOutputStream is created to write to the file "o.ser".
* The object a is serialized and written to the file.
* The ObjectOutputStream is closed.
* Deserialization:
* An ObjectInputStream is created to read from the file "o.ser".
* The program attempts to read an object from the file and cast it to an instance of class B.
* The ObjectInputStream is closed.
Analysis:
* Serialization Process:
* The object a is an instance of class A and is serialized into the file "o.ser".
* Deserialization Process:
* When deserializing, the program reads the object from the file and attempts to cast it to class B.
* However, the object in the file is of type A, not B.
* Since A and B are distinct classes with no inheritance relationship, casting an A instance to B is invalid.
Exception Details:
* Attempting to cast an object of type A to type B results in a ClassCastException.
* The exception message would be similar to:
pgsql
Exception in thread "main" java.lang.ClassCastException: class A cannot be cast to class B Conclusion:
The program compiles successfully but throws a ClassCastException at runtime when it attempts to cast the deserialized object to class B.

NEW QUESTION # 78
Given:
java
var _ = 3;
var $ = 7;
System.out.println(_ + $);
What is printed?

A. 0
B. It throws an exception.
C. Compilation fails.
D. _$

Answer: C

Explanation:
* The var keyword and identifier rules:
* The var keyword is used for local variable type inference introduced inJava 10.
* However,Java does not allow _ (underscore) as an identifiersinceJava 9.
* If we try to use _ as a variable name, the compiler will throw an error:
pgsql
error: as of release 9, '_' is a keyword, and may not be used as an identifier
* The $ symbol as an identifier:
* The $ characteris a valid identifierin Java.
* However, since _ is not allowed, the codefails to compile before even reaching $.
Thus,the correct answer is "Compilation fails."
References:
* Java SE 21 - var Local Variable Type Inference
* Java SE 9 - Restrictions on _ Identifier

NEW QUESTION # 79
......

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