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Oracle Java SE 21 Developer Professional Sample Questions (Q48-Q53):

NEW QUESTION # 48
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. fos.writeObject("Today");
B. fos.write("Today");
C. oos.writeObject("Today");
D. oos.write("Today");

Answer: C

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 # 49
Which two of the following aren't the correct ways to create a Stream?

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

Answer: C,G

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

NEW QUESTION # 50
Given:
java
double amount = 42_000.00;
NumberFormat format = NumberFormat.getCompactNumberInstance(Locale.FRANCE, NumberFormat.Style.
SHORT);
System.out.println(format.format(amount));
What is the output?

A. 42 k
B. 42000E
C. 0
D. 42 000,00 €

Answer: A

Explanation:
In this code, a double variable amount is initialized to 42,000.00. The NumberFormat.
getCompactNumberInstance(Locale.FRANCE, NumberFormat.Style.SHORT) method is used to obtain a compact number formatter for the French locale with the short style. The format method is then called to format the amount.
The compact number formatting is designed to represent numbers in a shorter form, based on the patterns provided for a given locale. In the French locale, the short style represents thousands with a lowercase 'k'.
Therefore, 42,000 is formatted as 42 k.
* Option Evaluations:
* A. 42000E: This format is not standard in the French locale for compact number formatting.
* B. 42 000,00 €: This represents the number as a currency with two decimal places, which is not the compact form.
* C. 42000: This is the plain number without any formatting, which does not match the compact number format.
* D. 42 k: This is the correct compact representation of 42,000 in the French locale with the short style.
Thus, option D (42 k) is the correct output.

NEW QUESTION # 51
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. Compilation fails
E. ClassCastException

Answer: E

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 # 52
Given:
java
var ceo = new HashMap<>();
ceo.put("Sundar Pichai", "Google");
ceo.put("Tim Cook", "Apple");
ceo.put("Mark Zuckerberg", "Meta");
ceo.put("Andy Jassy", "Amazon");
Does the code compile?

A. False
B. True

Answer: A

Explanation:
In this code, a HashMap is instantiated using the var keyword:
java
var ceo = new HashMap<>();
The diamond operator <> is used without explicit type arguments. While the diamond operatorallows the compiler to infer types in many cases, when using var, the compiler requires explicit type information to infer the variable's type.
Therefore, the code will not compile because the compiler cannot infer the type of the HashMap when both var and the diamond operator are used without explicit type parameters.
To fix this issue, provide explicit type parameters when creating the HashMap:
java
var ceo = new HashMap<String, String>();
Alternatively, you can specify the variable type explicitly:
java
Map<String, String>
contentReference[oaicite:0]{index=0}

NEW QUESTION # 53
......

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