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1z1-830 Flexible Testing Engine - 1z1-830 Latest Braindumps Ppt

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Oracle Java SE 21 Developer Professional Sample Questions (Q22-Q27):

NEW QUESTION # 22
Which of the followingisn'ta correct way to write a string to a file?

A. java
Path path = Paths.get("file.txt");
byte[] strBytes = "Hello".getBytes();
Files.write(path, strBytes);
B. java
try (FileOutputStream outputStream = new FileOutputStream("file.txt")) { byte[] strBytes = "Hello".getBytes(); outputStream.write(strBytes);
}
C. None of the suggestions
D. java
try (BufferedWriter writer = new BufferedWriter("file.txt")) {
writer.write("Hello");
}
E. java
try (FileWriter writer = new FileWriter("file.txt")) {
writer.write("Hello");
}
F. java
try (PrintWriter printWriter = new PrintWriter("file.txt")) {
printWriter.printf("Hello %s", "James");
}

Answer: D

Explanation:
(BufferedWriter writer = new BufferedWriter("file.txt") is incorrect.)
Theincorrect statementisoption Bbecause BufferedWriterdoes nothave a constructor that accepts a String (file name) directly. The correct way to use BufferedWriter is to wrap it around a FileWriter, like this:
java
try (BufferedWriter writer = new BufferedWriter(new FileWriter("file.txt"))) { writer.write("Hello");
}
Evaluation of Other Options:
Option A (Files.write)# Correct
* Uses Files.write() to write bytes to a file.
* Efficient and concise method for writing small text files.
Option C (FileOutputStream)# Correct
* Uses a FileOutputStream to write raw bytes to a file.
* Works for both text and binary data.
Option D (PrintWriter)# Correct
* Uses PrintWriter for formatted text output.
Option F (FileWriter)# Correct
* Uses FileWriter to write text data.
Option E (None of the suggestions)# Incorrect becauseoption Bis incorrect.

NEW QUESTION # 23
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. Compilation fails
C. ok the 2024-07-10T07:17:45.523939600
D. An exception is thrown

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 # 24
What do the following print?
java
public class Main {
int instanceVar = staticVar;
static int staticVar = 666;
public static void main(String args[]) {
System.out.printf("%d %d", new Main().instanceVar, staticVar);
}
static {
staticVar = 42;
}
}

A. 666 42
B. 42 42
C. Compilation fails
D. 666 666

Answer: B

Explanation:
In this code, the class Main contains both an instance variable instanceVar and a static variable staticVar. The sequence of initialization and execution is as follows:
* Static Variable Initialization:
* staticVar is declared and initialized to 666.
* Static Block Execution:
* The static block executes, updating staticVar to 42.
* Instance Variable Initialization:
* When a new instance of Main is created, instanceVar is initialized to the current value of staticVar, which is 42.
* main Method Execution:
* The main method creates a new instance of Main and prints the values of instanceVar and staticVar.
Therefore, the output of the program is 42 42.

NEW QUESTION # 25
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. Compilation fails
D. 1
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 # 26
Given:
java
var counter = 0;
do {
System.out.print(counter + " ");
} while (++counter < 3);
What is printed?

A. An exception is thrown.
B. 0 1 2 3
C. 1 2 3 4
D. Compilation fails.
E. 1 2 3
F. 0 1 2

Answer: F

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 # 27
......

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