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100% Pass Quiz Oracle - 1z0-830 Pass-Sure Latest Test Questions

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Oracle Java SE 21 Developer Professional Sample Questions (Q79-Q84):

NEW QUESTION # 79
Given:
java
public class BoomBoom implements AutoCloseable {
public static void main(String[] args) {
try (BoomBoom boomBoom = new BoomBoom()) {
System.out.print("bim ");
throw new Exception();
} catch (Exception e) {
System.out.print("boom ");
}
}
@Override
public void close() throws Exception {
System.out.print("bam ");
throw new RuntimeException();
}
}
What is printed?

A. Compilation fails.
B. bim boom bam
C. bim boom
D. bim bam boom
E. bim bam followed by an exception

Answer: D

Explanation:
* Understanding Try-With-Resources (AutoCloseable)
* BoomBoom implements AutoCloseable, meaning its close() method isautomatically calledat the end of the try block.
* Step-by-Step Execution
* Step 1: Enter Try Block
java
try (BoomBoom boomBoom = new BoomBoom()) {
System.out.print("bim ");
throw new Exception();
}
* "bim " is printed.
* Anexception (Exception) is thrown, butbefore it is handled, the close() method is executed.
* Step 2: close() is Called
java
@Override
public void close() throws Exception {
System.out.print("bam ");
throw new RuntimeException();
}
* "bam " is printed.
* A new RuntimeException is thrown, but it doesnot override the existing Exception yet.
* Step 3: Exception Handling
java
} catch (Exception e) {
System.out.print("boom ");
}
* The catch (Exception e)catches the original Exception from the try block.
* "boom " is printed.
* Final Output
nginx
bim bam boom
* Theoriginal Exception is caught, not the RuntimeException from close().
* TheRuntimeException from close() is ignoredbecause thecatch block is already handling Exception.
Thus, the correct answer is:bim bam boom
References:
* Java SE 21 - Try-With-Resources
* Java SE 21 - AutoCloseable Interface

NEW QUESTION # 80
Which of the following statements are correct?

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

Answer: B

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 # 81
Which methods compile?

A. ```java public List<? extends IOException> getListExtends() { return new ArrayList<Exception>(); } csharp
B. ```java
public List<? extends IOException> getListExtends() {
return new ArrayList<FileNotFoundException>();
}
C. ```java public List<? super IOException> getListSuper() { return new ArrayList<Exception>(); } csharp
D. ```java
public List<? super IOException> getListSuper() {
return new ArrayList<FileNotFoundException>();
}

Answer: B,C

Explanation:
In Java generics, wildcards are used to relax the type constraints of generic types. The extends wildcard (<?
extends Type>) denotes an upper bounded wildcard, allowing any type that is a subclass of Type. Conversely, the super wildcard (<? super Type>) denotes a lower bounded wildcard, allowing any type that is a superclass of Type.
Option A:
java
public List<? super IOException> getListSuper() {
return new ArrayList<Exception>();
}
Here, List<? super IOException> represents a list that can hold IOException objects and objects of its supertypes. Since Exception is a superclass of IOException, ArrayList<Exception> is compatible with List<?
super IOException>. Therefore, this method compiles successfully.
Option B:
java
public List<? extends IOException> getListExtends() {
return new ArrayList<FileNotFoundException>();
}
In this case, List<? extends IOException> represents a list that can hold objects of IOException and its subclasses. Since FileNotFoundException is a subclass of IOException, ArrayList<FileNotFoundException> is compatible with List<? extends IOException>. Thus, this method compiles successfully.
Option C:
java
public List<? extends IOException> getListExtends() {
return new ArrayList<Exception>();
}
Here, List<? extends IOException> expects a list of IOException or its subclasses. However, Exception is a superclass of IOException, not a subclass. Therefore, ArrayList<Exception> is not compatible with List<?
extends IOException>, and this method will not compile.
Option D:
java
public List<? super IOException> getListSuper() {
return new ArrayList<FileNotFoundException>();
}
In this scenario, List<? super IOException> expects a list that can hold IOException objects and objects of its supertypes. Since FileNotFoundException is a subclass of IOException, ArrayList<FileNotFoundException> is not compatible with List<? super IOException>, and this method will not compile.
Therefore, the methods in options A and B compile successfully, while those in options C and D do not.

NEW QUESTION # 82
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. 0
B. 1
C. ClassCastException
D. NotSerializableException
E. Compilation fails

Answer: C

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 # 83
Given:
java
var array1 = new String[]{ "foo", "bar", "buz" };
var array2[] = { "foo", "bar", "buz" };
var array3 = new String[3] { "foo", "bar", "buz" };
var array4 = { "foo", "bar", "buz" };
String array5[] = new String[]{ "foo", "bar", "buz" };
Which arrays compile? (Select 2)

A. array5
B. array3
C. array1
D. array2
E. array4

Answer: A,C

Explanation:
In Java, array initialization can be performed in several ways, but certain syntaxes are invalid and will cause compilation errors. Let's analyze each declaration:
* var array1 = new String[]{ "foo", "bar", "buz" };
This is a valid declaration. The var keyword allows the compiler to infer the type from the initializer. Here, new String[]{ "foo", "bar", "buz" } creates an anonymous array of String with three elements. The compiler infers array1 as String[]. This syntax is correct and compiles successfully.
* var array2[] = { "foo", "bar", "buz" };
This declaration is invalid. While var can be used for type inference, appending [] after var is not allowed.
The correct syntax would be either String[] array2 = { "foo", "bar", "buz" }; or var array2 = new String[]{
"foo", "bar", "buz" };. Therefore, this line will cause a compilation error.
* var array3 = new String[3] { "foo", "bar", "buz" };
This declaration is invalid. In Java, when specifying the size of the array (new String[3]), you cannot simultaneously provide an initializer. The correct approach is either to provide the size without an initializer (new String[3]) or to provide the initializer without specifying the size (new String[]{ "foo", "bar", "buz" }).
Therefore, this line will cause a compilation error.
* var array4 = { "foo", "bar", "buz" };
This declaration is invalid. The array initializer { "foo", "bar", "buz" } can only be used in an array declaration when the type is explicitly provided. Since var relies on type inference and there's no explicit type provided here, this will cause a compilation error. The correct syntax would be String[] array4 = { "foo",
"bar", "buz" };.
* String array5[] = new String[]{ "foo", "bar", "buz" };
This is a valid declaration. Here, String array5[] declares array5 as an array of String. The initializer new String[]{ "foo", "bar", "buz" } creates an array with three elements. This syntax is correct and compiles successfully.
Therefore, the declarations that compile successfully are array1 and array5.
References:
* Java SE 21 & JDK 21 - Local Variable Type Inference
* Java SE 21 & JDK 21 - Arrays

NEW QUESTION # 84
......

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