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Oracle Java SE 21 Developer Professional Sample Questions (Q80-Q85):

NEW QUESTION # 80
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. array2
B. array1
C. array4
D. array5
E. array3

Answer: B,D

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 # 81
Given:
java
interface SmartPhone {
boolean ring();
}
class Iphone15 implements SmartPhone {
boolean isRinging;
boolean ring() {
isRinging = !isRinging;
return isRinging;
}
}
Choose the right statement.

A. Iphone15 class does not compile
B. SmartPhone interface does not compile
C. An exception is thrown at running Iphone15.ring();
D. Everything compiles

Answer: A

Explanation:
In this code, the SmartPhone interface declares a method ring() with a boolean return type. The Iphone15 class implements the SmartPhone interface and provides an implementation for the ring() method.
However, in the Iphone15 class, the ring() method is declared without the public access modifier. In Java, when a class implements an interface, it must provide implementations for all the interface's methods with the same or a more accessible access level. Since interface methods are implicitly public, the implementing methods in the class must also be public. Failing to do so results in a compilation error.
Therefore, the Iphone15 class does not compile because the ring() method is not declared as public.

NEW QUESTION # 82
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. bim boom
B. bim boom bam
C. bim bam followed by an exception
D. bim bam boom
E. Compilation fails.

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 # 83
What do the following print?
java
public class DefaultAndStaticMethods {
public static void main(String[] args) {
WithStaticMethod.print();
}
}
interface WithDefaultMethod {
default void print() {
System.out.print("default");
}
}
interface WithStaticMethod extends WithDefaultMethod {
static void print() {
System.out.print("static");
}
}

A. Compilation fails
B. static
C. nothing
D. default

Answer: B

Explanation:
In this code, we have two interfaces and a class with a main method:
* WithDefaultMethod Interface:
* Declares a default method print() that outputs "default".
* WithStaticMethod Interface:
* Extends WithDefaultMethod.
* Declares a static method print() that outputs "static".
* DefaultAndStaticMethods Class:
* Contains the main method, which calls WithStaticMethod.print().
Key Points:
* Static Methods in Interfaces:
* Static methods in interfaces are not inherited by implementing or extending classes or interfaces.
They belong solely to the interface in which they are declared.
* Default Methods in Interfaces:
* Default methods can be inherited by implementing classes, but they cannot be overridden by static methods in subinterfaces.
Execution Flow:
* The main method calls WithStaticMethod.print().
* This invokes the static method print() defined in the WithStaticMethod interface, which outputs "static".
Therefore, the program compiles successfully and prints static.

NEW QUESTION # 84
Given:
java
Object myVar = 0;
String print = switch (myVar) {
case int i -> "integer";
case long l -> "long";
case String s -> "string";
default -> "";
};
System.out.println(print);
What is printed?

A. Compilation fails.
B. long
C. string
D. It throws an exception at runtime.
E. nothing
F. integer

Answer: A

Explanation:
* Why does the compilation fail?
* TheJava switch statement does not support primitive type pattern matchingin switch expressions as of Java 21.
* The case pattern case int i -> "integer"; isinvalidbecausepattern matching with primitive types (like int or long) is not yet supported in switch statements.
* The error occurs at case int i -> "integer";, leading to acompilation failure.
* Correcting the Code
* Since myVar is of type Object,autoboxing converts 0 into an Integer.
* To make the code compile, we should use Integer instead of int:
java
Object myVar = 0;
String print = switch (myVar) {
case Integer i -> "integer";
case Long l -> "long";
case String s -> "string";
default -> "";
};
System.out.println(print);
* Output:
bash
integer
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - Pattern Matching for switch
* Java SE 21 - switch Expressions

NEW QUESTION # 85
......

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