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Oracle Java SE 21 Developer Professional Sample Questions (Q68-Q73):
NEW QUESTION # 68
Which StringBuilder variable fails to compile?
java
public class StringBuilderInstantiations {
public static void main(String[] args) {
var stringBuilder1 = new StringBuilder();
var stringBuilder2 = new StringBuilder(10);
var stringBuilder3 = new StringBuilder("Java");
var stringBuilder4 = new StringBuilder(new char[]{'J', 'a', 'v', 'a'});
}
}
Answer: D
Explanation:
In the provided code, four StringBuilder instances are being created using different constructors:
* stringBuilder1: new StringBuilder()
* This constructor creates an empty StringBuilder with an initial capacity of 16 characters.
* stringBuilder2: new StringBuilder(10)
* This constructor creates an empty StringBuilder with a specified initial capacity of 10 characters.
* stringBuilder3: new StringBuilder("Java")
* This constructor creates a StringBuilder initialized to the contents of the specified string "Java".
* stringBuilder4: new StringBuilder(new char[]{'J', 'a', 'v', 'a'})
* This line attempts to create a StringBuilder using a char array. However, the StringBuilder class does not have a constructor that accepts a char array directly. The available constructors are:
* StringBuilder()
* StringBuilder(int capacity)
* StringBuilder(String str)
* StringBuilder(CharSequence seq)
Since a char array does not implement the CharSequence interface, and there is no constructor that directly accepts a char array, this line will cause a compilation error.
To initialize a StringBuilder with a char array, you can convert the char array to a String first:
java
var stringBuilder4 = new StringBuilder(new String(new char[]{'J', 'a', 'v', 'a'})); This approach utilizes the String constructor that accepts a char array, and then passes the resulting String to the StringBuilder constructor.
NEW QUESTION # 69
Given:
java
interface A {
default void ma() {
}
}
interface B extends A {
static void mb() {
}
}
interface C extends B {
void ma();
void mc();
}
interface D extends C {
void md();
}
interface E extends D {
default void ma() {
}
default void mb() {
}
default void mc() {
}
}
Which interface can be the target of a lambda expression?
Answer: C
Explanation:
In Java, a lambda expression can be used where a target type is a functional interface. A functional interface is an interface that contains exactly one abstract method. This concept is also known as a Single Abstract Method (SAM) type.
Analyzing each interface:
* Interface A: Contains a single default method ma(). Since default methods are not abstract, A has no abstract methods.
* Interface B: Extends A and adds a static method mb(). Static methods are also not abstract, so B has no abstract methods.
* Interface C: Extends B and declares two abstract methods: ma() (which overrides the default method from A) and mc(). Therefore, C has two abstract methods.
* Interface D: Extends C and adds another abstract method md(). Thus, D has three abstract methods.
* Interface E: Extends D and provides default implementations for ma(), mb(), and mc(). However, it does not provide an implementation for md(), leaving it as the only abstract method in E.
For an interface to be a functional interface, it must have exactly one abstract method. In this case, E has one abstract method (md()), so it qualifies as a functional interface. However, the question asks which interface can be the target of a lambda expression. Since E is a functional interface, it can be the target of a lambda expression.
Therefore, the correct answer is D (E).
NEW QUESTION # 70
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?
Answer: B
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 # 71
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?
Answer: D
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 # 72
Given:
var cabarets = new TreeMap<>();
cabarets.put(1, "Moulin Rouge");
cabarets.put(2, "Crazy Horse");
cabarets.put(3, "Paradis Latin");
cabarets.put(4, "Le Lido");
cabarets.put(5, "Folies Bergere");
System.out.println(cabarets.subMap(2, true, 5, false));
What is printed?
Answer: E
Explanation:
Understanding TreeMap.subMap(fromKey, fromInclusive, toKey, toInclusive)
* TreeMap.subMap(K fromKey, boolean fromInclusive, K toKey, boolean toInclusive) returns aportion of the mapthat falls within the specified key range.
* Thefirst boolean parameter(fromInclusive) determines if the fromKey should be included.
* Thesecond boolean parameter(toInclusive) determines if the toKey should be included.
Given TreeMap Contents
CopyEdit
{1=Moulin Rouge, 2=Crazy Horse, 3=Paradis Latin, 4=Le Lido, 5=Folies Bergere} Applying subMap(2, true, 5, false)
* Includeskey 2 ("Crazy Horse")#(fromInclusive = true)
* Includeskey 3 ("Paradis Latin")#
* Includeskey 4 ("Le Lido")#
* Excludes key 5 ("Folies Bergere")#(toInclusive = false)
Final Output
CopyEdit
{2=Crazy Horse, 3=Paradis Latin, 4=Le Lido}
Thus, the correct answer is:#{2=Crazy Horse, 3=Paradis Latin, 4=Le Lido} References:
* Java SE 21 - TreeMap.subMap()
* Java SE 21 - NavigableMap
NEW QUESTION # 73
......
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