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Oracle Java SE 21 Developer Professional Sample Questions (Q58-Q63):

NEW QUESTION # 58
Given:
java
Optional o1 = Optional.empty();
Optional o2 = Optional.of(1);
Optional o3 = Stream.of(o1, o2)
.filter(Optional::isPresent)
.findAny()
.flatMap(o -> o);
System.out.println(o3.orElse(2));
What is the given code fragment's output?

A. Optional.empty
B. Optional[1]
C. 0
D. 1
E. 2
F. An exception is thrown
G. Compilation fails

Answer: C

Explanation:
In this code, two Optional objects are created:
* o1 is an empty Optional.
* o2 is an Optional containing the integer 1.
A stream is created from o1 and o2. The filter method retains only the Optional instances that are present (i.e., non-empty). This results in a stream containing only o2.
The findAny method returns an Optional describing some element of the stream, or an empty Optional if the stream is empty. Since the stream contains o2, findAny returns Optional[Optional[1]].
The flatMap method is then used to flatten this nested Optional. It applies the provided mapping function (o -
> o) to the value, resulting in Optional[1].
Finally, o3.orElse(2) returns the value contained in o3 if it is present; otherwise, it returns 2. Since o3 contains
1, the output is 1.

NEW QUESTION # 59
Which two of the following aren't the correct ways to create a Stream?

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

Answer: A,D

NEW QUESTION # 60
Given:
java
var lyrics = """
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose
""";
for ( int i = 0, int j = 3; i < j; i++ ) {
System.out.println( lyrics.lines()
.toList()
.get( i ) );
}
What is printed?

A. vbnet
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose
B. Nothing
C. An exception is thrown at runtime.
D. Compilation fails.

Answer: D

Explanation:
* Error in for Loop Initialization
* The initialization part of a for loopcannot declare multiple variables with different types in a single statement.
* Error:
java
for (int i = 0, int j = 3; i < j; i++) {
* Fix:Declare variables separately:
java
for (int i = 0, j = 3; i < j; i++) {
* lyrics.lines() in Java 21
* The lines() method of String returns aStream<String>, splitting the string by line breaks.
* Calling .toList() on a streamconverts it to a list.
* Valid Code After Fixing the Loop:
java
var lyrics = """
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose
""";
for (int i = 0, j = 3; i < j; i++) {
System.out.println(lyrics.lines()
toList()
get(i));
}
* Expected Output After Fixing:
vbnet
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - String.lines()
* Java SE 21 - for Statement Rules

NEW QUESTION # 61
Given:
java
void verifyNotNull(Object input) {
boolean enabled = false;
assert enabled = true;
assert enabled;
System.out.println(input.toString());
assert input != null;
}
When does the given method throw a NullPointerException?

A. Only if assertions are enabled and the input argument isn't null
B. A NullPointerException is never thrown
C. Only if assertions are disabled and the input argument isn't null
D. Only if assertions are disabled and the input argument is null
E. Only if assertions are enabled and the input argument is null

Answer: D

Explanation:
In the verifyNotNull method, the following operations are performed:
* Assertion to Enable Assertions:
java
boolean enabled = false;
assert enabled = true;
assert enabled;
* The variable enabled is initially set to false.
* The first assertion assert enabled = true; assigns true to enabled if assertions are enabled. If assertions are disabled, this assignment does not occur.
* The second assertion assert enabled; checks if enabled is true. If assertions are enabled and the previous assignment occurred, this assertion passes. If assertions are disabled, this assertion is ignored.
* Dereferencing the input Object:
java
System.out.println(input.toString());
* This line attempts to call the toString() method on the input object. If input is null, this will throw a NullPointerException.
* Assertion to Check input for null:
java
assert input != null;
* This assertion checks that input is not null. If input is null and assertions are enabled, this assertion will fail, throwing an AssertionError. If assertions are disabled, this assertion is ignored.
Analysis:
* If Assertions Are Enabled:
* The enabled variable is set to true by the first assertion, and the second assertion passes.
* If input is null, calling input.toString() will throw a NullPointerException before the final assertion is reached.
* If input is not null, input.toString() executes without issue, and the final assertion assert input != null; passes.
* If Assertions Are Disabled:
* The enabled variable remains false, but the assertions are ignored, so this has no effect.
* If input is null, calling input.toString() will throw a NullPointerException.
* If input is not null, input.toString() executes without issue.
Conclusion:
A NullPointerException is thrown if input is null, regardless of whether assertions are enabled or disabled.
Therefore, the correct answer is:
C: Only if assertions are disabled and the input argument is null

NEW QUESTION # 62
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 2024-07-10T07:17:45.523939600
B. Compilation fails
C. An exception is thrown
D. ok the 2024-07-10

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 # 63
......

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