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Oracle Java SE 21 Developer Professional Sample Questions (Q68-Q73):
NEW QUESTION # 68
Which of the following methods of java.util.function.Predicate aredefault methods?
- A. and(Predicate<? super T> other)
- B. test(T t)
- C. not(Predicate<? super T> target)
- D. negate()
- E. isEqual(Object targetRef)
- F. or(Predicate<? super T> other)
Answer: A,D,F
Explanation:
* Understanding java.util.function.Predicate<T>
* The Predicate<T> interface represents a function thattakes an input and returns a boolean(true or false).
* It is often used for filtering operations in functional programming and streams.
* Analyzing the Methods:
* and(Predicate<? super T> other)#Default method
* Combines two predicates usinglogical AND(&&).
java
Predicate<String> startsWithA = s -> s.startsWith("A");
Predicate<String> hasLength3 = s -> s.length() == 3;
Predicate<String> combined = startsWithA.and(hasLength3);
* #isEqual(Object targetRef)#Static method
* Not a default method, because it doesnot operate on an instance.
java
Predicate<String> isEqualToHello = Predicate.isEqual("Hello");
* negate()#Default method
* Negates a predicate (! operator).
java
Predicate<String> notEmpty = s -> !s.isEmpty();
Predicate<String> isEmpty = notEmpty.negate();
* #not(Predicate<? super T> target)#Static method (introduced in Java 11)
* Not a default method, since it is static.
* or(Predicate<? super T> other)#Default method
* Combines two predicates usinglogical OR(||).
* #test(T t)#Abstract method
* Not a default method, because every predicatemust implement this method.
Thus, the correct answers are:and(Predicate<? super T> other), negate(), or(Predicate<? super T> other) References:
* Java SE 21 - Predicate Interface
* Java SE 21 - Functional Interfaces
NEW QUESTION # 69
Which of the following statements is correct about a final class?
- A. It cannot be extended by any other class.
- B. It must contain at least a final method.
- C. The final keyword in its declaration must go right before the class keyword.
- D. It cannot implement any interface.
- E. It cannot extend another class.
Answer: A
Explanation:
In Java, the final keyword can be applied to classes, methods, and variables to impose certain restrictions.
Final Classes:
* Definition:A class declared with the final keyword is known as a final class.
* Purpose:Declaring a class as final prevents it from being subclassed. This is useful when you want to ensure that the class's implementation remains unchanged and cannot be extended or modified through inheritance.
Option Evaluations:
* A. The final keyword in its declaration must go right before the class keyword.
* This is correct. The syntax for declaring a final class is:
java
public final class ClassName {
// class body
}
* However, this statement is about syntax rather than the core characteristic of a final class.
* B. It must contain at least a final method.
* Incorrect. A final class can have zero or more methods, and none of them are required to be declared as final. The final keyword at the class level prevents inheritance, regardless of the methods' finality.
* C. It cannot be extended by any other class.
* Correct. The primary characteristic of a final class is that it cannot be subclassed. Attempting to do so will result in a compilation error.
* D. It cannot implement any interface.
* Incorrect. A final class can implement interfaces. Declaring a class as final restricts inheritance but does not prevent the class from implementing interfaces.
* E. It cannot extend another class.
* Incorrect. A final class can extend another class. The final keyword prevents the class from being subclassed but does not prevent it from being a subclass itself.
Therefore, the correct statement about a final class is option C: "It cannot be extended by any other class."
NEW QUESTION # 70
Given:
java
var ceo = new HashMap<>();
ceo.put("Sundar Pichai", "Google");
ceo.put("Tim Cook", "Apple");
ceo.put("Mark Zuckerberg", "Meta");
ceo.put("Andy Jassy", "Amazon");
Does the code compile?
Answer: A
Explanation:
In this code, a HashMap is instantiated using the var keyword:
java
var ceo = new HashMap<>();
The diamond operator <> is used without explicit type arguments. While the diamond operatorallows the compiler to infer types in many cases, when using var, the compiler requires explicit type information to infer the variable's type.
Therefore, the code will not compile because the compiler cannot infer the type of the HashMap when both var and the diamond operator are used without explicit type parameters.
To fix this issue, provide explicit type parameters when creating the HashMap:
java
var ceo = new HashMap<String, String>();
Alternatively, you can specify the variable type explicitly:
java
Map<String, String>
contentReference[oaicite:0]{index=0}
NEW QUESTION # 71
Given:
java
List<String> abc = List.of("a", "b", "c");
abc.stream()
.forEach(x -> {
x = x.toUpperCase();
});
abc.stream()
.forEach(System.out::print);
What is the output?
- A. ABC
- B. abc
- C. Compilation fails.
- D. An exception is thrown.
Answer: B
Explanation:
In the provided code, a list abc is created containing the strings "a", "b", and "c". The first forEach operation attempts to convert each element to uppercase by assigning x = x.toUpperCase();. However, this assignment only changes the local variable x within the lambda expression and does not modify the elements in the original list abc. Strings in Java are immutable, meaning their values cannot be changed once created.
Therefore, the original list remains unchanged.
The second forEach operation iterates over the original list and prints each element. Since the list was not modified, the output will be the concatenation of the original elements: abc.
To achieve the output ABC, you would need to collect the transformed elements into a new list, as shown below:
java
List<String> abc = List.of("a", "b", "c");
List<String> upperCaseAbc = abc.stream()
map(String::toUpperCase)
collect(Collectors.toList());
upperCaseAbc.forEach(System.out::print);
In this corrected version, the map operation creates a new stream with the uppercase versions of the original elements, which are then collected into a new list upperCaseAbc. The forEach operation then prints ABC.
NEW QUESTION # 72
Given:
java
LocalDate localDate = LocalDate.of(2020, 8, 8);
Date date = java.sql.Date.valueOf(localDate);
DateFormat formatter = new SimpleDateFormat(/* pattern */);
String output = formatter.format(date);
System.out.println(output);
It's known that the given code prints out "August 08".
Which of the following should be inserted as the pattern?
- A. MM dd
- B. MMMM dd
- C. MMM dd
- D. MM d
Answer: B
Explanation:
To achieve the output "August 08", the SimpleDateFormat pattern must format the month in its full textual form and the day as a two-digit number.
* Pattern Analysis:
* MMMM: Represents the full name of the month (e.g., "August").
* dd: Represents the day of the month as a two-digit number, with leading zeros if necessary (e.g.,
"08").
Therefore, the correct pattern to produce the desired output is MMMM dd.
* Option Evaluations:
* A. MM d: Formats the month as a two-digit number and the day as a single or two-digit number without leading zeros. For example, "08 8".
* B. MM dd: Formats the month and day both as two-digit numbers. For example, "08 08".
* C. MMMM dd: Formats the month as its full name and the day as a two-digit number. For example, "August 08".
* D. MMM dd: Formats the month as its abbreviated name and the day as a two-digit number. For example, "Aug 08".
Thus, option C (MMMM dd) is the correct choice to match the output "August 08".
NEW QUESTION # 73
......
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