
[Full-Version] 2025 New 1z1-830 Actual Exam Dumps, Oracle Practice Test
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NEW QUESTION # 33
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. ClassCastException
- C. NotSerializableException
- D. 1
- E. Compilation fails
Answer: B
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 # 34
Given:
java
try (FileOutputStream fos = new FileOutputStream("t.tmp");
ObjectOutputStream oos = new ObjectOutputStream(fos)) {
fos.write("Today");
fos.writeObject("Today");
oos.write("Today");
oos.writeObject("Today");
} catch (Exception ex) {
// handle exception
}
Which statement compiles?
- A. oos.write("Today");
- B. oos.writeObject("Today");
- C. fos.writeObject("Today");
- D. fos.write("Today");
Answer: B
Explanation:
In Java, FileOutputStream and ObjectOutputStream are used for writing data to files, but they have different purposes and methods. Let's analyze each statement:
* fos.write("Today");
The FileOutputStream class is designed to write raw byte streams to files. The write method in FileOutputStream expects a parameter of type int or byte[]. Since "Today" is a String, passing it directly to fos.
write("Today"); will cause a compilation error because there is no write method in FileOutputStream that accepts a String parameter.
* fos.writeObject("Today");
The FileOutputStream class does not have a method named writeObject. The writeObject method is specific to ObjectOutputStream. Therefore, attempting to call fos.writeObject("Today"); will result in a compilation error.
* oos.write("Today");
The ObjectOutputStream class is used to write objects to an output stream. However, it does not have a write method that accepts a String parameter. The available write methods in ObjectOutputStream are for writing primitive data types and objects. Therefore, oos.write("Today"); will cause a compilation error.
* oos.writeObject("Today");
The ObjectOutputStream class provides the writeObject method, which is used to serialize objects and write them to the output stream. Since String implements the Serializable interface, "Today" can be serialized.
Therefore, oos.writeObject("Today"); is valid and compiles successfully.
In summary, the only statement that compiles without errors is oos.writeObject("Today");.
References:
* Java SE 21 & JDK 21 - ObjectOutputStream
* Java SE 21 & JDK 21 - FileOutputStream
NEW QUESTION # 35
Which of the following isn't a valid option of the jdeps command?
- A. --generate-open-module
- B. --list-reduced-deps
- C. --check-deps
- D. --list-deps
- E. --print-module-deps
- F. --generate-module-info
Answer: C
Explanation:
The jdeps tool is a Java class dependency analyzer that can be used to understand the static dependencies of applications and libraries. It provides several command-line options to customize its behavior.
Valid jdeps Options:
* --generate-open-module: Generates a module declaration (module-info.java) with open directives for the given JAR files or classes.
* --list-deps: Lists the immediate dependencies of the specified classes or JAR files.
* --generate-module-info: Generates a module declaration (module-info.java) for the given JAR files or classes.
* --print-module-deps: Prints the module dependencies of the specified modules or JAR files.
* --list-reduced-deps: Lists the reduced dependencies, showing only the packages that are directly depended upon.
Invalid Option:
* --check-deps: There is no --check-deps option in the jdeps tool.
Conclusion:
Option A (--check-deps) is not a valid option of the jdeps command.
NEW QUESTION # 36
Given:
java
public class ExceptionPropagation {
public static void main(String[] args) {
try {
thrower();
System.out.print("Dom Perignon, ");
} catch (Exception e) {
System.out.print("Chablis, ");
} finally {
System.out.print("Saint-Emilion");
}
}
static int thrower() {
try {
int i = 0;
return i / i;
} catch (NumberFormatException e) {
System.out.print("Rose");
return -1;
} finally {
System.out.print("Beaujolais Nouveau, ");
}
}
}
What is printed?
- A. Beaujolais Nouveau, Chablis, Dom Perignon, Saint-Emilion
- B. Saint-Emilion
- C. Rose
- D. Beaujolais Nouveau, Chablis, Saint-Emilion
Answer: D
Explanation:
* Analyzing the thrower() Method Execution
java
int i = 0;
return i / i;
* i / i evaluates to 0 / 0, whichthrows ArithmeticException (/ by zero).
* Since catch (NumberFormatException e) doesnot matchArithmeticException, it is skipped.
* The finally block always executes, printing:
nginx
Beaujolais Nouveau,
* The exceptionpropagates backto main().
* Handling the Exception in main()
java
try {
thrower();
System.out.print("Dom Perignon, ");
} catch (Exception e) {
System.out.print("Chablis, ");
} finally {
System.out.print("Saint-Emilion");
}
* Since thrower() throws ArithmeticException, it is caught by catch (Exception e).
* "Chablis, "is printed.
* Thefinally block always executes, printing "Saint-Emilion".
* Final Output
nginx
Beaujolais Nouveau, Chablis, Saint-Emilion
Thus, the correct answer is:Beaujolais Nouveau, Chablis, Saint-Emilion
References:
* Java SE 21 - Exception Handling
* Java SE 21 - finally Block Execution
NEW QUESTION # 37
Which of the following statements is correct about a final class?
- A. It cannot be extended by any other class.
- B. The final keyword in its declaration must go right before the class keyword.
- C. It must contain at least a final method.
- 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 # 38
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?
- A. An exception is thrown at runtime.
- B. {}
- C. CopyEdit{2=Crazy Horse, 3=Paradis Latin, 4=Le Lido, 5=Folies Bergere}
- D. Compilation fails.
- E. {2=Crazy Horse, 3=Paradis Latin, 4=Le Lido}
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 # 39
Given:
java
public class ThisCalls {
public ThisCalls() {
this(true);
}
public ThisCalls(boolean flag) {
this();
}
}
Which statement is correct?
- A. It compiles.
- B. It throws an exception at runtime.
- C. It does not compile.
Answer: C
Explanation:
In the provided code, the class ThisCalls has two constructors:
* No-Argument Constructor (ThisCalls()):
* This constructor calls the boolean constructor with this(true);.
* Boolean Constructor (ThisCalls(boolean flag)):
* This constructor attempts to call the no-argument constructor with this();.
This setup creates a circular call between the two constructors:
* The no-argument constructor calls the boolean constructor.
* The boolean constructor calls the no-argument constructor.
Such a circular constructor invocation leads to a compile-time error in Java, specifically "recursiveconstructor invocation." The Java Language Specification (JLS) states:
"It is a compile-time error for a constructor to directly or indirectly invoke itself through a series of one or more explicit constructor invocations involving this." Therefore, the code will not compile due to this recursive constructor invocation.
NEW QUESTION # 40
Given:
java
Object input = 42;
String result = switch (input) {
case String s -> "It's a string with value: " + s;
case Double d -> "It's a double with value: " + d;
case Integer i -> "It's an integer with value: " + i;
};
System.out.println(result);
What is printed?
- A. It's a double with value: 42
- B. It's a string with value: 42
- C. It's an integer with value: 42
- D. Compilation fails.
- E. It throws an exception at runtime.
- F. null
Answer: D
Explanation:
* Pattern Matching in switch
* The switch expression introduced inJava 21supportspattern matchingfor different types.
* However,a switch expression must be exhaustive, meaningit must cover all possible cases or provide a default case.
* Why does compilation fail?
* input is an Object, and the switch expression attempts to pattern-match it to String, Double, and Integer.
* If input had been of another type (e.g., Float or Long), there would beno matching case, leading to anon-exhaustive switch.
* Javarequires a default caseto ensure all possible inputs are covered.
* Corrected Code (Adding a default Case)
java
Object input = 42;
String result = switch (input) {
case String s -> "It's a string with value: " + s;
case Double d -> "It's a double with value: " + d;
case Integer i -> "It's an integer with value: " + i;
default -> "Unknown type";
};
System.out.println(result);
* With this change, the codecompiles and runs successfully.
* Output:
vbnet
It's an integer with value: 42
Thus, the correct answer is:Compilation failsdue to a missing default case.
References:
* Java SE 21 - Pattern Matching for switch
* Java SE 21 - switch Expressions
NEW QUESTION # 41
Given:
java
public class Test {
class A {
}
static class B {
}
public static void main(String[] args) {
// Insert here
}
}
Which three of the following are valid statements when inserted into the given program?
- A. A a = new Test.A();
- B. A a = new A();
- C. A a = new Test().new A();
- D. B b = new B();
- E. B b = new Test().new B();
- F. B b = new Test.B();
Answer: C,D,F
Explanation:
In the provided code, we have two inner classes within the Test class:
* Class A:
* An inner (non-static) class.
* Instances of A are associated with an instance of the enclosing Test class.
* Class B:
* A static nested class.
* Instances of B are not associated with any instance of the enclosing Test class and can be instantiated without an instance of Test.
Evaluation of Statements:
A: A a = new A();
* Invalid.Since A is a non-static inner class, it requires an instance of the enclosing class Test to be instantiated. Attempting to instantiate A without an instance of Test will result in a compilation error.
B: B b = new Test.B();
* Valid.B is a static nested class and can be instantiated without an instance of Test. This syntax is correct.
C: A a = new Test.A();
* Invalid.Even though A is referenced through Test, it is a non-static inner class and requires an instance of Test for instantiation. This will result in a compilation error.
D: B b = new Test().new B();
* Invalid.While this syntax is used for instantiating non-static inner classes, B is a static nested class and does not require an instance of Test. This will result in a compilation error.
E: B b = new B();
* Valid.Since B is a static nested class, it can be instantiated directly without referencing the enclosing class.
F: A a = new Test().new A();
* Valid.This is the correct syntax for instantiating a non-static inner class. An instance of Test is created, and then an instance of A is created associated with that Test instance.
Therefore, the valid statements are B, E, and F.
NEW QUESTION # 42
Given:
java
Optional<String> optionalName = Optional.ofNullable(null);
String bread = optionalName.orElse("Baguette");
System.out.print("bread:" + bread);
String dish = optionalName.orElseGet(() -> "Frog legs");
System.out.print(", dish:" + dish);
try {
String cheese = optionalName.orElseThrow(() -> new Exception());
System.out.println(", cheese:" + cheese);
} catch (Exception exc) {
System.out.println(", no cheese.");
}
What is printed?
- A. bread:bread, dish:dish, cheese.
- B. bread:Baguette, dish:Frog legs, cheese.
- C. bread:Baguette, dish:Frog legs, no cheese.
- D. Compilation fails.
Answer: C
Explanation:
Understanding Optional.ofNullable(null)
* Optional.ofNullable(null); creates an empty Optional (i.e., it contains no value).
* Optional.of(null); would throw a NullPointerException, but ofNullable(null); safely creates an empty Optional.
Execution of orElse, orElseGet, and orElseThrow
* orElse("Baguette")
* Since optionalName is empty, "Baguette" is returned.
* bread = "Baguette"
* Output:"bread:Baguette"
* orElseGet(() -> "Frog legs")
* Since optionalName is empty, "Frog legs" is returned from the lambda expression.
* dish = "Frog legs"
* Output:", dish:Frog legs"
* orElseThrow(() -> new Exception())
* Since optionalName is empty, an exception is thrown.
* The catch block catches this exception and prints ", no cheese.".
Thus, the final output is:
makefile
bread:Baguette, dish:Frog legs, no cheese.
References:
* Java SE 21 & JDK 21 - Optional
* Java SE 21 - Functional Interfaces
NEW QUESTION # 43
Given:
java
List<String> l1 = new ArrayList<>(List.of("a", "b"));
List<String> l2 = new ArrayList<>(Collections.singletonList("c"));
Collections.copy(l1, l2);
l2.set(0, "d");
System.out.println(l1);
What is the output of the given code fragment?
- A. An UnsupportedOperationException is thrown
- B. [c, b]
- C. An IndexOutOfBoundsException is thrown
- D. [d]
- E. [d, b]
- F. [a, b]
Answer: B
Explanation:
In this code, two lists l1 and l2 are created and initialized as follows:
* l1 Initialization:
* Created using List.of("a", "b"), which returns an immutable list containing the elements "a" and
"b".
* Wrapped with new ArrayList<>(...) to create a mutable ArrayList containing the same elements.
* l2 Initialization:
* Created using Collections.singletonList("c"), which returns an immutable list containing the single element "c".
* Wrapped with new ArrayList<>(...) to create a mutable ArrayList containing the same element.
State of Lists Before Collections.copy:
* l1: ["a", "b"]
* l2: ["c"]
Collections.copy(l1, l2):
The Collections.copy method copies elements from the source list (l2) into the destination list (l1). The destination list must have at least as many elements as the source list; otherwise, an IndexOutOfBoundsException is thrown.
In this case, l1 has two elements, and l2 has one element, so the copy operation is valid. After copying, the first element of l1 is replaced with the first element of l2:
* l1 after copy: ["c", "b"]
l2.set(0, "d"):
This line sets the first element of l2 to "d".
* l2 after set: ["d"]
Final State of Lists:
* l1: ["c", "b"]
* l2: ["d"]
The System.out.println(l1); statement outputs the current state of l1, which is ["c", "b"]. Therefore, the correct answer is C: [c, b].
NEW QUESTION # 44
Given:
java
DoubleSummaryStatistics stats1 = new DoubleSummaryStatistics();
stats1.accept(4.5);
stats1.accept(6.0);
DoubleSummaryStatistics stats2 = new DoubleSummaryStatistics();
stats2.accept(3.0);
stats2.accept(8.5);
stats1.combine(stats2);
System.out.println("Sum: " + stats1.getSum() + ", Max: " + stats1.getMax() + ", Avg: " + stats1.getAverage()); What is printed?
- A. An exception is thrown at runtime.
- B. Compilation fails.
- C. Sum: 22.0, Max: 8.5, Avg: 5.0
- D. Sum: 22.0, Max: 8.5, Avg: 5.5
Answer: D
Explanation:
The DoubleSummaryStatistics class in Java is part of the java.util package and is used to collect and summarize statistics for a stream of double values. Let's analyze how the methods work:
* Initialization and Data Insertion
* stats1.accept(4.5); # Adds 4.5 to stats1.
* stats1.accept(6.0); # Adds 6.0 to stats1.
* stats2.accept(3.0); # Adds 3.0 to stats2.
* stats2.accept(8.5); # Adds 8.5 to stats2.
* Combining stats1 and stats2
* stats1.combine(stats2); merges stats2 into stats1, resulting in one statistics summary containing all values {4.5, 6.0, 3.0, 8.5}.
* Calculating Output Values
* Sum= 4.5 + 6.0 + 3.0 + 8.5 = 22.0
* Max= 8.5
* Average= (22.0) / 4 = 5.5
Thus, the output is:
yaml
Sum: 22.0, Max: 8.5, Avg: 5.5
References:
* Java SE 21 & JDK 21 - DoubleSummaryStatistics
* Java SE 21 - Streams and Statistical Operations
NEW QUESTION # 45
Given:
java
var now = LocalDate.now();
var format1 = new DateTimeFormatter(ISO_WEEK_DATE);
var format2 = DateTimeFormatter.ISO_WEEK_DATE;
var format3 = new DateFormat(WEEK_OF_YEAR_FIELD);
var format4 = DateFormat.getDateInstance(WEEK_OF_YEAR_FIELD);
System.out.println(now.format(REPLACE_HERE));
Which variable prints 2025-W01-2 (present-day is 12/31/2024)?
- A. format1
- B. format2
- C. format3
- D. format4
Answer: B
Explanation:
In this code, now is assigned the current date using LocalDate.now(). The goal is to format this date to the ISO week date format, which represents dates in the YYYY-'W'WW-E pattern, where:
* YYYY: Week-based year
* 'W': Literal 'W' character
* WW: Week number
* E: Day of the week
Given that the present day is December 31, 2024, this date falls in the first week of the week-based year 2025.
Therefore, the ISO week date representation would be 2025-W01-2, where '2' denotes Tuesday.
Among the provided formatters:
* format1: This line attempts to create a DateTimeFormatter using a constructor, which is incorrect because DateTimeFormatter does not have a public constructor that accepts a pattern directly. This would result in a compilation error.
* format2: This is correctly assigned the predefined DateTimeFormatter.ISO_WEEK_DATE, which formats dates in the ISO week date format.
* format3: This line attempts to create a DateFormat instance using a field, which is incorrect because DateFormat does not have such a constructor. This would result in a compilation error.
* format4: This line attempts to get a DateFormat instance using an integer field, which is incorrect because DateFormat.getDateInstance() does not accept such parameters. This would result in a compilation error.
Therefore, the only correct and applicable formatter is format2. Using format2 in the now.format() method will produce the desired output: 2025-W01-2.
NEW QUESTION # 46
Which of the following methods of java.util.function.Predicate aredefault methods?
- A. or(Predicate<? super T> other)
- B. test(T t)
- C. negate()
- D. not(Predicate<? super T> target)
- E. and(Predicate<? super T> other)
- F. isEqual(Object targetRef)
Answer: A,C,E
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 # 47
Given:
java
var _ = 3;
var $ = 7;
System.out.println(_ + $);
What is printed?
- A. It throws an exception.
- B. _$
- C. Compilation fails.
- D. 0
Answer: C
Explanation:
* The var keyword and identifier rules:
* The var keyword is used for local variable type inference introduced inJava 10.
* However,Java does not allow _ (underscore) as an identifiersinceJava 9.
* If we try to use _ as a variable name, the compiler will throw an error:
pgsql
error: as of release 9, '_' is a keyword, and may not be used as an identifier
* The $ symbol as an identifier:
* The $ characteris a valid identifierin Java.
* However, since _ is not allowed, the codefails to compile before even reaching $.
Thus,the correct answer is "Compilation fails."
References:
* Java SE 21 - var Local Variable Type Inference
* Java SE 9 - Restrictions on _ Identifier
NEW QUESTION # 48
Given:
java
Stream<String> strings = Stream.of("United", "States");
BinaryOperator<String> operator = (s1, s2) -> s1.concat(s2.toUpperCase()); String result = strings.reduce("-", operator); System.out.println(result); What is the output of this code fragment?
- A. United-STATES
- B. UNITED-STATES
- C. -UnitedSTATES
- D. -UNITEDSTATES
- E. -UnitedStates
- F. UnitedStates
- G. United-States
Answer: C
Explanation:
In this code, a Stream of String elements is created containing "United" and "States". A BinaryOperator<String> named operator is defined to concatenate the first string (s1) with the uppercase version of the second string (s2). The reduce method is then used with "-" as the identity value and operator as the accumulator.
The reduce method processes the elements of the stream as follows:
* Initial Identity Value: "-"
* First Iteration:
* Accumulator Operation: "-".concat("United".toUpperCase())
* Result: "-UNITED"
* Second Iteration:
* Accumulator Operation: "-UNITED".concat("States".toUpperCase())
* Result: "-UNITEDSTATES"
Therefore, the final result stored in result is "-UNITEDSTATES", and the output of theSystem.out.println (result); statement is -UNITEDSTATES.
NEW QUESTION # 49
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. Compilation fails
- B. ok the 2024-07-10T07:17:45.523939600
- C. An exception is thrown
- D. ok the 2024-07-10
Answer: A
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 # 50
Given:
java
var sList = new CopyOnWriteArrayList<Customer>();
Which of the following statements is correct?
- A. The CopyOnWriteArrayList class is not thread-safe and does not prevent interference amongconcurrent threads.
- B. The CopyOnWriteArrayList class does not allow null elements.
- C. The CopyOnWriteArrayList class's iterator reflects all additions, removals, or changes to the list since the iterator was created.
- D. The CopyOnWriteArrayList class is a thread-safe variant of ArrayList where all mutative operations are implemented by making a fresh copy of the underlying array.
- E. Element-changing operations on iterators of CopyOnWriteArrayList, such as remove, set, and add, are supported and do not throw UnsupportedOperationException.
Answer: D
Explanation:
The CopyOnWriteArrayList is a thread-safe variant of ArrayList in which all mutative operations (such as add, set, and remove) are implemented by creating a fresh copy of the underlying array. This design allows for safe iteration over the list without requiring external synchronization, as iterators operate over a snapshot of the array at the time the iterator was created. Consequently, modifications made to the list after the creation of an iterator are not reflected in that iterator.
docs.oracle.com
Evaluation of Options:
* Option A:Correct. This statement accurately describes the behavior of CopyOnWriteArrayList.
* Option B:Incorrect. CopyOnWriteArrayList is thread-safe and is designed to prevent interference among concurrent threads.
* Option C:Incorrect. Iterators of CopyOnWriteArrayList do not reflect additions, removals, or changes made to the list after the iterator was created; they operate on a snapshot of the list's state at the time of their creation.
* Option D:Incorrect. CopyOnWriteArrayList allows null elements.
* Option E:Incorrect. Element-changing operations on iterators, such as remove, set, and add, are not supported in CopyOnWriteArrayList and will throw UnsupportedOperationException.
NEW QUESTION # 51
Given:
java
public class Versailles {
int mirrorsCount;
int gardensHectares;
void Versailles() { // n1
this.mirrorsCount = 17;
this.gardensHectares = 800;
System.out.println("Hall of Mirrors has " + mirrorsCount + " mirrors."); System.out.println("The gardens cover " + gardensHectares + " hectares.");
}
public static void main(String[] args) {
var castle = new Versailles(); // n2
}
}
What is printed?
- A. Nothing
- B. An exception is thrown at runtime.
- C. nginx
Hall of Mirrors has 17 mirrors.
The gardens cover 800 hectares. - D. Compilation fails at line n2.
- E. Compilation fails at line n1.
Answer: E
Explanation:
* Understanding Constructors vs. Methods in Java
* In Java, aconstructormustnot have a return type.
* The followingis NOT a constructorbut aregular method:
java
void Versailles() { // This is NOT a constructor!
* Correct way to define a constructor:
java
public Versailles() { // Constructor must not have a return type
* Since there isno constructor explicitly defined,Java provides a default no-argument constructor, which does nothing.
* Why Does Compilation Fail?
* void Versailles() is interpreted as amethod,not a constructor.
* This means the default constructor (which does nothing) is called.
* Since the method Versailles() is never called, the object fields remain uninitialized.
* If the constructor were correctly defined, the output would be:
nginx
Hall of Mirrors has 17 mirrors.
The gardens cover 800 hectares.
* How to Fix It
java
public Versailles() { // Corrected constructor
this.mirrorsCount = 17;
this.gardensHectares = 800;
System.out.println("Hall of Mirrors has " + mirrorsCount + " mirrors."); System.out.println("The gardens cover " + gardensHectares + " hectares.");
}
Thus, the correct answer is:Compilation fails at line n1.
References:
* Java SE 21 - Constructors
* Java SE 21 - Methods vs. Constructors
NEW QUESTION # 52
Given:
java
public class OuterClass {
String outerField = "Outer field";
class InnerClass {
void accessMembers() {
System.out.println(outerField);
}
}
public static void main(String[] args) {
System.out.println("Inner class:");
System.out.println("------------");
OuterClass outerObject = new OuterClass();
InnerClass innerObject = new InnerClass(); // n1
innerObject.accessMembers(); // n2
}
}
What is printed?
- A. Nothing
- B. An exception is thrown at runtime.
- C. Compilation fails at line n2.
- D. Compilation fails at line n1.
- E. markdown
Inner class:
------------
Outer field
Answer: D
Explanation:
* Understanding Inner Classes in Java
* Aninner class (non-static nested class)requires an instance of the outer classbefore it can be instantiated.
* Incorrect instantiationof the inner class at n1:
java
InnerClass innerObject = new InnerClass(); // Compilation error
* Since InnerClass is anon-staticinner class, itmust be created from an instance of OuterClass.
* Correct Way to Instantiate the Inner Class
java
OuterClass outerObject = new OuterClass();
OuterClass.InnerClass innerObject = outerObject.new InnerClass(); // Correct
* Thiscorrectly associatesthe inner class with an instance of OuterClass.
* Why Does Compilation Fail?
* The error occurs atline n1because InnerClass is beinginstantiated incorrectly.
Thus, the correct answer is:Compilation fails at line n1.
References:
* Java SE 21 - Nested and Inner Classes
* Java SE 21 - Accessing Outer Class Members
NEW QUESTION # 53
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[1]
- B. 0
- C. Optional.empty
- D. 1
- E. 2
- F. Compilation fails
- G. An exception is thrown
Answer: B
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 # 54
......
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