1z0-830 Java SE 21 Developer Professional Questions and Answers

1. private Access Modifier

The private access modifiercan only be used for inner classes(nested classes).

Top-level classes cannot be private.

Example ofinvaliduse:

java

private class MyClass {} // Compilation error

Example ofvaliduse (for inner class):

java

class Outer {

private class Inner {}

}

2. protected Access Modifier

Top-level classes cannot be protected.

protectedonly applies to members (fields, methods, and constructors).

Example ofinvaliduse:

java

protected class MyClass {} // Compilation error

Example ofvaliduse (for methods/fields):

java

class Parent {

protected void display() {}

}

3. public Access Modifier

Atop-level class can be public, butonly one public class per file is allowed.

Example ofvaliduse:

java

public class MyClass {}

Example ofinvaliduse:

java

public class A {}

public class B {} // Compilation error: Only one public class per file

4. final Modifier

finalcan be used with classes, but not all kinds of classes.

Interfaces cannot be final, because they are meant to be implemented.

Example ofinvaliduse:

java

final interface MyInterface {} // Compilation error

Thus,none of the statements are fully correct, making the correct answer:None

References:

Java SE 21 - Access Modifiers

Java SE 21 - Class Modifiers

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 Bergère}

Applying subMap(2, true, 5, false)

Includeskey 2 ("Crazy Horse")✅(fromInclusive = true)

Includeskey 3 ("Paradis Latin")✅

Includeskey 4 ("Le Lido")✅

Excludes key 5 ("Folies Bergère")❌(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

In this code, a text block is defined using the """ syntax introduced in Java 13. Text blocks allow for multiline string literals, preserving the format as written in the code.

Text Block Analysis:

The text block is defined as:

java

String textBlock = """

j \

a \t

contentReference[oaicite:0]{index=0}

1. Why Does Compilation Fail?

The error is in this line:

java

final Stream strings = Files.readAllLines(Paths.get("orders.csv"));

Files.readAllLines(Paths.get("orders.csv")) returns a List<String>,not a Stream<String>.

A List cannot be assigned to a Stream.

2. Correcting the Code

The correct way to create a stream from the file:

java

Stream strings = Files.lines(Paths.get("orders.csv"));

This correctly creates a Stream from the file.

3. Expected Output After Fixing

java

Files.lines(Paths.get("orders.csv"))

skip(1) // Skips the header row

limit(2) // Limits to first two data rows

forEach(System.out::println);

Output:

arduino

1,Kylian Mbappé,Keyboard,2,25.50

2,Teddy Riner,Mouse,1,15.99

Thus, the correct answer is:Compilation fails.

References:

Java SE 21 - Files.readAllLines

Java SE 21 - Files.lines

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

In this code, the Calculable interface defines a single abstract method calculate that takes an int parameter and returns a long. The main method contains three lambda expressions assigned to variables c1, c2, and c3 of type Calculable.

Line 1:Calculable c1 = i -> i + 1;

This lambda expression takes an integer i and returns the result of i + 1. Since the expression i + 1 results in an int, and Java allows implicit widening conversion from int to long, this line compiles successfully.

Line 2:Calculable c2 = i -> Long.valueOf(i);

Here, the lambda expression takes an integer i and returns the result of Long.valueOf(i). The Long.valueOf(int i) method returns a Long object. However, Java allows unboxing of the Long object to a long primitive type when necessary. Therefore, this line compiles successfully.

Line 3:Calculable c3 = i -> { throw new ArithmeticException(); };

This lambda expression takes an integer i and throws an ArithmeticException. Since the method calculate has a return type of long, and throwing an exception is a valid way to exit the method without returning a value, this line compiles successfully.

Since all three lines adhere to the method signature defined in the Calculable interface and there are no type mismatches or syntax errors, the program compiles successfully.

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".

Option A (sealed class Figure permits Rectangle {} and final class Rectangle extends Figure {}) - Valid

Why it compiles?

Figure issealed, meaning itmust explicitly declareits subclasses.

Rectangle ispermittedto extend Figure and isdeclared final, meaning itcannot be extended further.

This followsvalid sealed class rules.

Option B (sealed class Figure permits Rectangle {} and public class Rectangle extends Figure {}) -❌Invalid

Why it fails?

Rectangle extends Figure, but it doesnot specify if it is sealed, final, or non-sealed.

Fix:The correct declaration must be one of the following:

java

final class Rectangle extends Figure {} // OR

sealed class Rectangle permits OtherClass {} // OR

non-sealed class Rectangle extends Figure {}

Option C (final sealed class Circle extends Figure {}) -❌Invalid

Why it fails?

A class cannot be both final and sealedat the same time.

sealed meansit must have permitted subclasses, but final meansit cannot be extended.

Fix:Change final sealed to just final:

java

final class Circle extends Figure {}

Option D (public sealed class Figure permits Circle, Rectangle {} with final class Circle and non-sealed class Rectangle) - Valid

Why it compiles?

Figure issealed, meaning it mustdeclare its permitted subclasses(Circle and Rectangle).

Circle is declaredfinal, so itcannot have subclasses.

Rectangle is declarednon-sealed, meaningit can be subclassedfreely.

This correctly followsJava's sealed class rules.

Thus, the correct answers are:A, D

References:

Java SE 21 - Sealed Classes

Java SE 21 - Class Modifiers

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

Understanding java.util.function.Predicate<T>

The Predicate 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 startsWithA = s -> s.startsWith("A");

Predicate hasLength3 = s -> s.length() == 3;

Predicate combined = startsWithA.and(hasLength3);

❌isEqual(Object targetRef)→Static method

Not a default method, because it doesnot operate on an instance.

java

Predicate isEqualToHello = Predicate.isEqual("Hello");

negate()→Default method

Negates a predicate (! operator).

java

Predicate notEmpty = s -> !s.isEmpty();

Predicate 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

In this code, a Duration object day is created representing a duration of 2 days using the Duration.ofDays(2) method. The dividedBy(long divisor) method is then called on this Duration object with the argument 8.

The dividedBy(long divisor) method returns a copy of the original Duration divided by the specified value. In this case, dividing 2 days by 8 results in a duration of 0.25 days. In the ISO-8601 duration format used by Java's Duration class, this is represented as PT6H, which stands for a period of 6 hours.

Therefore, the output of the System.out.print statement is PT6H.

In this code, the try block throws an IOException. The catch block catches this exception and throws a new RuntimeException. Regardless of exceptions thrown in the try or catch blocks, the finally block is always executed. In this case, the finally block throws an ArithmeticException.

When an exception is thrown in a finally block, it overrides any previous exceptions that were thrown in the try or catch blocks. Therefore, the ArithmeticException thrown in the finally block is the exception that propagates out of the method. As a result, the program terminates with an ArithmeticException.

Understanding do-while Execution

A do-while loopexecutes at least oncebefore checking the condition.

++counter < 3 increments counterbeforeevaluating the condition.

Step-by-Step Execution

Iteration 1:counter = 0, print "0", then ++counter becomes 1, condition 1 < 3 istrue.

Iteration 2:counter = 1, print "1", then ++counter becomes 2, condition 2 < 3 istrue.

Iteration 3:counter = 2, print "2", then ++counter becomes 3, condition 3 < 3 isfalse, so loop exits.

Final Output

0 1 2

Thus, the correct answer is:0 1 2

References:

Java SE 21 - Control Flow Statements

Java SE 21 - do-while Loop

Understanding Sealed Classes in Java

Asealed classrestricts which other classes can extend it.

A sealed classmust explicitly declare its permitted subclassesusing the permits keyword.

Subclasses can be declared as:

sealed(restricts further extension).

non-sealed(removes the restriction, allowing unrestricted subclassing).

final(prevents further subclassing).

Analyzing the Given Code

Vehicle is declared as sealed with permits Car, Bike, meaning only Car and Bike can extend it.

Car is declared as non-sealed, which means itis no longer sealedand can have subclasses.

Bike is declared as final, meaningit cannot be subclassed.

Using isSealed() Method

vehicleClass.isSealed() →truebecause Vehicle is explicitly marked as sealed.

carClass.isSealed() →falsebecause Car is marked non-sealed.

bikeClass.isSealed() →falsebecause Bike is final, and a final class isnot considered sealed.

Final Output

csharp

Is Vehicle sealed? true; Is Car sealed? false; Is Bike sealed? false

Thus, the correct answer is:"Is Vehicle sealed? true; Is Car sealed? false; Is Bike sealed? false"

References:

Java SE 21 - Sealed Classes

Java SE 21 - isSealed() Method

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

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

In Java, the Stream API provides several methods to create streams. However, not all approaches are valid.

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

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 Pérignon, ");

} catch (Exception e) {

System.out.print("Chablis, ");

} finally {

System.out.print("Saint-Émilion");

}

Since thrower() throws ArithmeticException, it is caught by catch (Exception e).

"Chablis, "is printed.

Thefinally block always executes, printing "Saint-Émilion".

Final Output

nginx

Beaujolais Nouveau, Chablis, Saint-Émilion

Thus, the correct answer is:Beaujolais Nouveau, Chablis, Saint-Émilion

References:

Java SE 21 - Exception Handling

Java SE 21 - finally Block Execution

In this code, the strip() method is used to remove leading and trailing whitespace from strings. The strip() method, introduced in Java 11, is Unicode-aware and removes all leading and trailing characters that are considered whitespace according to the Unicode standard.

docs.oracle.com

Analysis of Each Statement:

First Statement:

java

String s = " ";

System.out.print("[" + s.strip());

The string s contains four spaces.

Applying s.strip() removes all leading and trailing spaces, resulting in an empty string.

The output is "[" followed by the empty string, so the printed result is "[".

Second Statement:

java

s = " hello ";

System.out.print("," + s.strip());

The string s is now " hello ".

Applying s.strip() removes all leading and trailing spaces, resulting in "hello".

The output is "," followed by "hello", so the printed result is ",hello".

Third Statement:

java

s = "h i ";

System.out.print("," + s.strip() + "]");

The string s is now "h i ".

Applying s.strip() removes the trailing spaces, resulting in "h i".

The output is "," followed by "h i" and then "]", so the printed result is ",h i]".

Combined Output:

Combining all parts, the final output is:

css

[,hello,h i]

Option A (Map<String, ArrayList<String>> authorsMap1 = new HashMap<>();)

❌Compilation Fails

frenchAuthors is declared as List,notArrayList.

The correct way to declare a Map that allows storing List is to use List as the generic type,notArrayList.

Fix:

java

Map> authorsMap1 = new HashMap<>();

authorsMap1.put("FR", frenchAuthors);

Reason:The type ArrayList is more specific than List, and this would cause atype mismatcherror.

Option B (Map<String, ? extends List<String>> authorsMap2 = new HashMap<String, ArrayList<String>>();)

❌Compilation Fails

? extends Listmakes the map read-onlyfor adding new elements.

The line authorsMap2.put("FR", frenchAuthors); causes acompilation errorbecause wildcard (? extends List) prevents modifying the map.

Fix:Remove the wildcard:

java

Map> authorsMap2 = new HashMap<>();

authorsMap2.put("FR", frenchAuthors);

Option C (var authorsMap3 = new HashMap<>();)

Compiles Successfully

The var keyword allows the compiler to infer the type.

However,the inferred type is HashMap<Object, Object>, which may cause issues when retrieving values.

Option D (Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>>();)

Compiles Successfully

Valid declaration:HashMap can be assigned to Map.

Using new HashMap>() with Map> isalloweddue to polymorphism.

Correct syntax:

java

Map> authorsMap4 = new HashMap>();

authorsMap4.put("FR", frenchAuthors);

Option E (Map<String, List<String>> authorsMap5 = new HashMap<String, List<String>>();)

Compiles Successfully

HashMap> isa valid instantiation.

Correct usage:

java

Map> authorsMap5 = new HashMap<>();

authorsMap5.put("FR", frenchAuthors);

Thus, the correct answers are:C, D, E

References:

Java SE 21 - Generics and Type Inference

Java SE 21 - var Keyword

In Java, records are a special kind of class designed to act as transparent carriers for immutabledata. They automatically provide implementations for equals(), hashCode(), and toString(), and their fields are final and private by default.

Option A: ExtendingClass

Analysis: Records in Java implicitly extend java.lang.Record and cannot extend any other class because Java does not support multiple inheritance. Attempting to extend another class, such as Exception, will result in a compilation error.

Conclusion: Does not compile.

Option B: WithInstanceField

Analysis: Records do not allow the declaration of instance fields outside of their components. The declaration of double fuz; is not permitted and will cause a compilation error.

Conclusion: Does not compile.

Option C: ImplementingInterface

Analysis: Records can implement interfaces. In this case, ImplementingInterface implements Cloneable, which is valid.

Conclusion: Compiles successfully.