1z0-830 Questions and Answers

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.

In this code, we have two interfaces and a class with a main method:

WithDefaultMethod Interface:

Declares a default method print() that outputs "default".

WithStaticMethod Interface:

Extends WithDefaultMethod.

Declares a static method print() that outputs "static".

DefaultAndStaticMethods Class:

Contains the main method, which calls WithStaticMethod.print().

Key Points:

Static Methods in Interfaces:

Static methods in interfaces are not inherited by implementing or extending classes or interfaces. They belong solely to the interface in which they are declared.

Default Methods in Interfaces:

Default methods can be inherited by implementing classes, but they cannot be overridden by static methods in subinterfaces.

Execution Flow:

The main method calls WithStaticMethod.print().

This invokes the static method print() defined in the WithStaticMethod interface, which outputs "static".

Therefore, the program compiles successfully and prints static.

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.

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.

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 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.

1. Why does the compilation fail?

The interface Special contains bar as int bar = 1;.

In Java, all interface fields are implicitly public, static, and final.

This means that bar is a constant (final variable).

The method add() contains bar--, which attempts to modify bar.

Since bar is final, it cannot be modified, causing acompilation error.

2. Correcting the Code

To make the code compile, bar must not be final. One way to fix this:

java

class SpecialImpl implements Special {

int bar = 1;

}

Or modify the add() method:

java

int add() {

return --foo + bar; // No modification of bar

}

Thus, the correct answer is:Compilation fails.

References:

Java SE 21 - Interfaces

Java SE 21 - Final Variables

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.

1. Understanding Supplier Functional Interfaces

The Supplier interface is part of java.util.function and provides valueswithout taking any arguments.

Java also provides primitive specializations of Supplier:

BooleanSupplier→ Returns a boolean. Exists

DoubleSupplier→ Returns a double. Exists

LongSupplier→ Returns a long. Exists

Supplier<T>→ Returns a generic T. Exists

2. What about BiSupplier<T, U, R>?

There is no BiSupplier<T, U, R> in Java.

In Java, suppliers donot take arguments, so abi-supplierdoes not exist.

If you need a function thattakes two arguments and returns a value, use BiFunction.

Thus, the correct answer is:BiSupplier<T, U, R> does not exist.

References:

Java SE 21 - Supplier

Java SE 21 - Functional Interfaces

java.io.Console is used for interactive input from the console.

Existing Methods in java.io.Console

reader() → Returns a Reader object.

readLine() → Reads a line of text from the console.

readLine(String fmt, Object... args) → Reads a formatted line.

readPassword() → Reads a password, returning a char[].

readPassword(String fmt, Object... args) → Reads a formatted password.

read() Does Not Exist

Consoledoes not have a read() method.

If character-by-character reading is required, use:

java

Console console = System.console();

Reader reader = console.reader();

int c = reader.read(); // Reads one character

read() is available inReader, butnot in Console.

Thus, the correct answer is:read() does not exist.

References:

Java SE 21 - Console API

Java SE 21 - Reader API

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

In this program, the following operations are performed:

Paths Initialization:

Path p1 is set to "f1.txt".

Path p2 is set to "f2.txt".

File Move Operation:

Files.move(p1, p2); attempts to move (or rename) f1.txt to f2.txt.

File Delete Operation:

Files.delete(p1); attempts to delete f1.txt.

Analysis:

If f1.txt Does Not Exist:

The Files.move(p1, p2); operation will throw a NoSuchFileException because the source file f1.txt is missing.

If f1.txt Exists and f2.txt Does Not Exist:

The Files.move(p1, p2); operation will successfully rename f1.txt to f2.txt.

Subsequently, the Files.delete(p1); operation will throw a NoSuchFileException because p1 (now f1.txt) no longer exists after the move.

If Both f1.txt and f2.txt Exist:

The Files.move(p1, p2); operation will throw a FileAlreadyExistsException because the target file f2.txt already exists.

If f2.txt Exists While f1.txt Does Not:

Similar to the first scenario, the Files.move(p1, p2); operation will throw a NoSuchFileException due to the absence of f1.txt.

In all possible scenarios, an exception is thrown during the execution of the program.

In this code, there is a type mismatch between the DoubleStream and the Predicate.

DoubleStream: A sequence of primitive double values.

Predicate<Double>: A functional interface that operates on objects of type Double (the wrapper class), not on primitive double values.

The DoubleStream class provides a method anyMatch(DoublePredicate predicate), where DoublePredicate is a functional interface that operates on primitive double values. However, in the code, a Predicate is used instead of a DoublePredicate. This mismatch leads to a compilation error because anyMatch cannot accept a Predicate when working with a DoubleStream.

To correct this, the predicate should be defined as a DoublePredicate to match the primitive double type:

java

DoubleStream doubleStream = DoubleStream.of(3.3, 4, 5.25, 6.66);

DoublePredicate doublePredicate = d -> d < 5;

System.out.println(doubleStream.anyMatch(doublePredicate));

With this correction, the code will compile and print true because there are elements in the stream (e.g., 3.3 and 4.0) that are less than 5.

Understanding LongStream.range(1, 1945).boxed().toList();

LongStream.range(1, 1945) generates a stream of numbersfrom 1 to 1944.

.boxed() converts the primitive long values to Long objects.

.toList() (introduced in Java 16)creates an immutable list.

Understanding Executors.newVirtualThreadPerTaskExecutor()

Java 21 introducedvirtual threadsto improve concurrency.

Executors.newVirtualThreadPerTaskExecutor()creates a new virtual thread per submitted task, allowing highly concurrent execution.

Execution Behavior

cannesFestivalfeatureFilms.stream().limit(25) → Limits the stream to thefirst 25 numbers(1 to 25).

.forEach(film -> executor.submit(() -> System.out.println(film)))

Each film is printed inside a virtual thread.

Virtual threads execute asynchronously, meaning numbers arenot guaranteed to print sequentially.

Output will contain numbers from 1 to 25, but their order is random due to concurrent execution.

Possible Output (Random Order)

python-repl

3

1

5

2

4

7

25

The ordermay differ in each rundue to concurrent execution.

Thus, the correct answer is:"Numbers from 1 to 25 randomly."

References:

Java SE 21 - Virtual Threads

Java SE 21 - Executors.newVirtualThreadPerTaskExecutor()

In this code, a Stream of String elements is created containing the characters "J", "a", "v", and "a". The reduce method is then used with String::concat as the accumulator function.

The reduce method with a single BinaryOperator parameter performs a reduction on the elements of the stream, using an associative accumulation function, and returns an Optional describing the reduced value, if any. In this case, it concatenates the strings in the stream.

Since the stream contains elements, the reduction operation concatenates them to form the string "Java". The result is wrapped in an Optional, resulting in Optional[Java]. The print statement outputs this Optional object, displaying Optional[Java].

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 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).

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].

The Java Platform Module System (JPMS), introduced in Java 9, modularizes the Java platform and applications. Understanding the behavior of named and unnamed modules is crucial.

B. The unnamed module exports all of its packages.

Correct. The unnamed module, which includes all code on the classpath, exports all of its packages. This means that any code can access the public types in these packages. However, the unnamed module cannot be explicitly required by named modules.

C. If a package is defined in both a named module and the unnamed module, then the package in the unnamed module is ignored.

Correct. In cases where a package is present in both a named module and the unnamed module, the version in the named module takes precedence. The package in the unnamed module is ignored to maintain module integrity and avoid conflicts.

F. If a request is made to load a type whose package is not defined in any known module, then the module system will attempt to load it from the classpath.

Correct. When the module system cannot find a requested type in any known module, it defaults to searching the classpath (i.e., the unnamed module) to locate the type.

Incorrect Options:

A. Code in an explicitly named module can access types in the unnamed module.

Incorrect. Named modules cannot access types in the unnamed module. The unnamed module can read from named modules, but the reverse is not allowed to ensure strong encapsulation.

D. We must add a module descriptor to make an application developed using a Java version prior to SE9 run on Java 11.

Incorrect. Adding a module descriptor (module-info.java) is not mandatory for applications developed before Java 9 to run on Java 11. Such applications can run in the unnamed module without modification.

E. The unnamed module can only access packages defined in the unnamed module.

Incorrect. The unnamed module can access all packages exported by all named modules, in addition to its own packages.

Why does the compilation fail?

TheJava switch statement does not support primitive type pattern matchingin switch expressions as of Java 21.

The case pattern case int i -> "integer"; isinvalidbecausepattern matching with primitive types (like int or long) is not yet supported in switch statements.

The error occurs at case int i -> "integer";, leading to acompilation failure.

Correcting the Code

Since myVar is of type Object,autoboxing converts 0 into an Integer.

To make the code compile, we should use Integer instead of int:

java

Object myVar = 0;

String print = switch (myVar) {

case Integer i -> "integer";

case Long l -> "long";

case String s -> "string";

default -> "";

};

System.out.println(print);

Output:

bash

integer

Thus, the correct answer is:Compilation fails.

References:

Java SE 21 - Pattern Matching for switch

Java SE 21 - switch Expressions