Oracle 1z0-830 Java SE 21 Developer Professional Exam Practice Test

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.

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

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, 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.

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".

Understanding post++ (Post-increment)

post++uses the value first, then increments it.

postResult = post++ + 10;

post starts as 5.

post++ returns 5, then post is incremented to 6.

postResult = 5 + 10 = 15.

Final value of post after this line is 6.

Understanding ++pre (Pre-increment)

++preincrements the value first, then uses it.

preResult = ++pre + 10;

pre starts as 5.

++pre increments pre to 6, then returns 6.

preResult = 6 + 10 = 16.

Final value of pre after this line is 6.

Thus, the final output is:

yaml

postResult: 15, preResult: 16, Final value of post: 6, Final value of pre: 6

References:

Java SE 21 - Operators and Expressions

Java SE 21 - Arithmetic Operators

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 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}

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();

Alternatively, you can specify the variable type explicitly:

java

Map

contentReference[oaicite:0]{index=0}

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.

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

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.

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].

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 abc = List.of("a", "b", "c");

List 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.

In Java's java.nio.file package, the Path class represents a file path in a file system. The Paths.get(String first, String... more) method is used to create a Path instance by converting a path string or URI.

In the provided code snippet, the Path object path is created with the string "C:\\home\\joe\\foo". This represents an absolute path on a Windows system.

The getName(int index) method of the Path class returns a name element of the path as a Path object. The index is zero-based, where index 0 corresponds to the first element in the path's name sequence. It's important to note that the root component (e.g., "C:\" on Windows) is not considered a name element and is not included in this sequence.

Therefore, for the path "C:\\home\\joe\\foo":

Root Component:"C:\"

Name Elements:

Index 0: "home"

Index 1: "joe"

Index 2: "foo"

When path.getName(0) is called, it returns the first name element, which is "home". Thus, the output of the System.out.println statement is home.

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

In this scenario, we have a Person class that extends ListResourceBundle and a properties file named Person.properties. Both define a resource with the key "name" but with different values:

Person class (ListResourceBundle):Defines the key "name" with the value "Jeanne".

Person.properties file:Defines the key "name" with the value "James".

When the ResourceBundle.getBundle("Person") method is called, the Java runtime searches for a resource bundle with the base name "Person". The search order is as follows:

Class-Based Resource Bundle:The runtime first looks for a class named Person (i.e., Person.class).

Properties File Resource Bundle:If the class is not found, it then looks for a properties file named Person.properties.

In this case, since the Person class is present and accessible, the runtime will load the Person class as the resource bundle. Therefore, the getBundle method returns an instance of the Person class.

Subsequently, when bundle.getString("name") is called, it retrieves the value associated with the key "name" from the Person class, which is "Jeanne".

Thus, the output of the program is:

nginx

Jeanne

(BufferedWriter writer = new BufferedWriter("file.txt") is incorrect.)

Theincorrect statementisoption Bbecause BufferedWriterdoes nothave a constructor that accepts a String (file name) directly. The correct way to use BufferedWriter is to wrap it around a FileWriter, like this:

java

try (BufferedWriter writer = new BufferedWriter(new FileWriter("file.txt"))) {

writer.write("Hello");

}

Evaluation of Other Options:

Option A (Files.write)→ Correct

Uses Files.write() to write bytes to a file.

Efficient and concise method for writing small text files.

Option C (FileOutputStream)→ Correct

Uses a FileOutputStream to write raw bytes to a file.

Works for both text and binary data.

Option D (PrintWriter)→ Correct

Uses PrintWriter for formatted text output.

Option F (FileWriter)→ Correct

Uses FileWriter to write text data.

Option E (None of the suggestions)→ Incorrect becauseoption Bis incorrect.

In this code, a Stream containing the elements "a" and "b" is processed using the collect method. The collect method is a terminal operation that performs a mutable reduction on the elements of the stream using a Collector. In this case, custom implementations for the supplier, accumulator, and combiner are provided.

Components of the collect Method:

Supplier:

() -> new StringBuilder("c")

This supplier creates a new StringBuilder initialized with the string "c".

Accumulator:

StringBuilder::append

This accumulator appends each element of the stream to the StringBuilder.

Combiner:

(a, b) -> b.append(a)

This combiner is used in parallel stream operations to merge two StringBuilder instances. It appends the contents of a to b.

Execution Flow:

Stream Elements:"a", "b"

Initial StringBuilder:"c"

Accumulation:

The first element "a" is appended to "c", resulting in "ca".

The second element "b" is appended to "ca", resulting in "cab".

Combiner:

In this sequential stream, the combiner is not utilized. The combiner is primarily used in parallel streams to merge partial results.

Final Result:

The StringBuilder contains "cab". Therefore, the output of the program is:

nginx

cab

Understanding Multi-line Strings in Java (""" Text Blocks)

Java 13 introducedtext blocks ("""), allowing multi-line stringswithout needing explicit \n for new lines.

In a text block,each line is preserved as it appears in the source code.

Analyzing the Options

Option A: \ (Backslash Continuation)

The backslash (\) at the end of a lineprevents a new line from being added, meaning:

nginx

red green blue

Incorrect.

Option B: \s (Whitespace Escape)

\s represents asingle space,not a new line.

The output would be:

nginx

red green blue

Incorrect.

Option C: \t (Tab Escape)

\t inserts atab, not a new line.

The output would be:

nginx

red green blue

Incorrect.

Option D: Correct Text Block

java

String colors = """

red

green

blue

""";

Thispreserves the new lines, producing:

nginx

red

green

blue

Correct.

Thus, the correct answer is:"String colors = """ red green blue """."

References:

Java SE 21 - Text Blocks

Java SE 21 - String Formatting

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

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

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.

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.

In this code, an ArrayDeque named deque is created, and the integers 1 and 2 are added to it using the offer method. The offer method inserts the specified element at the end of the deque.

State of deque after offers:[1, 2]

The peek method retrieves, but does not remove, the head of the deque, returning 1. Therefore, i1 is assigned the value 1.

State of deque after peek:[1, 2]

Value of i1:1

The poll method retrieves and removes the head of the deque, returning 1. Therefore, i2 is assigned the value 1.

State of deque after poll:[2]

Value of i2:1

Another peek operation retrieves the current head of the deque, which is now 2, without removing it. Therefore, i3 is assigned the value 2.

State of deque after second peek:[2]

Value of i3:2

The System.out.println statement then outputs the values of i1, i2, and i3, resulting in 1 1 2.