1z0-830 Exam Dumps - Oracle Java SE Questions and Answers

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

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

(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 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.

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.

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

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.

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

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 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].