Oracle 1Z0-830 Online Practice Questions and Exam Preparation

Oracle 1Z0-830 Online Questions & Answers

• Question 11:

  Given:

  java

  var array1 = new String[]{ "foo", "bar", "buz" };

  var array2[] = { "foo", "bar", "buz" };

  var array3 = new String[3] { "foo", "bar", "buz" };

  var array4 = { "foo", "bar", "buz" };

  String array5[] = new String[]{ "foo", "bar", "buz" };

  Which arrays compile? (Select 2)

  A. array1
  B. array2
  C. array3
  D. array4
  E. array5
  A. array1
  E. array5

  ---

  Explanation

  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 and JDK 21 - Local Variable Type Inference Java SE 21 and JDK 21 - Arrays

• Question 12:

  What is the output?

  record R(int x) implements AutoCloseable {

  public void close() { System.out.print("close "); }

  }

  try (R r = new R(1)) {

  System.out.print("run ");

  } finally {

  System.out.print("finally ");

  }

  A. run finally
  B. close run finally
  C. run close finally
  D. run finally close
  C. run close finally

  ---

  Explanation

  The try-with-resources block invokes AutoCloseable.close() after the try body completes but before finally. Output order: run close finally.

• Question 13:

  Given:

  java

  interface Calculable {

  long calculate(int i);

  }

  public class Test {

  public static void main(String[] args) {

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

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

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

  }

  Which lines fail to compile?

  A. Line 1 and line 3
  B. Line 2 only
  C. Line 1 only
  D. Line 1 and line 2
  E. Line 2 and line 3
  F. Line 3 only
  G. The program successfully compiles
  G. The program successfully compiles

  ---

  Explanation

  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.

• Question 14:

  What do the following print?

  java

  public class DefaultAndStaticMethods {

  public static void main(String[] args) {

  WithStaticMethod.print();

  }

  }

  interface WithDefaultMethod {

  default void print() {

  System.out.print("default");

  } } interface WithStaticMethod extends WithDefaultMethod { static void print() { System.out.print("static"); } }

  A. nothing
  B. default
  C. Compilation fails
  D. static
  D. static

  ---

  Explanation

  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.

• Question 15:

  Given:

  java

  Integer frenchRevolution = 1789;

  Object o1 = new String("1789");

  Object o2 = frenchRevolution;

  frenchRevolution = null;

  Object o3 = o2.toString();

  System.out.println(o1.equals(o3));

  What is printed?

  A. true
  B. false
  C. A NullPointerException is thrown.
  D. A ClassCastException is thrown.
  E. Compilation fails.
  A. true

  ---

  Explanation

  Understanding Variable Assignments

  java

  Integer frenchRevolution = 1789;

  Object o1 = new String("1789");

  Object o2 = frenchRevolution;

  frenchRevolution = null;

  frenchRevolution is an Integer with value 1789 .

  o1 is a String with value "1789" .

  o2 stores a reference to frenchRevolution , which is an Integer (1789).

  frenchRevolution = null; only nullifies the reference , but o2 still holds the Integer 1789.

  Calling toString() on o2

  java

  Object o3 = o2.toString();

  o2 refers to an Integer (1789).

  Integer.toString() returns the String representation "1789" .

  o3 is assigned "1789" (String).

  Evaluating o1.equals(o3)

  java

  System.out.println(o1.equals(o3));

  o1.equals(o3) isequivalent to:

  java

  "1789".equals("1789")

  Since both are equal strings , the output is:

  arduino

  true

  Thus, the correct answer is:true

  References:

  Java SE 21 - Integer.toString()

  Java SE 21 - String.equals()

• Question 16:

  Given:

  java

  package vehicule.parent;

  public class Car {

  protected String brand = "Peugeot";

  }

  and

  java

  package vehicule.child;

  import vehicule.parent.Car;

  public class MiniVan extends Car {

  public static void main(String[] args) {

  Car car = new Car();

  car.brand = "Peugeot 807";

  System.out.println(car.brand);

  }

  }

  What is printed?

  A. Peugeot
  B. Peugeot 807
  C. An exception is thrown at runtime.
  D. Compilation fails.
  D. Compilation fails.

  ---

  Explanation

  In Java, protected members can only be accessed within the same package or by subclasses , but there is a key restriction:

  A protected member of a superclass is only accessible through inheritance in a subclass but not through an instance of the superclass that is declared outside the package.

  Why does compilation fail?

  In the MiniVan class, the following line causes a compilation error :

  java

  Car car = new Car();

  car.brand = "Peugeot 807";

  The brand field is protected in Car, which means it is not accessible via an instance of Car outside the vehicule.parent package.

  Even though MiniVan extends Car, it cannot access brand using a Car instance (car.brand) because car is declared as an instance of Car, not MiniVan.

  The correct way to access brand inside MiniVan is through inheritance (this.brand or super.brand).

  Corrected Code

  If we change the MiniVan class like this, it will compile and run successfully:

  java

  package vehicule.child;

  import vehicule.parent.Car;

  public class MiniVan extends Car {

  public static void main(String[] args) {

  MiniVan minivan = new MiniVan(); // Access via inheritance

  minivan.brand = "Peugeot 807";

  System.out.println(minivan.brand);

  }

  }

  This would output:

  nginx

  Peugeot 807

  Key Rule from Oracle Java Documentation

  Protected membersof a class are accessible within the same package and to subclasses , but only through inheritance, not through a superclass instance declared outside the package.

  References:

  Java SE 21 and JDK 21 - Controlling Access to Members of a Class

  Java SE 21 and JDK 21 - Inheritance Rules

• Question 17:

  Given:

  java

  var deque = new ArrayDeque<>();

  deque.add(1);

  deque.add(2);

  deque.add(3);

  deque.add(4);

  deque.add(5);

  System.out.print(deque.peek() + " ");

  System.out.print(deque.poll() + " ");

  System.out.print(deque.pop() + " ");

  System.out.print(deque.element() + " ");

  What is printed?

  A. 1 1 1 1
  B. 1 5 5 1
  C. 1 1 2 3
  D. 1 1 2 2
  E. 5 5 2 3
  C. 1 1 2 3

  ---

  Explanation

  Understanding ArrayDeque Behavior

  ArrayDeque is a double-ended queue (deque) , working as a FIFO (queue) and LIFO (stack) .

  The default behavior queue-like (FIFO) is unless explicitly used as a stack.

  Step-by-Step Execution

  java

  var deque = new ArrayDeque<>();

  deque.add(1);

  deque.add(2);

  deque.add(3);

  deque.add(4);

  deque.add(5);

  Deque after additions# [1, 2, 3, 4, 5]

  Operations Breakdown

  deque.peek()# Returns the head (first element) without removal .

  makefile

  Output: 1

  deque.poll()# Removes and returns the head .

  go

  Output: 1, Deque after poll # `[2, 3, 4, 5]`

  deque.pop()# Same as removeFirst() ; removes and returns the head .

  perl

  Output: 2, Deque after pop # `[3, 4, 5]`

  deque.element()# Returns the head (same as peek(), but throws an exception if empty).

  makefile

  Output: 3

  Final Output

  1 1 2 3

  Thus, the correct answer is:1 1 2 3

  References:

  Java SE 21 - ArrayDeque

  Java SE 21 - Queue Operations

• Question 18:

  Given:

  java

  String colors = "red\n" +

  "green\n" +

  "blue\n";

  Which text block can replace the above code?

  A. java String colors = """ red \ green\ blue \ """;
  B. java String colors = """ red \s green\s blue \s """;
  C. java String colors = """ red \t green\t blue \t """;
  D. java String colors = """ red green blue """;
  E. None of the propositions
  D. java String colors = """ red green blue """;

  ---

  Explanation

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

  Java 13 introduced text blocks (""") , allowing multi-line strings without 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 line prevents a new line from being added , meaning:

  nginx

  red green blue

  Incorrect.

  Option B: \s (Whitespace Escape)

  \s represents a single space not a new line , .

  The output would be:

  nginx

  red green blue

  Incorrect.

  Option C: \t (Tab Escape)

  \t inserts a tab , not a new line.

  The output would be:

  nginx

  red green blue

  Incorrect.

  Option D: Correct Text Block

  java

  String colors = """

  red

  green

  blue

  """;

  This preserves 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

• Question 19:

  You are working on a module named perfumery.shop that depends on another module named perfumery.

  provider.

  The perfumery.shop module should also make its package perfumery.shop.eaudeparfum available to other modules.

  Which of the following is the correct file to declare the perfumery.shop module?

  A. File name: module-info.perfumery.shop.java java module perfumery.shop { requires perfumery.provider; exports perfumery.shop.eaudeparfum.*; }
  B. File name: module-info.java java module perfumery.shop { requires perfumery.provider; exports perfumery.shop.eaudeparfum; }
  C. File name: module.java java module shop.perfumery { requires perfumery.provider; exports perfumery.shop.eaudeparfum; }
  B. File name: module-info.java java module perfumery.shop { requires perfumery.provider; exports perfumery.shop.eaudeparfum; }

  ---

  Explanation

  Correct module descriptor file name

  A module declaration must be placed inside a file named module-info.java .

  The incorrect filename module-info.perfumery.shop.java is invalid (Option A).

  The incorrect filename module.java is invalid (Option C).

  Correct module declaration

  The module declaration must match the name of the module (perfumery.shop).

  The requires perfumery.provider; directive specifies that perfumery.shop depends on perfumery provider.

  The exports perfumery.shop.eaudeparfum; statement allows the perfumery.shop.eaudeparfum package to be accessible by other modules .

  The incorrect syntax exports perfumery.shop.eaudeparfum.*; in Option A is invalid , as wildcards (*) are not allowed in module exports.

  Thus, the correct answer is:File name: module-info.java

  References:

  Java SE 21 - Modules Java SE 21 - module-info.java File

• Question 20:

  Given:

  java

  final Stream strings =

  Files.readAllLines(Paths.get("orders.csv"));

  strings.skip(1)

  .limit(2)

  .forEach(System.out::println);

  And that the orders.csv file contains:

  mathematica

  OrderID,Customer,Product,Quantity,Price

  2,Teddy Riner,Mouse,1,15.99

  3,Sebastien Loeb,Monitor,1,199.99

  4,Antoine Griezmann,Headset,3,45.00

  What is printed?

  A. arduino 2,Teddy Riner,Mouse,1,15.99 3,Sebastien Loeb,Monitor,1,199.99 4,Antoine Griezmann,Headset,3,45.00
  B. arduino 2,Teddy Riner,Mouse,1,15.99
  C. arduino 2,Teddy Riner,Mouse,1,15.99 3,Sebastien Loeb,Monitor,1,199.99
  D. An exception is thrown at runtime.
  E. Compilation fails.
  D. An exception is thrown at runtime.
  E. Compilation fails.

  ---

  Explanation

  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, not a Stream .

  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

  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