Oracle 1Z0-830 Online Practice Questions and Exam Preparation

Oracle 1Z0-830 Online Questions & Answers

• Question 21:

  Given:

  java

  Stream strings = Stream.of("United", "States");

  BinaryOperator operator = (s1, s2) -> s1.concat(s2.toUpperCase());

  String result = strings.reduce("-", operator);

  System.out.println(result);

  What is the output of this code fragment?

  A. United-States
  B. United-STATES
  C. UNITED-STATES
  D. -UnitedStates
  E. -UNITEDSTATES
  F. -UnitedSTATES
  G. UnitedStates
  F. -UnitedSTATES

  ---

  Explanation

  In this code, a Stream of String elements is created containing "United" and "States". A BinaryOperator named operator is defined to concatenate the first string (s1) with the uppercase version of the second string (s2). The reduce

  method is then used with "-" as the identity value and operator as the accumulator.

  The reduce method processes the elements of the stream as follows:

  Initial Identity Value: "-"

  First Iteration:

  Accumulator Operation: "-".concat("United".toUpperCase())

  Result: "-UNITED"

  Second Iteration:

  Accumulator Operation: "-UNITED".concat("States".toUpperCase())

  Result: "-UNITEDSTATES"

  Therefore, the final result stored in result is "-UNITEDSTATES", and the output of theSystem.out.println (result); statement is -UNITEDSTATES.

• Question 22:

  Given:

  java

  List frenchAuthors = new ArrayList<>();

  frenchAuthors.add("Victor Hugo");

  frenchAuthors.add("Gustave Flaubert");

  Which compiles?

  A. Map authorsMap1 = new HashMap(); java authorsMap1.put("FR", frenchAuthors);
  B. Map authorsMap2 = new HashMap (); java authorsMap2.put("FR", frenchAuthors);
  C. var authorsMap3 = new HashMap(); java authorsMap3.put("FR", frenchAuthors);
  D. Map authorsMap4 = new HashMap(); java authorsMap4.put("FR", frenchAuthors);
  E. Map authorsMap5 = new HashMap(); java authorsMap5.put("FR", frenchAuthors);
  C. var authorsMap3 = new HashMap(); java authorsMap3.put("FR", frenchAuthors);
  D. Map authorsMap4 = new HashMap(); java authorsMap4.put("FR", frenchAuthors);
  E. Map authorsMap5 = new HashMap(); java authorsMap5.put("FR", frenchAuthors);

  ---

  Explanation

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

  #Compilation Fails

  frenchAuthors is declared as List, not ArrayList.

  The correct way to declare a Map that allows storing List is to use List as the generic type, not ArrayList.

  Fix:

  java

  Map> authorsMap1 = new HashMap<>();

  authorsMap1.put("FR", frenchAuthors);

  Reason:The type ArrayList is more specific than List, and this would cause a type mismatch error.

  Option B (Map> authorsMap2 = new HashMap>();)

  #Compilation Fails

  extends List makes the map read-only for adding new elements.

  The line authorsMap2.put("FR", frenchAuthors); causes a compilation error because 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 , which may cause issues when retrieving values.

  Option D (Map> authorsMap4 = new HashMap >();)

  Compiles Successfully

  Valid declaration:HashMap can be assigned to Map.

  Using new HashMap>() with Map> isallowed due to polymorphism.

  Correct syntax:

  java

  Map> authorsMap4 = new HashMap>();

  authorsMap4.put("FR", frenchAuthors);

  Option E (Map> authorsMap5 = new HashMap>();)

  Compiles Successfully

  HashMap> is a 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

• Question 23:

  Given:

  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;

  };

  System.out.println(result);

  What is printed?

  A. null
  B. It's a string with value: 42
  C. It's a double with value: 42
  D. It's an integer with value: 42
  E. It throws an exception at runtime.
  F. Compilation fails.
  F. Compilation fails.

  ---

  Explanation

  Pattern Matching in switch

  The switch expression introduced in Java 21 supports pattern matching for different types.

  However, a switch expression must be exhaustive , meaning it 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 be no matching case , leading to a non-exhaustive switch .

  Java requires a default case to 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 code compiles and runs successfully .

  Output:

  vbnet

  It's an integer with value: 42

  Thus, the correct answer is: Compilation fails due to a missing default case.

  References:

  Java SE 21 - Pattern Matching for switch

  Java SE 21 - switch Expressions

• Question 24:

  Given:

  java

  int post = 5;

  int pre = 5;

  int postResult = post++ + 10;

  int preResult = ++pre + 10;

  System.out.println("postResult: " + postResult +

  ", preResult: " + preResult +

  ", Final value of post: " + post + ", Final value of pre: " + pre); What is printed?

  A. postResult: 15, preResult: 16, Final value of post: 6, Final value of pre: 6
  B. postResult: 16, preResult: 16, Final value of post: 6, Final value of pre: 6
  C. postResult: 15, preResult: 16, Final value of post: 5, Final value of pre: 6
  D. postResult: 16, preResult: 15, Final value of post: 6, Final value of pre: 5
  A. postResult: 15, preResult: 16, Final value of post: 6, Final value of pre: 6

  ---

  Explanation

  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)

  ++pre increments 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

• Question 25:

  Given:

  java

  Map map = Map.of("b", 1, "a", 3, "c", 2);

  TreeMap treeMap = new TreeMap<>(map);

  System.out.println(treeMap);

  What is the output of the given code fragment?

  A. {c=2, a=3, b=1}
  B. {b=1, a=3, c=2}
  C. {b=1, c=2, a=3}
  D. {c=1, b=2, a=3}
  E. Compilation fails
  F. {a=3, b=1, c=2}
  G. {a=1, b=2, c=3}
  F. {a=3, b=1, c=2}

  ---

  Explanation

  In this code, a Map named map is created using Map.of with the following key-value pairs:

  "b": 1

  "a": 3

  "c": 2

  The Map.of method returns an immutable map containing these mappings.

  Next, a TreeMap named treeMap is instantiated by passing the map to its constructor:

  java

  TreeMap treeMap = new TreeMap<>(map);

  The TreeMap constructor with a Map parameter creates a new tree map containing the same mappings as the given map, ordered according to the natural ordering of its keys. In Java, the natural ordering for String keys is lexicographical

  order.

  Therefore, the TreeMap will store the entries in the following order:

  "a": 3

  "b": 1

  "c": 2

  When System.out.println(treeMap); is executed, it outputs the TreeMap in its natural order, resulting in:

  r

  {a=3, b=1, c=2}

  Thus, the correct answer is option F: {a=3, b=1, c=2}.

• Question 26:

  Which StringBuilder variable fails to compile?

  java

  public class StringBuilderInstantiations {

  public static void main(String[] args) {

  var stringBuilder1 = new StringBuilder();

  var stringBuilder2 = new StringBuilder(10);

  var stringBuilder3 = new StringBuilder("Java");

  var stringBuilder4 = new StringBuilder(new char[]{'J', 'a', 'v', 'a'});

  }

  }

  A. None of them
  B. stringBuilder4
  C. stringBuilder1
  D. stringBuilder3
  E. stringBuilder2
  B. stringBuilder4

  ---

  Explanation

  In the provided code, four StringBuilder instances are being created using different constructors:

  stringBuilder1: new StringBuilder()

  This constructor creates an empty StringBuilder with an initial capacity of 16 characters.

  stringBuilder2: new StringBuilder(10)

  This constructor creates an empty StringBuilder with a specified initial capacity of 10 characters.

  stringBuilder3: new StringBuilder("Java")

  This constructor creates a StringBuilder initialized to the contents of the specified string "Java".

  stringBuilder4: new StringBuilder(new char[]{'J', 'a', 'v', 'a'})

  This line attempts to create a StringBuilder using a char array. However, the StringBuilder class does not have a constructor that accepts a char array directly. The available constructors are:

  StringBuilder()

  StringBuilder(int capacity)

  StringBuilder(String str)

  StringBuilder(CharSequence seq)

  Since a char array does not implement the CharSequence interface, and there is no constructor that directly accepts a char array, this line will cause a compilation error.

  To initialize a StringBuilder with a char array, you can convert the char array to a String first:

  java

  var stringBuilder4 = new StringBuilder(new String(new char[]{'J', 'a', 'v', 'a'}));

  This approach utilizes the String constructor that accepts a char array, and then passes the resulting String to the StringBuilder constructor.

• Question 27:

  Given:

  java

  LocalDate localDate = LocalDate.of(2020, 8, 8);

  Date date = java.sql.Date.valueOf(localDate);

  DateFormat formatter = new SimpleDateFormat(/* pattern */);

  String output = formatter.format(date);

  System.out.println(output);

  It's known that the given code prints out "August 08".

  Which of the following should be inserted as the pattern?

  A. MM d
  B. MM dd
  C. MMMM dd
  D. MMM dd
  C. MMMM dd

  ---

  Explanation

  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".

• Question 28:

  Which methods compile?

  A. ```java public List
  B. ```java public List
  C. ```java public List
  D. ```java public List
  A. ```java public List
  B. ```java public List

  ---

  Explanation

  In Java generics, wildcards are used to relax the type constraints of generic types. The extends wildcard () denotes an upper bounded wildcard, allowing any type that is a subclass of Type. Conversely, the super wildcard

  () denotes a lower bounded wildcard, allowing any type that is a superclass of Type.

  Option A:

  java

  public List getListSuper() {

  return new ArrayList();

  }

  Here, List represents a list that can hold IOException objects and objects of its supertypes. Since Exception is a superclass of IOException, ArrayList is compatible with List.

  Therefore, this method compiles successfully.

  Option B:

  java

  public List getListExtends() {

  return new ArrayList();

  }

  In this case, List represents a list that can hold objects of IOException and its subclasses. Since FileNotFoundException is a subclass of IOException, ArrayList is compatible with List

  extends IOException>. Thus, this method compiles successfully.

  Option C:

  java

  public List getListExtends() {

  return new ArrayList();

  }

  Here, List expects a list of IOException or its subclasses. However, Exception is a superclass of IOException, not a subclass. Therefore, ArrayList is not compatible with List, and

  this method will not compile.

  Option D:

  java

  public List getListSuper() {

  return new ArrayList();

  }

  In this scenario, List expects a list that can hold IOException objects and objects of its supertypes. Since FileNotFoundException is a subclass of IOException, ArrayList is not compatible with

  List, and this method will not compile.

  Therefore, the methods in options A and B compile successfully, while those in options C and D do not.

• Question 29:

  Given:

  java

  record WithInstanceField(String foo, int bar) {

  double fuz;

  }

  record WithStaticField(String foo, int bar) {

  static double wiz;

  }

  record ExtendingClass(String foo) extends Exception {}

  record ImplementingInterface(String foo) implements Cloneable {}

  Which records compile? (Select 2)

  A. ExtendingClass
  B. WithInstanceField
  C. ImplementingInterface
  D. WithStaticField
  C. ImplementingInterface
  D. WithStaticField

  ---

  Explanation

  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.

• Question 30:

  Given:

  java

  public class Versailles {

  int mirrorsCount;

  int gardensHectares;

  void Versailles() { // n1

  this.mirrorsCount = 17;

  this.gardensHectares = 800;

  System.out.println("Hall of Mirrors has " + mirrorsCount + " mirrors.");

  System.out.println("The gardens cover " + gardensHectares + " hectares.");

  }

  public static void main(String[] args) {

  var castle = new Versailles(); // n2

  }

  }

  What is printed?

  A. nginx Hall of Mirrors has 17 mirrors. The gardens cover 800 hectares.
  B. Nothing
  C. An exception is thrown at runtime.
  D. Compilation fails at line n1.
  E. Compilation fails at line n2.
  D. Compilation fails at line n1.

  ---

  Explanation

  Understanding Constructors vs. Methods in Java

  In Java, a constructor must not have a return type .

  The following is NOT a constructor but a regular method :

  java

  void Versailles() { // This is NOT a constructor!

  Correct way to define a constructor:

  java

  public Versailles() { // Constructor must not have a return type

  Since there is no constructor explicitly defined Java provides a default no-argument , constructor , which does nothing.

  Why Does Compilation Fail?

  void Versailles() is interpreted as a method not a constructor , .

  This means the default constructor (which does nothing) is called.

  Since the method Versailles() is never called, the object fields remain uninitialized.

  If the constructor were correctly defined, the output would be:

  nginx

  Hall of Mirrors has 17 mirrors.

  The gardens cover 800 hectares.

  How to Fix It

  java

  public Versailles() { // Corrected constructor

  this.mirrorsCount = 17;

  this.gardensHectares = 800;

  System.out.println("Hall of Mirrors has " + mirrorsCount + " mirrors.");

  System.out.println("The gardens cover " + gardensHectares + " hectares.");

  }

  Thus, the correct answer is:Compilation fails at line n1.

  References:

  Java SE 21 - Constructors

  Java SE 21 - Methods vs. Constructors