Oracle 1Z0-830 Online Practice Questions and Exam Preparation

Oracle 1Z0-830 Online Questions & Answers

• Question 31:

  Given:

  java

  String s = " ";

  System.out.print("[" + s.strip());

  s = " hello ";

  System.out.print("," + s.strip());

  s = "h i ";

  System.out.print("," + s.strip() + "]");

  What is printed?

  A. [ ,hello,h i]
  B. [,hello,h i]
  C. [,hello,hi]
  D. [ , hello ,hi ]
  B. [,hello,h i]

  ---

  Explanation

  In this code, the strip() method is used to remove leading and trailing whitespace from strings. The strip() method, introduced in Java 11, is Unicode-aware and removes all leading and trailing characters that are considered whitespace

  according to the Unicode standard.

  docs.oracle.com

  Analysis of Each Statement:

  First Statement:

  java

  String s = " ";

  System.out.print("[" + s.strip());

  The string s contains four spaces.

  Applying s.strip() removes all leading and trailing spaces, resulting in an empty string.

  The output is "[" followed by the empty string, so the printed result is "[".

  Second Statement:

  java

  s = " hello ";

  System.out.print("," + s.strip());

  The string s is now " hello ".

  Applying s.strip() removes all leading and trailing spaces, resulting in "hello".

  The output is "," followed by "hello", so the printed result is ",hello".

  Third Statement:

  java

  s = "h i ";

  System.out.print("," + s.strip() + "]");

  The string s is now "h i ".

  Applying s.strip() removes the trailing spaces, resulting in "h i".

  The output is "," followed by "h i" and then "]", so the printed result is ",h i]".

  Combined Output:

  Combining all parts, the final output is:

  css

  [,hello,h i]

• Question 32:

  What does the following code print?

  java

  import java.util.stream.Stream;

  public class StreamReduce {

  public static void main(String[] args) {

  Stream stream = Stream.of("J", "a", "v", "a");

  System.out.print(stream.reduce(String::concat));

  }

  }

  A. Optional[Java]
  B. Java
  C. null
  D. Compilation fails
  A. Optional[Java]

  ---

  Explanation

  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].

• Question 33:

  Given:

  java

  String textBlock = """

  j \

  a \t

  v \s

  a \

  """;

  System.out.println(textBlock.length());

  What is the output?

  A. 11
  B. 12
  C. 14
  D. 10
  B. 12

  ---

  Explanation

  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}

• Question 34:

  What is the output of the following snippet? (Assume the file exists)

  java

  Path path = Paths.get("C:\\home\\joe\\foo");

  System.out.println(path.getName(0));

  A. C:
  B. IllegalArgumentException
  C. C
  D. Compilation error
  E. home
  E. home

  ---

  Explanation

  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.

• Question 35:

  Given:

  java

  DoubleSummaryStatistics stats1 = new DoubleSummaryStatistics();

  stats1.accept(4.5);

  stats1.accept(6.0);

  DoubleSummaryStatistics stats2 = new DoubleSummaryStatistics();

  stats2.accept(3.0);

  stats2.accept(8.5);

  stats1.combine(stats2);

  System.out.println("Sum: " + stats1.getSum() + ", Max: " + stats1.getMax() + ", Avg: " + stats1.getAverage());

  What is printed?

  A. Sum: 22.0, Max: 8.5, Avg: 5.5
  B. Sum: 22.0, Max: 8.5, Avg: 5.0
  C. An exception is thrown at runtime.
  D. Compilation fails.
  A. Sum: 22.0, Max: 8.5, Avg: 5.5

  ---

  Explanation

  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 and JDK 21 - DoubleSummaryStatistics

  Java SE 21 - Streams and Statistical Operations

• Question 36:

  Given:

  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, int j = 3; i < j; i++ ) {

  System.out.println( lyrics.lines()

  .toList()

  .get( i ) );

  }

  What is printed?

  A. vbnet Quand il me prend dans ses bras Qu'il me parle tout bas Je vois la vie en rose
  B. Nothing
  C. An exception is thrown at runtime.
  D. Compilation fails.
  D. Compilation fails.

  ---

  Explanation

  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 a Stream , splitting the string by line breaks.

  Calling .toList() on a stream converts 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

• Question 37:

  Given:

  java

  public class BoomBoom implements AutoCloseable {

  public static void main(String[] args) {

  try (BoomBoom boomBoom = new BoomBoom()) {

  System.out.print("bim ");

  throw new Exception();

  } catch (Exception e) {

  System.out.print("boom ");

  }

  }

  @Override

  public void close() throws Exception {

  System.out.print("bam ");

  throw new RuntimeException();

  }

  }

  What is printed?

  A. bim boom bam
  B. bim bam boom
  C. bim boom
  D. bim bam followed by an exception
  E. Compilation fails.
  B. bim bam boom

  ---

  Explanation

  Understanding Try-With-Resources (AutoCloseable)

  BoomBoom implements AutoCloseable, meaning its close() method is automatically called at the end of the try block.

  Step-by-Step Execution

  Step 1: Enter Try Block

  java

  try (BoomBoom boomBoom = new BoomBoom()) {

  System.out.print("bim ");

  throw new Exception();

  }

  "bim " is printed.

  An exception (Exception) is thrown , but before it is handled, the close() method is executed .

  Step 2: close() is Called

  java

  @Override

  public void close() throws Exception {

  System.out.print("bam ");

  throw new RuntimeException();

  }

  "bam " is printed.

  A new RuntimeException is thrown, but it doesnot override the existing Exception yet.

  Step 3: Exception Handling

  java

  } catch (Exception e) {

  System.out.print("boom ");

  }

  The catch (Exception e)catches the original Exception from the try block.

  "boom " is printed.

  Final Output

  nginx

  bim bam boom

  The original Exception is caught, not the RuntimeException from close() .

  The RuntimeException from close() is ignored because the catch block is already handling Exception .

  Thus, the correct answer is:bim bam boom

  References: Java SE 21 - Try-With-Resources Java SE 21 - AutoCloseable Interface

• Question 38:

  Given:

  java

  Runnable task1 = () -> System.out.println("Executing Task-1");

  Callable task2 = () -> {

  System.out.println("Executing Task-2");

  return "Task-2 Finish.";

  };

  ExecutorService execService = Executors.newCachedThreadPool();

  // INSERT CODE HERE

  execService.awaitTermination(3, TimeUnit.SECONDS);

  execService.shutdownNow();

  Which of the following statements, inserted in the code above, prints both :

  "Executing Task-2" and "Executing Task-1"?

  A. execService.call(task1);
  B. execService.call(task2);
  C. execService.execute(task1);
  D. execService.execute(task2);
  E. execService.run(task1);
  F. execService.run(task2);
  G. execService.submit(task1);
  H. execService.submit(task2);
  G. execService.submit(task1);
  H. execService.submit(task2);

  ---

  Explanation

  Understanding ExecutorService Methods

  execute(Runnable command)

  Runs the task but only supports Runnable (not Callable).

  #execService.execute(task2); fails because task2 is Callable.

  submit(Runnable task)

  Submits a Runnable task for execution.

  execService.submit(task1); executes "Executing Task-1".

  submit(Callable task)

  Submits a Callable task for execution.

  execService.submit(task2); executes "Executing Task-2".

  call() Does Not Exist in ExecutorService

  #execService.call(task1); and execService.call(task2); are invalid.

  run() Does Not Exist in ExecutorService

  #execService.run(task1); and execService.run(task2); are invalid.

  Correct Code to Print Both Messages:

  java

  execService.submit(task1);

  execService.submit(task2);

  Thus, the correct answer is:execService.submit(task1); execService.submit(task2);

  References:

  Java SE 21 - ExecutorService

  Java SE 21 - Callable and Runnable

• Question 39:

  Given a properties file on the classpath named Person.properties with the content:

  ini

  name=James

  And:

  java

  public class Person extends ListResourceBundle {

  protected Object[][] getContents() {

  return new Object[][]{

  {"name", "Jeanne"}

  };

  }

  }

  And:

  java

  public class Test {

  public static void main(String[] args) {

  ResourceBundle bundle = ResourceBundle.getBundle("Person");

  String name = bundle.getString("name");

  System.out.println(name);

  }

  }

  What is the given program's output?

  A. MissingResourceException
  B. Compilation fails
  C. James
  D. JeanneJames
  E. JamesJeanne
  F. Jeanne
  F. Jeanne

  ---

  Explanation

  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

• Question 40:

  Given:

  java

  public class ThisCalls {

  public ThisCalls() {

  this(true);

  }

  public ThisCalls(boolean flag) {

  this();

  }

  }

  Which statement is correct?

  A. It does not compile.
  B. It throws an exception at runtime.
  C. It compiles.
  A. It does not compile.

  ---

  Explanation

  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.