2025 Best 1z1-830 Exam Preparation Material with New Dumps Questions
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NEW QUESTION # 30
Given:
java
try (FileOutputStream fos = new FileOutputStream("t.tmp");
ObjectOutputStream oos = new ObjectOutputStream(fos)) {
fos.write("Today");
fos.writeObject("Today");
oos.write("Today");
oos.writeObject("Today");
} catch (Exception ex) {
// handle exception
}
Which statement compiles?
- A. fos.writeObject("Today");
- B. fos.write("Today");
- C. oos.write("Today");
- D. oos.writeObject("Today");
Answer: D
Explanation:
In Java, FileOutputStream and ObjectOutputStream are used for writing data to files, but they have different purposes and methods. Let's analyze each statement:
* fos.write("Today");
The FileOutputStream class is designed to write raw byte streams to files. The write method in FileOutputStream expects a parameter of type int or byte[]. Since "Today" is a String, passing it directly to fos.
write("Today"); will cause a compilation error because there is no write method in FileOutputStream that accepts a String parameter.
* fos.writeObject("Today");
The FileOutputStream class does not have a method named writeObject. The writeObject method is specific to ObjectOutputStream. Therefore, attempting to call fos.writeObject("Today"); will result in a compilation error.
* oos.write("Today");
The ObjectOutputStream class is used to write objects to an output stream. However, it does not have a write method that accepts a String parameter. The available write methods in ObjectOutputStream are for writing primitive data types and objects. Therefore, oos.write("Today"); will cause a compilation error.
* oos.writeObject("Today");
The ObjectOutputStream class provides the writeObject method, which is used to serialize objects and write them to the output stream. Since String implements the Serializable interface, "Today" can be serialized.
Therefore, oos.writeObject("Today"); is valid and compiles successfully.
In summary, the only statement that compiles without errors is oos.writeObject("Today");.
References:
* Java SE 21 & JDK 21 - ObjectOutputStream
* Java SE 21 & JDK 21 - FileOutputStream
NEW QUESTION # 31
Which of the following java.io.Console methods doesnotexist?
- A. read()
- B. readLine(String fmt, Object... args)
- C. reader()
- D. readLine()
- E. readPassword(String fmt, Object... args)
- F. readPassword()
Answer: A
Explanation:
* java.io.Console is used for interactive input from the console.
* Existing Methods in java.io.Console
* reader() # Returns a Reader object.
* readLine() # Reads a line of text from the console.
* readLine(String fmt, Object... args) # Reads a formatted line.
* readPassword() # Reads a password, returning a char[].
* readPassword(String fmt, Object... args) # Reads a formatted password.
* read() Does Not Exist
* Consoledoes not have a read() method.
* If character-by-character reading is required, use:
java
Console console = System.console();
Reader reader = console.reader();
int c = reader.read(); // Reads one character
* read() is available inReader, butnot in Console.
Thus, the correct answer is:read() does not exist.
References:
* Java SE 21 - Console API
* Java SE 21 - Reader API
NEW QUESTION # 32
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. ImplementingInterface
- B. WithInstanceField
- C. WithStaticField
- D. ExtendingClass
Answer: A,C
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.
NEW QUESTION # 33
Given:
java
Period p = Period.between(
LocalDate.of(2023, Month.MAY, 4),
LocalDate.of(2024, Month.MAY, 4));
System.out.println(p);
Duration d = Duration.between(
LocalDate.of(2023, Month.MAY, 4),
LocalDate.of(2024, Month.MAY, 4));
System.out.println(d);
What is the output?
- A. P1Y
PT8784H - B. P1Y
UnsupportedTemporalTypeException - C. PT8784H
P1Y - D. UnsupportedTemporalTypeException
Answer: B
Explanation:
In this code, two LocalDate instances are created representing May 4, 2023, and May 4, 2024. The Period.
between() method is used to calculate the period between these two dates, and the Duration.between() method is used to calculate the duration between them.
Period Calculation:
The Period.between() method calculates the amount of time between two LocalDate objects in terms of years, months, and days. In this case, the period between May 4, 2023, and May 4, 2024, is exactly one year.
Therefore, p is P1Y, which stands for a period of one year. Printing p will output P1Y.
Duration Calculation:
The Duration.between() method is intended to calculate the duration between two temporal objects that have time components, such as LocalDateTime or Instant. However, LocalDate represents a date without a time component. Attempting to use Duration.between() with LocalDate instances will result in an UnsupportedTemporalTypeException because Duration requires time-based units, which LocalDate does not support.
Exception Details:
The UnsupportedTemporalTypeException is thrown when an unsupported unit is used. In this case, Duration.
between() internally attempts to access time-based fields (like seconds), which are not supported by LocalDate. This behavior is documented in the Java Bug System underJDK-8170275.
Correct Usage:
To calculate the duration between two dates, including time components, you should use LocalDateTime or Instant. For example:
java
LocalDateTime start = LocalDateTime.of(2023, Month.MAY, 4, 0, 0);
LocalDateTime end = LocalDateTime.of(2024, Month.MAY, 4, 0, 0);
Duration d = Duration.between(start, end);
System.out.println(d); // Outputs: PT8784H
This will correctly calculate the duration as PT8784H, representing 8,784 hours (which is 366 days, accounting for a leap year).
Conclusion:
The output of the given code will be:
pgsql
P1Y
Exception in thread "main" java.time.temporal.UnsupportedTemporalTypeException: Unsupported unit:
Seconds
Therefore, the correct answer is D:
nginx
P1Y
UnsupportedTemporalTypeException
NEW QUESTION # 34
Given:
java
double amount = 42_000.00;
NumberFormat format = NumberFormat.getCompactNumberInstance(Locale.FRANCE, NumberFormat.Style.
SHORT);
System.out.println(format.format(amount));
What is the output?
- A. 0
- B. 42 k
- C. 42 000,00 €
- D. 42000E
Answer: B
Explanation:
In this code, a double variable amount is initialized to 42,000.00. The NumberFormat.
getCompactNumberInstance(Locale.FRANCE, NumberFormat.Style.SHORT) method is used to obtain a compact number formatter for the French locale with the short style. The format method is then called to format the amount.
The compact number formatting is designed to represent numbers in a shorter form, based on the patterns provided for a given locale. In the French locale, the short style represents thousands with a lowercase 'k'.
Therefore, 42,000 is formatted as 42 k.
* Option Evaluations:
* A. 42000E: This format is not standard in the French locale for compact number formatting.
* B. 42 000,00 €: This represents the number as a currency with two decimal places, which is not the compact form.
* C. 42000: This is the plain number without any formatting, which does not match the compact number format.
* D. 42 k: This is the correct compact representation of 42,000 in the French locale with the short style.
Thus, option D (42 k) is the correct output.
NEW QUESTION # 35
Given:
var cabarets = new TreeMap<>();
cabarets.put(1, "Moulin Rouge");
cabarets.put(2, "Crazy Horse");
cabarets.put(3, "Paradis Latin");
cabarets.put(4, "Le Lido");
cabarets.put(5, "Folies Bergere");
System.out.println(cabarets.subMap(2, true, 5, false));
What is printed?
- A. An exception is thrown at runtime.
- B. Compilation fails.
- C. {2=Crazy Horse, 3=Paradis Latin, 4=Le Lido}
- D. {}
- E. CopyEdit{2=Crazy Horse, 3=Paradis Latin, 4=Le Lido, 5=Folies Bergere}
Answer: C
Explanation:
Understanding TreeMap.subMap(fromKey, fromInclusive, toKey, toInclusive)
* TreeMap.subMap(K fromKey, boolean fromInclusive, K toKey, boolean toInclusive) returns aportion of the mapthat falls within the specified key range.
* Thefirst boolean parameter(fromInclusive) determines if the fromKey should be included.
* Thesecond boolean parameter(toInclusive) determines if the toKey should be included.
Given TreeMap Contents
CopyEdit
{1=Moulin Rouge, 2=Crazy Horse, 3=Paradis Latin, 4=Le Lido, 5=Folies Bergere} Applying subMap(2, true, 5, false)
* Includeskey 2 ("Crazy Horse")#(fromInclusive = true)
* Includeskey 3 ("Paradis Latin")#
* Includeskey 4 ("Le Lido")#
* Excludes key 5 ("Folies Bergere")#(toInclusive = false)
Final Output
CopyEdit
{2=Crazy Horse, 3=Paradis Latin, 4=Le Lido}
Thus, the correct answer is:#{2=Crazy Horse, 3=Paradis Latin, 4=Le Lido} References:
* Java SE 21 - TreeMap.subMap()
* Java SE 21 - NavigableMap
NEW 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. Compilation fails.
- B. vbnet
Quand il me prend dans ses bras
Qu'il me parle tout bas
Je vois la vie en rose - C. Nothing
- D. An exception is thrown at runtime.
Answer: A
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 aStream<String>, splitting the string by line breaks.
* Calling .toList() on a streamconverts 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
NEW QUESTION # 37
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 \t
green\t
blue \t
"""; - C. java
String colors = """
red \s
green\s
blue \s
"""; - D. java
String colors = """
red
green
blue
"""; - E. None of the propositions
Answer: D
Explanation:
* 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
NEW QUESTION # 38
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: 16, preResult: 16, Final value of post: 6, Final value of pre: 6
- B. postResult: 15, preResult: 16, Final value of post: 5, Final value of pre: 6
- C. postResult: 16, preResult: 15, Final value of post: 6, Final value of pre: 5
- D. postResult: 15, preResult: 16, Final value of post: 6, Final value of pre: 6
Answer: D
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)
* ++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
NEW QUESTION # 39
Given:
java
Runnable task1 = () -> System.out.println("Executing Task-1");
Callable<String> 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, printsboth:
"Executing Task-2" and "Executing Task-1"?
- A. execService.call(task2);
- B. execService.submit(task1);
- C. execService.run(task2);
- D. execService.execute(task1);
- E. execService.call(task1);
- F. execService.execute(task2);
- G. execService.run(task1);
- H. execService.submit(task2);
Answer: B,H
Explanation:
* Understanding ExecutorService Methods
* execute(Runnable command)
* Runs the task but only supports Runnable (not Callable).
* #execService.execute(task2); fails because task2 is Callable<String>.
* submit(Runnable task)
* Submits a Runnable task for execution.
* execService.submit(task1); executes "Executing Task-1".
* submit(Callable<T> task)
* Submits a Callable<T> 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
NEW QUESTION # 40
Given:
java
Optional o1 = Optional.empty();
Optional o2 = Optional.of(1);
Optional o3 = Stream.of(o1, o2)
.filter(Optional::isPresent)
.findAny()
.flatMap(o -> o);
System.out.println(o3.orElse(2));
What is the given code fragment's output?
- A. An exception is thrown
- B. Compilation fails
- C. Optional.empty
- D. Optional[1]
- E. 0
- F. 1
- G. 2
Answer: G
Explanation:
In this code, two Optional objects are created:
* o1 is an empty Optional.
* o2 is an Optional containing the integer 1.
A stream is created from o1 and o2. The filter method retains only the Optional instances that are present (i.e., non-empty). This results in a stream containing only o2.
The findAny method returns an Optional describing some element of the stream, or an empty Optional if the stream is empty. Since the stream contains o2, findAny returns Optional[Optional[1]].
The flatMap method is then used to flatten this nested Optional. It applies the provided mapping function (o -
> o) to the value, resulting in Optional[1].
Finally, o3.orElse(2) returns the value contained in o3 if it is present; otherwise, it returns 2. Since o3 contains
1, the output is 1.
NEW QUESTION # 41
Given:
java
Optional<String> optionalName = Optional.ofNullable(null);
String bread = optionalName.orElse("Baguette");
System.out.print("bread:" + bread);
String dish = optionalName.orElseGet(() -> "Frog legs");
System.out.print(", dish:" + dish);
try {
String cheese = optionalName.orElseThrow(() -> new Exception());
System.out.println(", cheese:" + cheese);
} catch (Exception exc) {
System.out.println(", no cheese.");
}
What is printed?
- A. bread:bread, dish:dish, cheese.
- B. bread:Baguette, dish:Frog legs, cheese.
- C. bread:Baguette, dish:Frog legs, no cheese.
- D. Compilation fails.
Answer: C
Explanation:
Understanding Optional.ofNullable(null)
* Optional.ofNullable(null); creates an empty Optional (i.e., it contains no value).
* Optional.of(null); would throw a NullPointerException, but ofNullable(null); safely creates an empty Optional.
Execution of orElse, orElseGet, and orElseThrow
* orElse("Baguette")
* Since optionalName is empty, "Baguette" is returned.
* bread = "Baguette"
* Output:"bread:Baguette"
* orElseGet(() -> "Frog legs")
* Since optionalName is empty, "Frog legs" is returned from the lambda expression.
* dish = "Frog legs"
* Output:", dish:Frog legs"
* orElseThrow(() -> new Exception())
* Since optionalName is empty, an exception is thrown.
* The catch block catches this exception and prints ", no cheese.".
Thus, the final output is:
makefile
bread:Baguette, dish:Frog legs, no cheese.
References:
* Java SE 21 & JDK 21 - Optional
* Java SE 21 - Functional Interfaces
NEW QUESTION # 42
Given:
java
public class ExceptionPropagation {
public static void main(String[] args) {
try {
thrower();
System.out.print("Dom Perignon, ");
} catch (Exception e) {
System.out.print("Chablis, ");
} finally {
System.out.print("Saint-Emilion");
}
}
static int thrower() {
try {
int i = 0;
return i / i;
} catch (NumberFormatException e) {
System.out.print("Rose");
return -1;
} finally {
System.out.print("Beaujolais Nouveau, ");
}
}
}
What is printed?
- A. Beaujolais Nouveau, Chablis, Dom Perignon, Saint-Emilion
- B. Beaujolais Nouveau, Chablis, Saint-Emilion
- C. Rose
- D. Saint-Emilion
Answer: B
Explanation:
* Analyzing the thrower() Method Execution
java
int i = 0;
return i / i;
* i / i evaluates to 0 / 0, whichthrows ArithmeticException (/ by zero).
* Since catch (NumberFormatException e) doesnot matchArithmeticException, it is skipped.
* The finally block always executes, printing:
nginx
Beaujolais Nouveau,
* The exceptionpropagates backto main().
* Handling the Exception in main()
java
try {
thrower();
System.out.print("Dom Perignon, ");
} catch (Exception e) {
System.out.print("Chablis, ");
} finally {
System.out.print("Saint-Emilion");
}
* Since thrower() throws ArithmeticException, it is caught by catch (Exception e).
* "Chablis, "is printed.
* Thefinally block always executes, printing "Saint-Emilion".
* Final Output
nginx
Beaujolais Nouveau, Chablis, Saint-Emilion
Thus, the correct answer is:Beaujolais Nouveau, Chablis, Saint-Emilion
References:
* Java SE 21 - Exception Handling
* Java SE 21 - finally Block Execution
NEW QUESTION # 43
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 bam boom
- B. bim boom bam
- C. bim bam followed by an exception
- D. bim boom
- E. Compilation fails.
Answer: A
Explanation:
* Understanding Try-With-Resources (AutoCloseable)
* BoomBoom implements AutoCloseable, meaning its close() method isautomatically calledat 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.
* Anexception (Exception) is thrown, butbefore 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
* Theoriginal Exception is caught, not the RuntimeException from close().
* TheRuntimeException from close() is ignoredbecause thecatch 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
NEW QUESTION # 44
Given:
java
List<String> abc = List.of("a", "b", "c");
abc.stream()
.forEach(x -> {
x = x.toUpperCase();
});
abc.stream()
.forEach(System.out::print);
What is the output?
- A. abc
- B. An exception is thrown.
- C. Compilation fails.
- D. ABC
Answer: A
Explanation:
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<String> abc = List.of("a", "b", "c");
List<String> 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.
NEW QUESTION # 45
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. false
- B. true
- C. A NullPointerException is thrown.
- D. A ClassCastException is thrown.
- E. Compilation fails.
Answer: B
Explanation:
* Understanding Variable Assignments
java
Integer frenchRevolution = 1789;
Object o1 = new String("1789");
Object o2 = frenchRevolution;
frenchRevolution = null;
* frenchRevolution is an Integer with value1789.
* o1 is aString with value "1789".
* o2 storesa 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 theString 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 areequal strings, the output is:
arduino
true
Thus, the correct answer is:true
References:
* Java SE 21 - Integer.toString()
* Java SE 21 - String.equals()
NEW QUESTION # 46
Given:
java
Deque<Integer> deque = new ArrayDeque<>();
deque.offer(1);
deque.offer(2);
var i1 = deque.peek();
var i2 = deque.poll();
var i3 = deque.peek();
System.out.println(i1 + " " + i2 + " " + i3);
What is the output of the given code fragment?
- A. 2 1 2
- B. 1 2 2
- C. 2 1 1
- D. 1 1 2
- E. 1 2 1
- F. An exception is thrown.
- G. 1 1 1
- H. 2 2 2
- I. 2 2 1
Answer: B
Explanation:
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.
NEW QUESTION # 47
Given:
java
public static void main(String[] args) {
try {
throw new IOException();
} catch (IOException e) {
throw new RuntimeException();
} finally {
throw new ArithmeticException();
}
}
What is the output?
- A. Compilation fails
- B. ArithmeticException
- C. IOException
- D. RuntimeException
Answer: B
Explanation:
In this code, the try block throws an IOException. The catch block catches this exception and throws a new RuntimeException. Regardless of exceptions thrown in the try or catch blocks, the finally block is always executed. In this case, the finally block throws an ArithmeticException.
When an exception is thrown in a finally block, it overrides any previous exceptions that were thrown in the try or catch blocks. Therefore, the ArithmeticException thrown in the finally block is the exception that propagates out of the method. As a result, the program terminates with an ArithmeticException.
NEW QUESTION # 48
Which three of the following are correct about the Java module system?
- A. 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.
- B. The unnamed module can only access packages defined in the unnamed module.
- C. If a package is defined in both a named module and the unnamed module, then the package in the unnamed module is ignored.
- D. The unnamed module exports all of its packages.
- E. Code in an explicitly named module can access types in the unnamed module.
- F. We must add a module descriptor to make an application developed using a Java version prior to SE9 run on Java 11.
Answer: A,C,D
Explanation:
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.
NEW QUESTION # 49
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. Compilation fails at line n1.
- C. Nothing
- D. An exception is thrown at runtime.
- E. Compilation fails at line n2.
Answer: B
Explanation:
* Understanding Constructors vs. Methods in Java
* In Java, aconstructormustnot have a return type.
* The followingis NOT a constructorbut aregular 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 isno constructor explicitly defined,Java provides a default no-argument constructor, which does nothing.
* Why Does Compilation Fail?
* void Versailles() is interpreted as amethod,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
NEW QUESTION # 50
Which of the following statements are correct?
- A. You can use 'final' modifier with all kinds of classes
- B. You can use 'private' access modifier with all kinds of classes
- C. You can use 'protected' access modifier with all kinds of classes
- D. None
- E. You can use 'public' access modifier with all kinds of classes
Answer: D
Explanation:
1. private Access Modifier
* The private access modifiercan only be used for inner classes(nested classes).
* Top-level classes cannot be private.
* Example ofinvaliduse:
java
private class MyClass {} // Compilation error
* Example ofvaliduse (for inner class):
java
class Outer {
private class Inner {}
}
2. protected Access Modifier
* Top-level classes cannot be protected.
* protectedonly applies to members (fields, methods, and constructors).
* Example ofinvaliduse:
java
protected class MyClass {} // Compilation error
* Example ofvaliduse (for methods/fields):
java
class Parent {
protected void display() {}
}
3. public Access Modifier
* Atop-level class can be public, butonly one public class per file is allowed.
* Example ofvaliduse:
java
public class MyClass {}
* Example ofinvaliduse:
java
public class A {}
public class B {} // Compilation error: Only one public class per file
4. final Modifier
* finalcan be used with classes, but not all kinds of classes.
* Interfaces cannot be final, because they are meant to be implemented.
* Example ofinvaliduse:
java
final interface MyInterface {} // Compilation error
Thus,none of the statements are fully correct, making the correct answer:None References:
* Java SE 21 - Access Modifiers
* Java SE 21 - Class Modifiers
NEW QUESTION # 51
Which of the followingisn'ta correct way to write a string to a file?
- A. java
try (PrintWriter printWriter = new PrintWriter("file.txt")) {
printWriter.printf("Hello %s", "James");
} - B. java
try (FileOutputStream outputStream = new FileOutputStream("file.txt")) { byte[] strBytes = "Hello".getBytes(); outputStream.write(strBytes);
} - C. None of the suggestions
- D. java
Path path = Paths.get("file.txt");
byte[] strBytes = "Hello".getBytes();
Files.write(path, strBytes); - E. java
try (BufferedWriter writer = new BufferedWriter("file.txt")) {
writer.write("Hello");
} - F. java
try (FileWriter writer = new FileWriter("file.txt")) {
writer.write("Hello");
}
Answer: E
Explanation:
(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.
NEW QUESTION # 52
......
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