Latest [Feb 22, 2026] 1z1-830 Exam Questions – Valid 1z1-830 Dumps Pdf
1z1-830 Practice Test Questions Answers Updated 85 Questions
NEW QUESTION # 16
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. UnsupportedTemporalTypeException
- C. P1Y
UnsupportedTemporalTypeException - D. PT8784H
P1Y
Answer: C
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 # 17
What does the following code print?
java
import java.util.stream.Stream;
public class StreamReduce {
public static void main(String[] args) {
Stream<String> stream = Stream.of("J", "a", "v", "a");
System.out.print(stream.reduce(String::concat));
}
}
- A. Java
- B. Compilation fails
- C. Optional[Java]
- D. null
Answer: C
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].
NEW QUESTION # 18
Given:
java
var frenchCities = new TreeSet<String>();
frenchCities.add("Paris");
frenchCities.add("Marseille");
frenchCities.add("Lyon");
frenchCities.add("Lille");
frenchCities.add("Toulouse");
System.out.println(frenchCities.headSet("Marseille"));
What will be printed?
- A. [Paris, Toulouse]
- B. Compilation fails
- C. [Paris]
- D. [Lille, Lyon]
- E. [Lyon, Lille, Toulouse]
Answer: D
Explanation:
In this code, a TreeSet named frenchCities is created and populated with the following cities: "Paris",
"Marseille", "Lyon", "Lille", and "Toulouse". The TreeSet class in Java stores elements in a sorted order according to their natural ordering, which, for strings, is lexicographical order.
Sorted Order of Elements:
When the elements are added to the TreeSet, they are stored in the following order:
* "Lille"
* "Lyon"
* "Marseille"
* "Paris"
* "Toulouse"
headSet Method:
The headSet(E toElement) method of the TreeSet class returns a view of the portion of this set whose elements are strictly less than toElement. In this case, frenchCities.headSet("Marseille") will return a subset of frenchCities containing all elements that are lexicographically less than "Marseille".
Elements Less Than "Marseille":
From the sorted order, the elements that are less than "Marseille" are:
* "Lille"
* "Lyon"
Therefore, the output of the System.out.println statement will be [Lille, Lyon].
Option Evaluations:
* A. [Paris]: Incorrect. "Paris" is lexicographically greater than "Marseille".
* B. [Paris, Toulouse]: Incorrect. Both "Paris" and "Toulouse" are lexicographically greater than
"Marseille".
* C. [Lille, Lyon]: Correct. These are the elements less than "Marseille".
* D. Compilation fails: Incorrect. The code compiles successfully.
* E. [Lyon, Lille, Toulouse]: Incorrect. "Toulouse" is lexicographically greater than "Marseille".
NEW QUESTION # 19
Given:
java
var ceo = new HashMap<>();
ceo.put("Sundar Pichai", "Google");
ceo.put("Tim Cook", "Apple");
ceo.put("Mark Zuckerberg", "Meta");
ceo.put("Andy Jassy", "Amazon");
Does the code compile?
- A. False
- B. True
Answer: A
Explanation:
In this code, a HashMap is instantiated using the var keyword:
java
var ceo = new HashMap<>();
The diamond operator <> is used without explicit type arguments. While the diamond operatorallows the compiler to infer types in many cases, when using var, the compiler requires explicit type information to infer the variable's type.
Therefore, the code will not compile because the compiler cannot infer the type of the HashMap when both var and the diamond operator are used without explicit type parameters.
To fix this issue, provide explicit type parameters when creating the HashMap:
java
var ceo = new HashMap<String, String>();
Alternatively, you can specify the variable type explicitly:
java
Map<String, String>
contentReference[oaicite:0]{index=0}
NEW QUESTION # 20
Which of the following doesnotexist?
- A. They all exist.
- B. Supplier<T>
- C. LongSupplier
- D. BiSupplier<T, U, R>
- E. DoubleSupplier
- F. BooleanSupplier
Answer: D
Explanation:
1. Understanding Supplier Functional Interfaces
* The Supplier<T> interface is part of java.util.function and provides valueswithout taking any arguments.
* Java also provides primitive specializations of Supplier<T>:
* BooleanSupplier# Returns a boolean. Exists
* DoubleSupplier# Returns a double. Exists
* LongSupplier# Returns a long. Exists
* Supplier<T># Returns a generic T. Exists
2. What about BiSupplier<T, U, R>?
* There is no BiSupplier<T, U, R> in Java.
* In Java, suppliers donot take arguments, so abi-supplierdoes not exist.
* If you need a function thattakes two arguments and returns a value, use BiFunction<T, U, R>.
Thus, the correct answer is:BiSupplier<T, U, R> does not exist.
References:
* Java SE 21 - Supplier<T>
* Java SE 21 - Functional Interfaces
NEW QUESTION # 21
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.run(task1);
- B. execService.run(task2);
- C. execService.submit(task2);
- D. execService.execute(task2);
- E. execService.submit(task1);
- F. execService.execute(task1);
- G. execService.call(task1);
- H. execService.call(task2);
Answer: C,E
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 # 22
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. ABC
- D. Compilation fails.
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 # 23
Which of the following methods of java.util.function.Predicate aredefault methods?
- A. negate()
- B. or(Predicate<? super T> other)
- C. and(Predicate<? super T> other)
- D. not(Predicate<? super T> target)
- E. isEqual(Object targetRef)
- F. test(T t)
Answer: A,B,C
Explanation:
* Understanding java.util.function.Predicate<T>
* The Predicate<T> interface represents a function thattakes an input and returns a boolean(true or false).
* It is often used for filtering operations in functional programming and streams.
* Analyzing the Methods:
* and(Predicate<? super T> other)#Default method
* Combines two predicates usinglogical AND(&&).
java
Predicate<String> startsWithA = s -> s.startsWith("A");
Predicate<String> hasLength3 = s -> s.length() == 3;
Predicate<String> combined = startsWithA.and(hasLength3);
* #isEqual(Object targetRef)#Static method
* Not a default method, because it doesnot operate on an instance.
java
Predicate<String> isEqualToHello = Predicate.isEqual("Hello");
* negate()#Default method
* Negates a predicate (! operator).
java
Predicate<String> notEmpty = s -> !s.isEmpty();
Predicate<String> isEmpty = notEmpty.negate();
* #not(Predicate<? super T> target)#Static method (introduced in Java 11)
* Not a default method, since it is static.
* or(Predicate<? super T> other)#Default method
* Combines two predicates usinglogical OR(||).
* #test(T t)#Abstract method
* Not a default method, because every predicatemust implement this method.
Thus, the correct answers are:and(Predicate<? super T> other), negate(), or(Predicate<? super T> other) References:
* Java SE 21 - Predicate Interface
* Java SE 21 - Functional Interfaces
NEW QUESTION # 24
Given:
java
var hauteCouture = new String[]{ "Chanel", "Dior", "Louis Vuitton" };
var i = 0;
do {
System.out.print(hauteCouture[i] + " ");
} while (i++ > 0);
What is printed?
- A. Chanel Dior Louis Vuitton
- B. Compilation fails.
- C. Chanel
- D. An ArrayIndexOutOfBoundsException is thrown at runtime.
Answer: C
Explanation:
* Understanding the do-while Loop
* The do-while loopexecutes at least oncebefore checking the condition.
* The condition i++ > 0 increments iafterchecking.
* Step-by-Step Execution
* Iteration 1:
* i = 0
* Prints: "Chanel"
* i++ updates i to 1
* Condition 1 > 0is true, so the loop exits.
* Why Doesn't the Loop Continue?
* Since i starts at 0, the conditioni++ > 0 is false after the first iteration.
* The loopexits immediately after printing "Chanel".
* Final Output
nginx
Chanel
Thus, the correct answer is:Chanel
References:
* Java SE 21 - do-while Loop
* Java SE 21 - Post-Increment Behavior
NEW QUESTION # 25
Which of the following statements oflocal variables declared with varareinvalid?(Choose 4)
- A. var b = 2, c = 3.0;
- B. var h = (g = 7);
- C. var a = 1;(Valid: var correctly infers int)
- D. var d[] = new int[4];
- E. var f = { 6 };
- F. var e;
Answer: A,D,E,F
Explanation:
1. Valid Use Cases of var
* var is alocal variable type inferencefeature.
* The compilerinfers the type from the assigned value.
* Example of valid use:
java
var a = 10; // Type inferred as int
var str = "Hello"; // Type inferred as String
2. Analyzing the Given Statements
Statement
Valid/Invalid
Reason
var a = 1;
Valid
Type inferred as int.
var b = 2, c = 3.0;
#Invalid
var doesnot allow multiple declarationsin one statement.
var d[] = new int[4];
#Invalid
Array brackets []are not allowedwith var.
var e;
#Invalid
varrequires an initializer(cannot be declared without assignment).
var f = { 6 };
#Invalid
{ 6 } is anarray initializer, which must have an explicit type.
var h = (g = 7);
Valid
g is assigned 7, and h gets its value.
Thus, the correct answers are:B, C, D, E
References:
* Java SE 21 - Local Variable Type Inference (var)
* Java SE 21 - var Restrictions
NEW QUESTION # 26
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. CopyEdit{2=Crazy Horse, 3=Paradis Latin, 4=Le Lido, 5=Folies Bergere}
- B. An exception is thrown at runtime.
- C. {}
- D. Compilation fails.
- E. {2=Crazy Horse, 3=Paradis Latin, 4=Le Lido}
Answer: E
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 # 27
Given:
java
String textBlock = """
j \
a \t
v \s
a \
""";
System.out.println(textBlock.length());
What is the output?
- A. 0
- B. 1
- C. 2
- D. 3
Answer: D
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}
NEW QUESTION # 28
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. 5 5 2 3
- D. 1 1 2 2
- E. 1 1 2 3
Answer: E
Explanation:
* Understanding ArrayDeque Behavior
* ArrayDeque<E>is a double-ended queue (deque), working as aFIFO (queue) and LIFO (stack).
* Thedefault behaviorisqueue-like (FIFO)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 thehead(first element)without removal.
makefile
Output: 1
* deque.poll()# Removes and returns thehead.
go
Output: 1, Deque after poll # `[2, 3, 4, 5]`
* deque.pop()#Same as removeFirst(); removes and returns thehead.
perl
Output: 2, Deque after pop # `[3, 4, 5]`
* deque.element()# Returns thehead(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
NEW QUESTION # 29
Given:
java
import java.io.*;
class A implements Serializable {
int number = 1;
}
class B implements Serializable {
int number = 2;
}
public class Test {
public static void main(String[] args) throws Exception {
File file = new File("o.ser");
A a = new A();
var oos = new ObjectOutputStream(new FileOutputStream(file));
oos.writeObject(a);
oos.close();
var ois = new ObjectInputStream(new FileInputStream(file));
B b = (B) ois.readObject();
ois.close();
System.out.println(b.number);
}
}
What is the given program's output?
- A. 0
- B. Compilation fails
- C. NotSerializableException
- D. ClassCastException
- E. 1
Answer: D
Explanation:
In this program, we have two classes, A and B, both implementing the Serializable interface, and a Test class with the main method.
Program Flow:
* Serialization:
* An instance of class A is created and assigned to the variable a.
* An ObjectOutputStream is created to write to the file "o.ser".
* The object a is serialized and written to the file.
* The ObjectOutputStream is closed.
* Deserialization:
* An ObjectInputStream is created to read from the file "o.ser".
* The program attempts to read an object from the file and cast it to an instance of class B.
* The ObjectInputStream is closed.
Analysis:
* Serialization Process:
* The object a is an instance of class A and is serialized into the file "o.ser".
* Deserialization Process:
* When deserializing, the program reads the object from the file and attempts to cast it to class B.
* However, the object in the file is of type A, not B.
* Since A and B are distinct classes with no inheritance relationship, casting an A instance to B is invalid.
Exception Details:
* Attempting to cast an object of type A to type B results in a ClassCastException.
* The exception message would be similar to:
pgsql
Exception in thread "main" java.lang.ClassCastException: class A cannot be cast to class B Conclusion:
The program compiles successfully but throws a ClassCastException at runtime when it attempts to cast the deserialized object to class B.
NEW QUESTION # 30
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. Compilation fails.
- B. Peugeot
- C. Peugeot 807
- D. An exception is thrown at runtime.
Answer: A
Explanation:
In Java,protected memberscan only be accessedwithin the same packageor bysubclasses, 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 acompilation error:
java
Car car = new Car();
car.brand = "Peugeot 807";
* The brand field isprotectedin Car, which means it isnot accessible via an instance of Car outside the vehicule.parent package.
* Even though MiniVan extends Car, itcannotaccess 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 withinthe same packageand tosubclasses, butonly through inheritance, not through a superclass instance declared outside the package.
References:
* Java SE 21 & JDK 21 - Controlling Access to Members of a Class
* Java SE 21 & JDK 21 - Inheritance Rules
NEW QUESTION # 31
A module com.eiffeltower.shop with the related sources in the src directory.
That module requires com.eiffeltower.membership, available in a JAR located in the lib directory.
What is the command to compile the module com.eiffeltower.shop?
- A. css
CopyEdit
javac --module-source-path src -p lib/com.eiffel.membership.jar -d out -m com.eiffeltower.shop - B. css
CopyEdit
javac -path src -p lib/com.eiffel.membership.jar -d out -m com.eiffeltower.shop - C. css
CopyEdit
javac --module-source-path src -p lib/com.eiffel.membership.jar -s out -m com.eiffeltower.shop - D. bash
CopyEdit
javac -source src -p lib/com.eiffel.membership.jar -d out -m com.eiffeltower.shop
Answer: A
Explanation:
Comprehensive and Detailed In-Depth Explanation:
Understanding Java Module Compilation (javac)
Java modules are compiled using the javac command with specific options to specify:
* Where the source files are located (--module-source-path)
* Where required dependencies (external modules) are located (-p / --module-path)
* Where the compiled output should be placed (-d)
Breaking Down the Correct Compilation Command
css
CopyEdit
javac --module-source-path src -p lib/com.eiffel.membership.jar -d out -m com.eiffeltower.shop
* --module-source-path src # Specifies the directory where module sources are located.
* -p lib/com.eiffel.membership.jar # Specifies the module path (JAR dependency in lib).
* -d out # Specifies the output directory for compiled .class files.
* -m com.eiffeltower.shop # Specifies the module to compile (com.eiffeltower.shop).
NEW QUESTION # 32
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 dd
- B. MM d
- C. MMM dd
- D. MMMM dd
Answer: D
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".
NEW QUESTION # 33
......
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