2025 1z1-830 exam torrent 1z1-830 Study Guide
Easily pass 1z1-830 Exam with our Dumps & PDF Test Engine
NEW QUESTION # 29
What do the following print?
java
import java.time.Duration;
public class DividedDuration {
public static void main(String[] args) {
var day = Duration.ofDays(2);
System.out.print(day.dividedBy(8));
}
}
- A. PT0H
- B. PT0D
- C. PT6H
- D. Compilation fails
- E. It throws an exception
Answer: C
Explanation:
In this code, a Duration object day is created representing a duration of 2 days using the Duration.ofDays(2) method. The dividedBy(long divisor) method is then called on this Duration object with the argument 8.
The dividedBy(long divisor) method returns a copy of the original Duration divided by the specified value. In this case, dividing 2 days by 8 results in a duration of 0.25 days. In the ISO-8601 duration format used by Java's Duration class, this is represented as PT6H, which stands for a period of 6 hours.
Therefore, the output of the System.out.print statement is PT6H.
NEW QUESTION # 30
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:Baguette, dish:Frog legs, no cheese.
- B. bread:Baguette, dish:Frog legs, cheese.
- C. bread:bread, dish:dish, cheese.
- D. Compilation fails.
Answer: A
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 # 31
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 'public' access modifier with all kinds of classes
- D. None
- E. You can use 'protected' 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 # 32
Given:
java
var now = LocalDate.now();
var format1 = new DateTimeFormatter(ISO_WEEK_DATE);
var format2 = DateTimeFormatter.ISO_WEEK_DATE;
var format3 = new DateFormat(WEEK_OF_YEAR_FIELD);
var format4 = DateFormat.getDateInstance(WEEK_OF_YEAR_FIELD);
System.out.println(now.format(REPLACE_HERE));
Which variable prints 2025-W01-2 (present-day is 12/31/2024)?
- A. format1
- B. format3
- C. format2
- D. format4
Answer: C
Explanation:
In this code, now is assigned the current date using LocalDate.now(). The goal is to format this date to the ISO week date format, which represents dates in the YYYY-'W'WW-E pattern, where:
* YYYY: Week-based year
* 'W': Literal 'W' character
* WW: Week number
* E: Day of the week
Given that the present day is December 31, 2024, this date falls in the first week of the week-based year 2025.
Therefore, the ISO week date representation would be 2025-W01-2, where '2' denotes Tuesday.
Among the provided formatters:
* format1: This line attempts to create a DateTimeFormatter using a constructor, which is incorrect because DateTimeFormatter does not have a public constructor that accepts a pattern directly. This would result in a compilation error.
* format2: This is correctly assigned the predefined DateTimeFormatter.ISO_WEEK_DATE, which formats dates in the ISO week date format.
* format3: This line attempts to create a DateFormat instance using a field, which is incorrect because DateFormat does not have such a constructor. This would result in a compilation error.
* format4: This line attempts to get a DateFormat instance using an integer field, which is incorrect because DateFormat.getDateInstance() does not accept such parameters. This would result in a compilation error.
Therefore, the only correct and applicable formatter is format2. Using format2 in the now.format() method will produce the desired output: 2025-W01-2.
NEW QUESTION # 33
Consider the following methods to load an implementation of MyService using ServiceLoader. Which of the methods are correct? (Choose all that apply)
- A. MyService service = ServiceLoader.services(MyService.class).getFirstInstance();
- B. MyService service = ServiceLoader.load(MyService.class).iterator().next();
- C. MyService service = ServiceLoader.load(MyService.class).findFirst().get();
- D. MyService service = ServiceLoader.getService(MyService.class);
Answer: B,C
Explanation:
The ServiceLoader class in Java is used to load service providers implementing a given service interface. The following methods are evaluated for their correctness in loading an implementation of MyService:
* A. MyService service = ServiceLoader.load(MyService.class).iterator().next(); This method uses the ServiceLoader.load(MyService.class) to create a ServiceLoader instance for MyService.
Calling iterator().next() retrieves the next available service provider. If no providers are available, a NoSuchElementException will be thrown. This approach is correct but requires handling the potential exception if no providers are found.
* B. MyService service = ServiceLoader.load(MyService.class).findFirst().get(); This method utilizes the findFirst() method introduced in Java 9, which returns an Optional describing the first available service provider. Calling get() on the Optional retrieves the service provider if present; otherwise, a NoSuchElementException is thrown. This approach is correct and provides a more concise way to obtain the first service provider.
* C. MyService service = ServiceLoader.getService(MyService.class);
The ServiceLoader class does not have a method named getService. Therefore, this method is incorrect and will result in a compilation error.
* D. MyService service = ServiceLoader.services(MyService.class).getFirstInstance(); The ServiceLoader class does not have a method named services or getFirstInstance. Therefore, this method is incorrect and will result in a compilation error.
In summary, options A and B are correct methods to load an implementation of MyService using ServiceLoader.
NEW QUESTION # 34
Given:
java
var sList = new CopyOnWriteArrayList<Customer>();
Which of the following statements is correct?
- A. The CopyOnWriteArrayList class does not allow null elements.
- B. Element-changing operations on iterators of CopyOnWriteArrayList, such as remove, set, and add, are supported and do not throw UnsupportedOperationException.
- C. The CopyOnWriteArrayList class is a thread-safe variant of ArrayList where all mutative operations are implemented by making a fresh copy of the underlying array.
- D. The CopyOnWriteArrayList class's iterator reflects all additions, removals, or changes to the list since the iterator was created.
- E. The CopyOnWriteArrayList class is not thread-safe and does not prevent interference amongconcurrent threads.
Answer: C
Explanation:
The CopyOnWriteArrayList is a thread-safe variant of ArrayList in which all mutative operations (such as add, set, and remove) are implemented by creating a fresh copy of the underlying array. This design allows for safe iteration over the list without requiring external synchronization, as iterators operate over a snapshot of the array at the time the iterator was created. Consequently, modifications made to the list after the creation of an iterator are not reflected in that iterator.
docs.oracle.com
Evaluation of Options:
* Option A:Correct. This statement accurately describes the behavior of CopyOnWriteArrayList.
* Option B:Incorrect. CopyOnWriteArrayList is thread-safe and is designed to prevent interference among concurrent threads.
* Option C:Incorrect. Iterators of CopyOnWriteArrayList do not reflect additions, removals, or changes made to the list after the iterator was created; they operate on a snapshot of the list's state at the time of their creation.
* Option D:Incorrect. CopyOnWriteArrayList allows null elements.
* Option E:Incorrect. Element-changing operations on iterators, such as remove, set, and add, are not supported in CopyOnWriteArrayList and will throw UnsupportedOperationException.
NEW QUESTION # 35
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. array3
- B. array5
- C. array1
- D. array2
- E. array4
Answer: B,C
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 & JDK 21 - Local Variable Type Inference
* Java SE 21 & JDK 21 - Arrays
NEW QUESTION # 36
Given:
java
CopyOnWriteArrayList<String> list = new CopyOnWriteArrayList<>();
list.add("A");
list.add("B");
list.add("C");
// Writing in one thread
new Thread(() -> {
list.add("D");
System.out.println("Element added: D");
}).start();
// Reading in another thread
new Thread(() -> {
for (String element : list) {
System.out.println("Read element: " + element);
}
}).start();
What is printed?
- A. It throws an exception.
- B. It prints all elements, but changes made during iteration may not be visible.
- C. Compilation fails.
- D. It prints all elements, including changes made during iteration.
Answer: B
Explanation:
* Understanding CopyOnWriteArrayList
* CopyOnWriteArrayList is a thread-safe variant of ArrayList whereall mutative operations (add, set, remove, etc.) create a new copy of the underlying array.
* This meansiterations will not reflect modifications made after the iterator was created.
* Instead of modifying the existing array, a new copy is created for modifications, ensuring that readers always see a consistent snapshot.
* Thread Execution Behavior
* Thread 1 (Writer Thread)adds "D" to the list.
* Thread 2 (Reader Thread)iterates over the list.
* The reader thread gets a snapshot of the listbefore"D" is added.
* The output may look like:
mathematica
Read element: A
Read element: B
Read element: C
Element added: D
* "D" may not appear in the output of the reader threadbecause the iteration occurs on a snapshot before the modification.
* Why doesn't it print all elements including changes?
* Since CopyOnWriteArrayList doesnot allow changes to be visible during iteration, the reader threadwill not see "D"if it started iterating before "D" was added.
Thus, the correct answer is:"It prints all elements, but changes made during iteration may not be visible." References:
* Java SE 21 - CopyOnWriteArrayList
NEW QUESTION # 37
Which of the following isn't a valid option of the jdeps command?
- A. --list-deps
- B. --generate-module-info
- C. --print-module-deps
- D. --generate-open-module
- E. --list-reduced-deps
- F. --check-deps
Answer: F
Explanation:
The jdeps tool is a Java class dependency analyzer that can be used to understand the static dependencies of applications and libraries. It provides several command-line options to customize its behavior.
Valid jdeps Options:
* --generate-open-module: Generates a module declaration (module-info.java) with open directives for the given JAR files or classes.
* --list-deps: Lists the immediate dependencies of the specified classes or JAR files.
* --generate-module-info: Generates a module declaration (module-info.java) for the given JAR files or classes.
* --print-module-deps: Prints the module dependencies of the specified modules or JAR files.
* --list-reduced-deps: Lists the reduced dependencies, showing only the packages that are directly depended upon.
Invalid Option:
* --check-deps: There is no --check-deps option in the jdeps tool.
Conclusion:
Option A (--check-deps) is not a valid option of the jdeps command.
NEW QUESTION # 38
Given:
java
interface A {
default void ma() {
}
}
interface B extends A {
static void mb() {
}
}
interface C extends B {
void ma();
void mc();
}
interface D extends C {
void md();
}
interface E extends D {
default void ma() {
}
default void mb() {
}
default void mc() {
}
}
Which interface can be the target of a lambda expression?
- A. A
- B. B
- C. E
- D. None of the above
- E. C
- F. D
Answer: D
Explanation:
In Java, a lambda expression can be used where a target type is a functional interface. A functional interface is an interface that contains exactly one abstract method. This concept is also known as a Single Abstract Method (SAM) type.
Analyzing each interface:
* Interface A: Contains a single default method ma(). Since default methods are not abstract, A has no abstract methods.
* Interface B: Extends A and adds a static method mb(). Static methods are also not abstract, so B has no abstract methods.
* Interface C: Extends B and declares two abstract methods: ma() (which overrides the default method from A) and mc(). Therefore, C has two abstract methods.
* Interface D: Extends C and adds another abstract method md(). Thus, D has three abstract methods.
* Interface E: Extends D and provides default implementations for ma(), mb(), and mc(). However, it does not provide an implementation for md(), leaving it as the only abstract method in E.
For an interface to be a functional interface, it must have exactly one abstract method. In this case, E has one abstract method (md()), so it qualifies as a functional interface. However, the question asks which interface can be the target of a lambda expression. Since E is a functional interface, it can be the target of a lambda expression.
Therefore, the correct answer is D (E).
NEW QUESTION # 39
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. MMMM dd
- D. MMM dd
Answer: C
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 # 40
Given:
java
List<Long> cannesFestivalfeatureFilms = LongStream.range(1, 1945)
.boxed()
.toList();
try (var executor = Executors.newVirtualThreadPerTaskExecutor()) {
cannesFestivalfeatureFilms.stream()
.limit(25)
.forEach(film -> executor.submit(() -> {
System.out.println(film);
}));
}
What is printed?
- A. An exception is thrown at runtime
- B. Numbers from 1 to 25 randomly
- C. Numbers from 1 to 25 sequentially
- D. Compilation fails
- E. Numbers from 1 to 1945 randomly
Answer: B
Explanation:
* Understanding LongStream.range(1, 1945).boxed().toList();
* LongStream.range(1, 1945) generates a stream of numbersfrom 1 to 1944.
* .boxed() converts the primitive long values to Long objects.
* .toList() (introduced in Java 16)creates an immutable list.
* Understanding Executors.newVirtualThreadPerTaskExecutor()
* Java 21 introducedvirtual threadsto improve concurrency.
* Executors.newVirtualThreadPerTaskExecutor()creates a new virtual thread per submitted task
, allowing highly concurrent execution.
* Execution Behavior
* cannesFestivalfeatureFilms.stream().limit(25) # Limits the stream to thefirst 25 numbers(1 to
25).
* .forEach(film -> executor.submit(() -> System.out.println(film)))
* Each film is printed inside a virtual thread.
* Virtual threads execute asynchronously, meaning numbers arenot guaranteed to print sequentially.
* Output will contain numbers from 1 to 25, but their order is random due to concurrent execution.
* Possible Output (Random Order)
python-repl
3
1
5
2
4
7
25
* The ordermay differ in each rundue to concurrent execution.
Thus, the correct answer is:"Numbers from 1 to 25 randomly."
References:
* Java SE 21 - Virtual Threads
* Java SE 21 - Executors.newVirtualThreadPerTaskExecutor()
NEW QUESTION # 41
Given:
java
StringBuilder result = Stream.of("a", "b")
.collect(
() -> new StringBuilder("c"),
StringBuilder::append,
(a, b) -> b.append(a)
);
System.out.println(result);
What is the output of the given code fragment?
- A. cba
- B. cacb
- C. cbca
- D. bca
- E. bac
- F. abc
- G. acb
Answer: A
Explanation:
In this code, a Stream containing the elements "a" and "b" is processed using the collect method. The collect method is a terminal operation that performs a mutable reduction on the elements of the stream using a Collector. In this case, custom implementations for the supplier, accumulator, and combiner are provided.
Components of the collect Method:
* Supplier:
* () -> new StringBuilder("c")
* This supplier creates a new StringBuilder initialized with the string "c".
* Accumulator:
* StringBuilder::append
* This accumulator appends each element of the stream to the StringBuilder.
* Combiner:
* (a, b) -> b.append(a)
* This combiner is used in parallel stream operations to merge two StringBuilder instances. It appends the contents of a to b.
Execution Flow:
* Stream Elements:"a", "b"
* Initial StringBuilder:"c"
* Accumulation:
* The first element "a" is appended to "c", resulting in "ca".
* The second element "b" is appended to "ca", resulting in "cab".
* Combiner:
* In this sequential stream, the combiner is not utilized. The combiner is primarily used in parallel streams to merge partial results.
Final Result:
The StringBuilder contains "cab". Therefore, the output of the program is:
nginx
cab
NEW QUESTION # 42
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 ,hi ]
- B. [,hello,h i]
- C. [ ,hello,h i]
- D. [,hello,hi]
Answer: B
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]
NEW QUESTION # 43
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. Optional[Java]
- B. Java
- C. Compilation fails
- D. null
Answer: A
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 # 44
Given:
java
public class Test {
class A {
}
static class B {
}
public static void main(String[] args) {
// Insert here
}
}
Which three of the following are valid statements when inserted into the given program?
- A. B b = new Test().new B();
- B. A a = new Test.A();
- C. B b = new Test.B();
- D. A a = new Test().new A();
- E. A a = new A();
- F. B b = new B();
Answer: C,D,F
Explanation:
In the provided code, we have two inner classes within the Test class:
* Class A:
* An inner (non-static) class.
* Instances of A are associated with an instance of the enclosing Test class.
* Class B:
* A static nested class.
* Instances of B are not associated with any instance of the enclosing Test class and can be instantiated without an instance of Test.
Evaluation of Statements:
A: A a = new A();
* Invalid.Since A is a non-static inner class, it requires an instance of the enclosing class Test to be instantiated. Attempting to instantiate A without an instance of Test will result in a compilation error.
B: B b = new Test.B();
* Valid.B is a static nested class and can be instantiated without an instance of Test. This syntax is correct.
C: A a = new Test.A();
* Invalid.Even though A is referenced through Test, it is a non-static inner class and requires an instance of Test for instantiation. This will result in a compilation error.
D: B b = new Test().new B();
* Invalid.While this syntax is used for instantiating non-static inner classes, B is a static nested class and does not require an instance of Test. This will result in a compilation error.
E: B b = new B();
* Valid.Since B is a static nested class, it can be instantiated directly without referencing the enclosing class.
F: A a = new Test().new A();
* Valid.This is the correct syntax for instantiating a non-static inner class. An instance of Test is created, and then an instance of A is created associated with that Test instance.
Therefore, the valid statements are B, E, and F.
NEW QUESTION # 45
Given:
java
List<Integer> integers = List.of(0, 1, 2);
integers.stream()
.peek(System.out::print)
.limit(2)
.forEach(i -> {});
What is the output of the given code fragment?
- A. 01
- B. Nothing
- C. Compilation fails
- D. An exception is thrown
- E. 012
Answer: A
Explanation:
In this code, a list of integers integers is created containing the elements 0, 1, and 2. A stream is then created from this list, and the following operations are performed in sequence:
* peek(System.out::print):
* The peek method is an intermediate operation that allows performing an action on each element as it is encountered in the stream. In this case, System.out::print is used to print each element.
However, since peek is intermediate, the printing occurs only when a terminal operation is executed.
* limit(2):
* The limit method is another intermediate operation that truncates the stream to contain no more than the specified number of elements. Here, it limits the stream to the first 2 elements.
* forEach(i -> {}):
* The forEach method is a terminal operation that performs the given action on each element of the stream. In this case, the action is an empty lambda expression (i -> {}), which does nothing for each element.
The sequence of operations can be visualized as follows:
* Original Stream Elements: 0, 1, 2
* After peek(System.out::print): Elements are printed as they are encountered.
* After limit(2): Stream is truncated to 0, 1.
* After forEach(i -> {}): No additional action; serves to trigger the processing.
Therefore, the output of the code is 01, corresponding to the first two elements of the list being printed due to the peek operation.
NEW QUESTION # 46
Given:
java
var _ = 3;
var $ = 7;
System.out.println(_ + $);
What is printed?
- A. It throws an exception.
- B. _$
- C. 0
- D. Compilation fails.
Answer: D
Explanation:
* The var keyword and identifier rules:
* The var keyword is used for local variable type inference introduced inJava 10.
* However,Java does not allow _ (underscore) as an identifiersinceJava 9.
* If we try to use _ as a variable name, the compiler will throw an error:
pgsql
error: as of release 9, '_' is a keyword, and may not be used as an identifier
* The $ symbol as an identifier:
* The $ characteris a valid identifierin Java.
* However, since _ is not allowed, the codefails to compile before even reaching $.
Thus,the correct answer is "Compilation fails."
References:
* Java SE 21 - var Local Variable Type Inference
* Java SE 9 - Restrictions on _ Identifier
NEW QUESTION # 47
......
1z1-830 PDF Pass Leader, 1z1-830 Latest Real Test: https://www.lead2passexam.com/Oracle/valid-1z1-830-exam-dumps.html
Valid 1z1-830 Test Answers & 1z1-830 Exam PDF: https://drive.google.com/open?id=1y5ruNMuc9Qju1j-z04titQRWhMqA8u5X