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NEW QUESTION # 19
What do the following print?
java
public class Main {
int instanceVar = staticVar;
static int staticVar = 666;
public static void main(String args[]) {
System.out.printf("%d %d", new Main().instanceVar, staticVar);
}
static {
staticVar = 42;
}
}
- A. Compilation fails
- B. 666 666
- C. 666 42
- D. 42 42
Answer: D
Explanation:
In this code, the class Main contains both an instance variable instanceVar and a static variable staticVar. The sequence of initialization and execution is as follows:
* Static Variable Initialization:
* staticVar is declared and initialized to 666.
* Static Block Execution:
* The static block executes, updating staticVar to 42.
* Instance Variable Initialization:
* When a new instance of Main is created, instanceVar is initialized to the current value of staticVar, which is 42.
* main Method Execution:
* The main method creates a new instance of Main and prints the values of instanceVar and staticVar.
Therefore, the output of the program is 42 42.
NEW QUESTION # 20
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. UnsupportedTemporalTypeException
- B. PT8784H
P1Y - C. P1Y
PT8784H - D. P1Y
UnsupportedTemporalTypeException
Answer: D
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 # 21
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, cheese.
- B. bread:bread, dish:dish, 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 # 22
What do the following print?
java
public class DefaultAndStaticMethods {
public static void main(String[] args) {
WithStaticMethod.print();
}
}
interface WithDefaultMethod {
default void print() {
System.out.print("default");
}
}
interface WithStaticMethod extends WithDefaultMethod {
static void print() {
System.out.print("static");
}
}
- A. default
- B. Compilation fails
- C. static
- D. nothing
Answer: C
Explanation:
In this code, we have two interfaces and a class with a main method:
* WithDefaultMethod Interface:
* Declares a default method print() that outputs "default".
* WithStaticMethod Interface:
* Extends WithDefaultMethod.
* Declares a static method print() that outputs "static".
* DefaultAndStaticMethods Class:
* Contains the main method, which calls WithStaticMethod.print().
Key Points:
* Static Methods in Interfaces:
* Static methods in interfaces are not inherited by implementing or extending classes or interfaces.
They belong solely to the interface in which they are declared.
* Default Methods in Interfaces:
* Default methods can be inherited by implementing classes, but they cannot be overridden by static methods in subinterfaces.
Execution Flow:
* The main method calls WithStaticMethod.print().
* This invokes the static method print() defined in the WithStaticMethod interface, which outputs "static".
Therefore, the program compiles successfully and prints static.
NEW QUESTION # 23
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. Compilation fails
- B. Numbers from 1 to 25 randomly
- C. An exception is thrown at runtime
- D. Numbers from 1 to 1945 randomly
- E. Numbers from 1 to 25 sequentially
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 # 24
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 # 25
Which of the following statements oflocal variables declared with varareinvalid?(Choose 4)
- A. var a = 1;(Valid: var correctly infers int)
- B. var h = (g = 7);
- C. var f = { 6 };
- D. var d[] = new int[4];
- E. var b = 2, c = 3.0;
- F. var e;
Answer: C,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:
java
Map<String, Integer> map = Map.of("b", 1, "a", 3, "c", 2);
TreeMap<String, Integer> treeMap = new TreeMap<>(map);
System.out.println(treeMap);
What is the output of the given code fragment?
- A. {c=2, a=3, b=1}
- B. {c=1, b=2, a=3}
- C. {b=1, a=3, c=2}
- D. Compilation fails
- E. {a=3, b=1, c=2}
- F. {a=1, b=2, c=3}
- G. {b=1, c=2, a=3}
Answer: E
Explanation:
In this code, a Map named map is created using Map.of with the following key-value pairs:
* "b": 1
* "a": 3
* "c": 2
The Map.of method returns an immutable map containing these mappings.
Next, a TreeMap named treeMap is instantiated by passing the map to its constructor:
java
TreeMap<String, Integer> treeMap = new TreeMap<>(map);
The TreeMap constructor with a Map parameter creates a new tree map containing the same mappings as the given map, ordered according to the natural ordering of its keys. In Java, the natural ordering for String keys is lexicographical order.
Therefore, the TreeMap will store the entries in the following order:
* "a": 3
* "b": 1
* "c": 2
When System.out.println(treeMap); is executed, it outputs the TreeMap in its natural order, resulting in:
r
{a=3, b=1, c=2}
Thus, the correct answer is option F: {a=3, b=1, c=2}.
NEW QUESTION # 27
Given:
java
var _ = 3;
var $ = 7;
System.out.println(_ + $);
What is printed?
- A. 0
- B. It throws an exception.
- C. Compilation fails.
- D. _$
Answer: C
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 # 28
Given:
java
Object myVar = 0;
String print = switch (myVar) {
case int i -> "integer";
case long l -> "long";
case String s -> "string";
default -> "";
};
System.out.println(print);
What is printed?
- A. long
- B. string
- C. nothing
- D. It throws an exception at runtime.
- E. Compilation fails.
- F. integer
Answer: E
Explanation:
* Why does the compilation fail?
* TheJava switch statement does not support primitive type pattern matchingin switch expressions as of Java 21.
* The case pattern case int i -> "integer"; isinvalidbecausepattern matching with primitive types (like int or long) is not yet supported in switch statements.
* The error occurs at case int i -> "integer";, leading to acompilation failure.
* Correcting the Code
* Since myVar is of type Object,autoboxing converts 0 into an Integer.
* To make the code compile, we should use Integer instead of int:
java
Object myVar = 0;
String print = switch (myVar) {
case Integer i -> "integer";
case Long l -> "long";
case String s -> "string";
default -> "";
};
System.out.println(print);
* Output:
bash
integer
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - Pattern Matching for switch
* Java SE 21 - switch Expressions
NEW QUESTION # 29
Which of the following doesnotexist?
- A. DoubleSupplier
- B. LongSupplier
- C. Supplier<T>
- D. BiSupplier<T, U, R>
- E. BooleanSupplier
- F. They all exist.
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 # 30
Which of the following can be the body of a lambda expression?
- A. A statement block
- B. Two statements
- C. An expression and a statement
- D. Two expressions
- E. None of the above
Answer: A
Explanation:
In Java, a lambda expression can have two forms for its body:
* Single Expression:A concise form where the body consists of a single expression. The result of this expression is implicitly returned.
Example:
java
(a, b) -> a + b
In this example, (a, b) are the parameters, and a + b is the single expression that adds them together.
* Statement Block:A more detailed form where the body consists of a block of statements enclosed in braces {}. Within this block, you can have multiple statements, and if a return value is expected, you must explicitly use the return statement.
Example:
java
(a, b) -> {
int sum = a + b;
System.out.println("Sum is: " + sum);
return sum;
}
In this example, the lambda body is a statement block that performs multiple actions: it calculates the sum, prints it, and then returns the sum.
Given the options:
* A. Two statements:While a lambda body can contain multiple statements, they must be enclosed within a statement block {}. Simply having two statements without braces is not valid syntax for a lambda expression.
* B. An expression and a statement:Similar to option A, if a lambda body contains more than one element (be it expressions or statements), they need to be enclosed in a statement block.
* C. A statement block:This is correct. A lambda expression can have a body that is a statement block, allowing multiple statements enclosed in braces.
* D. None of the above:This is incorrect since option C is valid.
* E. Two expressions:As with options A and B, multiple expressions must be enclosed in a statement block to form a valid lambda body.
Therefore, the correct answer is C: A statement block.
NEW QUESTION # 31
Given:
java
var counter = 0;
do {
System.out.print(counter + " ");
} while (++counter < 3);
What is printed?
- A. 0 1 2
- B. 1 2 3
- C. Compilation fails.
- D. 0 1 2 3
- E. 1 2 3 4
- F. An exception is thrown.
Answer: A
Explanation:
* Understanding do-while Execution
* A do-while loopexecutes at least oncebefore checking the condition.
* ++counter < 3 increments counterbeforeevaluating the condition.
* Step-by-Step Execution
* Iteration 1:counter = 0, print "0", then ++counter becomes 1, condition 1 < 3 istrue.
* Iteration 2:counter = 1, print "1", then ++counter becomes 2, condition 2 < 3 istrue.
* Iteration 3:counter = 2, print "2", then ++counter becomes 3, condition 3 < 3 isfalse, so loop exits.
* Final Output
0 1 2
Thus, the correct answer is:0 1 2
References:
* Java SE 21 - Control Flow Statements
* Java SE 21 - do-while Loop
NEW QUESTION # 32
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]
- B. [Lille, Lyon]
- C. [Lyon, Lille, Toulouse]
- D. Compilation fails
- E. [Paris, Toulouse]
Answer: B
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 # 33
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 followed by an exception
- C. bim boom
- D. Compilation fails.
- E. bim bam boom
Answer: E
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 # 34
Given:
java
interface Calculable {
long calculate(int i);
}
public class Test {
public static void main(String[] args) {
Calculable c1 = i -> i + 1; // Line 1
Calculable c2 = i -> Long.valueOf(i); // Line 2
Calculable c3 = i -> { throw new ArithmeticException(); }; // Line 3
}
}
Which lines fail to compile?
- A. Line 1 only
- B. Line 1 and line 2
- C. Line 1 and line 3
- D. Line 3 only
- E. The program successfully compiles
- F. Line 2 only
- G. Line 2 and line 3
Answer: E
Explanation:
In this code, the Calculable interface defines a single abstract method calculate that takes an int parameter and returns a long. The main method contains three lambda expressions assigned to variables c1, c2, and c3 of type Calculable.
* Line 1:Calculable c1 = i -> i + 1;
This lambda expression takes an integer i and returns the result of i + 1. Since the expression i + 1 results in an int, and Java allows implicit widening conversion from int to long, this line compiles successfully.
* Line 2:Calculable c2 = i -> Long.valueOf(i);
Here, the lambda expression takes an integer i and returns the result of Long.valueOf(i). The Long.valueOf (int i) method returns a Long object. However, Java allows unboxing of the Long object to a long primitive type when necessary. Therefore, this line compiles successfully.
* Line 3:Calculable c3 = i -> { throw new ArithmeticException(); };
This lambda expression takes an integer i and throws an ArithmeticException. Since the method calculate has a return type of long, and throwing an exception is a valid way to exit the method without returning a value, this line compiles successfully.
Since all three lines adhere to the method signature defined in the Calculable interface and there are no type mismatches or syntax errors, the program compiles successfully.
NEW QUESTION # 35
Given:
java
DoubleStream doubleStream = DoubleStream.of(3.3, 4, 5.25, 6.66);
Predicate<Double> doublePredicate = d -> d < 5;
System.out.println(doubleStream.anyMatch(doublePredicate));
What is printed?
- A. Compilation fails
- B. false
- C. An exception is thrown at runtime
- D. 3.3
- E. true
Answer: A
Explanation:
In this code, there is a type mismatch between the DoubleStream and the Predicate<Double>.
* DoubleStream: A sequence of primitive double values.
* Predicate<Double>: A functional interface that operates on objects of type Double (the wrapper class), not on primitive double values.
The DoubleStream class provides a method anyMatch(DoublePredicate predicate), where DoublePredicate is a functional interface that operates on primitive double values. However, in the code, a Predicate<Double> is used instead of a DoublePredicate. This mismatch leads to a compilation error because anyMatch cannot accept a Predicate<Double> when working with a DoubleStream.
To correct this, the predicate should be defined as a DoublePredicate to match the primitive double type:
java
DoubleStream doubleStream = DoubleStream.of(3.3, 4, 5.25, 6.66);
DoublePredicate doublePredicate = d -> d < 5;
System.out.println(doubleStream.anyMatch(doublePredicate));
With this correction, the code will compile and print true because there are elements in the stream (e.g., 3.3 and 4.0) that are less than 5.
NEW QUESTION # 36
Given:
java
public class Test {
public static void main(String[] args) throws IOException {
Path p1 = Path.of("f1.txt");
Path p2 = Path.of("f2.txt");
Files.move(p1, p2);
Files.delete(p1);
}
}
In which case does the given program throw an exception?
- A. An exception is always thrown
- B. File f2.txt exists while file f1.txt doesn't
- C. File f1.txt exists while file f2.txt doesn't
- D. Both files f1.txt and f2.txt exist
- E. Neither files f1.txt nor f2.txt exist
Answer: A
Explanation:
In this program, the following operations are performed:
* Paths Initialization:
* Path p1 is set to "f1.txt".
* Path p2 is set to "f2.txt".
* File Move Operation:
* Files.move(p1, p2); attempts to move (or rename) f1.txt to f2.txt.
* File Delete Operation:
* Files.delete(p1); attempts to delete f1.txt.
Analysis:
* If f1.txt Does Not Exist:
* The Files.move(p1, p2); operation will throw a NoSuchFileException because the source file f1.
txt is missing.
* If f1.txt Exists and f2.txt Does Not Exist:
* The Files.move(p1, p2); operation will successfully rename f1.txt to f2.txt.
* Subsequently, the Files.delete(p1); operation will throw a NoSuchFileException because p1 (now f1.txt) no longer exists after the move.
* If Both f1.txt and f2.txt Exist:
* The Files.move(p1, p2); operation will throw a FileAlreadyExistsException because the target file f2.txt already exists.
* If f2.txt Exists While f1.txt Does Not:
* Similar to the first scenario, the Files.move(p1, p2); operation will throw a NoSuchFileException due to the absence of f1.txt.
In all possible scenarios, an exception is thrown during the execution of the program.
NEW QUESTION # 37
Given:
java
Stream<String> strings = Stream.of("United", "States");
BinaryOperator<String> operator = (s1, s2) -> s1.concat(s2.toUpperCase()); String result = strings.reduce("-", operator); System.out.println(result); What is the output of this code fragment?
- A. United-STATES
- B. -UnitedSTATES
- C. UnitedStates
- D. UNITED-STATES
- E. -UNITEDSTATES
- F. -UnitedStates
- G. United-States
Answer: B
Explanation:
In this code, a Stream of String elements is created containing "United" and "States". A BinaryOperator<String> named operator is defined to concatenate the first string (s1) with the uppercase version of the second string (s2). The reduce method is then used with "-" as the identity value and operator as the accumulator.
The reduce method processes the elements of the stream as follows:
* Initial Identity Value: "-"
* First Iteration:
* Accumulator Operation: "-".concat("United".toUpperCase())
* Result: "-UNITED"
* Second Iteration:
* Accumulator Operation: "-UNITED".concat("States".toUpperCase())
* Result: "-UNITEDSTATES"
Therefore, the final result stored in result is "-UNITEDSTATES", and the output of theSystem.out.println (result); statement is -UNITEDSTATES.
NEW QUESTION # 38
Given:
java
interface SmartPhone {
boolean ring();
}
class Iphone15 implements SmartPhone {
boolean isRinging;
boolean ring() {
isRinging = !isRinging;
return isRinging;
}
}
Choose the right statement.
- A. An exception is thrown at running Iphone15.ring();
- B. SmartPhone interface does not compile
- C. Everything compiles
- D. Iphone15 class does not compile
Answer: D
Explanation:
In this code, the SmartPhone interface declares a method ring() with a boolean return type. The Iphone15 class implements the SmartPhone interface and provides an implementation for the ring() method.
However, in the Iphone15 class, the ring() method is declared without the public access modifier. In Java, when a class implements an interface, it must provide implementations for all the interface's methods with the same or a more accessible access level. Since interface methods are implicitly public, the implementing methods in the class must also be public. Failing to do so results in a compilation error.
Therefore, the Iphone15 class does not compile because the ring() method is not declared as public.
NEW QUESTION # 39
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