VALID TEST 1Z1-830 EXPERIENCE | 1Z1-830 EXAM COLLECTION PDF

Valid Test 1z1-830 Experience | 1z1-830 Exam Collection Pdf

Valid Test 1z1-830 Experience | 1z1-830 Exam Collection Pdf

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Oracle Java SE 21 Developer Professional Sample Questions (Q70-Q75):

NEW QUESTION # 70
Given:
java
Object input = 42;
String result = switch (input) {
case String s -> "It's a string with value: " + s;
case Double d -> "It's a double with value: " + d;
case Integer i -> "It's an integer with value: " + i;
};
System.out.println(result);
What is printed?

  • A. It's an integer with value: 42
  • B. It's a string with value: 42
  • C. Compilation fails.
  • D. It's a double with value: 42
  • E. null
  • F. It throws an exception at runtime.

Answer: C

Explanation:
* Pattern Matching in switch
* The switch expression introduced inJava 21supportspattern matchingfor different types.
* However,a switch expression must be exhaustive, meaningit must cover all possible cases or provide a default case.
* Why does compilation fail?
* input is an Object, and the switch expression attempts to pattern-match it to String, Double, and Integer.
* If input had been of another type (e.g., Float or Long), there would beno matching case, leading to anon-exhaustive switch.
* Javarequires a default caseto ensure all possible inputs are covered.
* Corrected Code (Adding a default Case)
java
Object input = 42;
String result = switch (input) {
case String s -> "It's a string with value: " + s;
case Double d -> "It's a double with value: " + d;
case Integer i -> "It's an integer with value: " + i;
default -> "Unknown type";
};
System.out.println(result);
* With this change, the codecompiles and runs successfully.
* Output:
vbnet
It's an integer with value: 42
Thus, the correct answer is:Compilation failsdue to a missing default case.
References:
* Java SE 21 - Pattern Matching for switch
* Java SE 21 - switch Expressions


NEW QUESTION # 71
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 3 only
  • B. Line 1 only
  • C. Line 2 and line 3
  • D. Line 2 only
  • E. The program successfully compiles
  • F. Line 1 and line 2
  • G. Line 1 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 # 72
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. WithStaticField
  • C. ExtendingClass
  • D. WithInstanceField

Answer: A,B

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 # 73
Given:
java
final Stream<String> strings =
Files.readAllLines(Paths.get("orders.csv"));
strings.skip(1)
.limit(2)
.forEach(System.out::println);
And that the orders.csv file contains:
mathematica
OrderID,Customer,Product,Quantity,Price
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99
3,Sebastien Loeb,Monitor,1,199.99
4,Antoine Griezmann,Headset,3,45.00
What is printed?

  • A. arduino
    1,Kylian Mbappe,Keyboard,2,25.50
    2,Teddy Riner,Mouse,1,15.99
    3,Sebastien Loeb,Monitor,1,199.99
    4,Antoine Griezmann,Headset,3,45.00
  • B. An exception is thrown at runtime.
  • C. arduino
    1,Kylian Mbappe,Keyboard,2,25.50
    2,Teddy Riner,Mouse,1,15.99
  • D. Compilation fails.
  • E. arduino
    2,Teddy Riner,Mouse,1,15.99
    3,Sebastien Loeb,Monitor,1,199.99

Answer: B,D

Explanation:
1. Why Does Compilation Fail?
* The error is in this line:
java
final Stream<String> strings = Files.readAllLines(Paths.get("orders.csv"));
* Files.readAllLines(Paths.get("orders.csv")) returns a List<String>,not a Stream<String>.
* A List<String> cannot be assigned to a Stream<String>.
2. Correcting the Code
* The correct way to create a stream from the file:
java
Stream<String> strings = Files.lines(Paths.get("orders.csv"));
* This correctly creates a Stream<String> from the file.
3. Expected Output After Fixing
java
Files.lines(Paths.get("orders.csv"))
skip(1) // Skips the header row
limit(2) // Limits to first two data rows
forEach(System.out::println);
* Output:
arduino
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - Files.readAllLines
* Java SE 21 - Files.lines


NEW QUESTION # 74
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 sequentially
  • C. An exception is thrown at runtime
  • D. Numbers from 1 to 1945 randomly
  • E. Numbers from 1 to 25 randomly

Answer: E

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 # 75
......

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