1Z0-830 VALID VCE DUMPS - LATEST 1Z0-830 TEST PDF

1z0-830 Valid Vce Dumps - Latest 1z0-830 Test Pdf

1z0-830 Valid Vce Dumps - Latest 1z0-830 Test Pdf

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Oracle Java SE 21 Developer Professional Sample Questions (Q30-Q35):

NEW QUESTION # 30
Given:
java
record WithInstanceField(String foo, int bar) {
double fuz;
}
record WithStaticField(String foo, int bar) {
static double wiz;
}
record ExtendingClass(String foo) extends Exception {}
record ImplementingInterface(String foo) implements Cloneable {}
Which records compile? (Select 2)

  • A. ImplementingInterface
  • B. WithInstanceField
  • C. ExtendingClass
  • D. WithStaticField

Answer: A,D

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 # 31
Given:
java
void verifyNotNull(Object input) {
boolean enabled = false;
assert enabled = true;
assert enabled;
System.out.println(input.toString());
assert input != null;
}
When does the given method throw a NullPointerException?

  • A. Only if assertions are enabled and the input argument is null
  • B. Only if assertions are enabled and the input argument isn't null
  • C. A NullPointerException is never thrown
  • D. Only if assertions are disabled and the input argument is null
  • E. Only if assertions are disabled and the input argument isn't null

Answer: D

Explanation:
In the verifyNotNull method, the following operations are performed:
* Assertion to Enable Assertions:
java
boolean enabled = false;
assert enabled = true;
assert enabled;
* The variable enabled is initially set to false.
* The first assertion assert enabled = true; assigns true to enabled if assertions are enabled. If assertions are disabled, this assignment does not occur.
* The second assertion assert enabled; checks if enabled is true. If assertions are enabled and the previous assignment occurred, this assertion passes. If assertions are disabled, this assertion is ignored.
* Dereferencing the input Object:
java
System.out.println(input.toString());
* This line attempts to call the toString() method on the input object. If input is null, this will throw a NullPointerException.
* Assertion to Check input for null:
java
assert input != null;
* This assertion checks that input is not null. If input is null and assertions are enabled, this assertion will fail, throwing an AssertionError. If assertions are disabled, this assertion is ignored.
Analysis:
* If Assertions Are Enabled:
* The enabled variable is set to true by the first assertion, and the second assertion passes.
* If input is null, calling input.toString() will throw a NullPointerException before the final assertion is reached.
* If input is not null, input.toString() executes without issue, and the final assertion assert input != null; passes.
* If Assertions Are Disabled:
* The enabled variable remains false, but the assertions are ignored, so this has no effect.
* If input is null, calling input.toString() will throw a NullPointerException.
* If input is not null, input.toString() executes without issue.
Conclusion:
A NullPointerException is thrown if input is null, regardless of whether assertions are enabled or disabled.
Therefore, the correct answer is:
C: Only if assertions are disabled and the input argument is null


NEW QUESTION # 32
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. It throws an exception at runtime.
  • B. Compilation fails.
  • C. string
  • D. nothing
  • E. long
  • F. integer

Answer: B

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 # 33
Which methods compile?

  • A. ```java
    public List<? super IOException> getListSuper() {
    return new ArrayList<FileNotFoundException>();
    }
  • B. ```java public List<? super IOException> getListSuper() { return new ArrayList<Exception>(); } csharp
  • C. ```java
    public List<? extends IOException> getListExtends() {
    return new ArrayList<FileNotFoundException>();
    }
  • D. ```java public List<? extends IOException> getListExtends() { return new ArrayList<Exception>(); } csharp

Answer: B,C

Explanation:
In Java generics, wildcards are used to relax the type constraints of generic types. The extends wildcard (<?
extends Type>) denotes an upper bounded wildcard, allowing any type that is a subclass of Type. Conversely, the super wildcard (<? super Type>) denotes a lower bounded wildcard, allowing any type that is a superclass of Type.
Option A:
java
public List<? super IOException> getListSuper() {
return new ArrayList<Exception>();
}
Here, List<? super IOException> represents a list that can hold IOException objects and objects of its supertypes. Since Exception is a superclass of IOException, ArrayList<Exception> is compatible with List<?
super IOException>. Therefore, this method compiles successfully.
Option B:
java
public List<? extends IOException> getListExtends() {
return new ArrayList<FileNotFoundException>();
}
In this case, List<? extends IOException> represents a list that can hold objects of IOException and its subclasses. Since FileNotFoundException is a subclass of IOException, ArrayList<FileNotFoundException> is compatible with List<? extends IOException>. Thus, this method compiles successfully.
Option C:
java
public List<? extends IOException> getListExtends() {
return new ArrayList<Exception>();
}
Here, List<? extends IOException> expects a list of IOException or its subclasses. However, Exception is a superclass of IOException, not a subclass. Therefore, ArrayList<Exception> is not compatible with List<?
extends IOException>, and this method will not compile.
Option D:
java
public List<? super IOException> getListSuper() {
return new ArrayList<FileNotFoundException>();
}
In this scenario, List<? super IOException> expects a list that can hold IOException objects and objects of its supertypes. Since FileNotFoundException is a subclass of IOException, ArrayList<FileNotFoundException> is not compatible with List<? super IOException>, and this method will not compile.
Therefore, the methods in options A and B compile successfully, while those in options C and D do not.


NEW QUESTION # 34
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. Numbers from 1 to 25 randomly
  • B. An exception is thrown at runtime
  • C. Numbers from 1 to 1945 randomly
  • D. Compilation fails
  • E. Numbers from 1 to 25 sequentially

Answer: A

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 # 35
......

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