FREE 1Z0-830 PDF GUIDE - TOP 1Z0-830 QUESTIONS

Free 1z0-830 Pdf Guide - Top 1z0-830 Questions

Free 1z0-830 Pdf Guide - Top 1z0-830 Questions

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Oracle Java SE 21 Developer Professional Sample Questions (Q43-Q48):

NEW QUESTION # 43
You are working on a module named perfumery.shop that depends on another module named perfumery.
provider.
The perfumery.shop module should also make its package perfumery.shop.eaudeparfum available to other modules.
Which of the following is the correct file to declare the perfumery.shop module?

  • A. File name: module-info.java
    java
    module perfumery.shop {
    requires perfumery.provider;
    exports perfumery.shop.eaudeparfum;
    }
  • B. File name: module.java
    java
    module shop.perfumery {
    requires perfumery.provider;
    exports perfumery.shop.eaudeparfum;
    }
  • C. File name: module-info.perfumery.shop.java
    java
    module perfumery.shop {
    requires perfumery.provider;
    exports perfumery.shop.eaudeparfum.*;
    }

Answer: A

Explanation:
* Correct module descriptor file name
* A module declaration must be placed inside a file namedmodule-info.java.
* The incorrect filename module-info.perfumery.shop.javais invalid(Option A).
* The incorrect filename module.javais invalid(Option C).
* Correct module declaration
* The module declaration must match the name of the module (perfumery.shop).
* The requires perfumery.provider; directive specifies that perfumery.shop depends on perfumery.
provider.
* The exports perfumery.shop.eaudeparfum; statement allows the perfumery.shop.eaudeparfum package to beaccessible by other modules.
* The incorrect syntax exports perfumery.shop.eaudeparfum.*; in Option A isinvalid, as wildcards (*) arenot allowedin module exports.
Thus, the correct answer is:File name: module-info.java
References:
* Java SE 21 - Modules
* Java SE 21 - module-info.java File


NEW QUESTION # 44
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. null
  • B. Optional[Java]
  • C. Compilation fails
  • D. Java

Answer: B

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 # 45
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. ClassCastException
  • B. 0
  • C. Compilation fails
  • D. 1
  • E. NotSerializableException

Answer: A

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 # 46
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. 3.3
  • B. Compilation fails
  • C. An exception is thrown at runtime
  • D. true
  • E. false

Answer: B

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 # 47
Given:
java
List<String> frenchAuthors = new ArrayList<>();
frenchAuthors.add("Victor Hugo");
frenchAuthors.add("Gustave Flaubert");
Which compiles?

  • A. var authorsMap3 = new HashMap<>();
    java
    authorsMap3.put("FR", frenchAuthors);
  • B. Map<String, ArrayList<String>> authorsMap1 = new HashMap<>();
    java
    authorsMap1.put("FR", frenchAuthors);
  • C. Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>>(); java authorsMap4.put("FR", frenchAuthors);
  • D. Map<String, List<String>> authorsMap5 = new HashMap<String, List<String>>(); java authorsMap5.put("FR", frenchAuthors);
  • E. Map<String, ? extends List<String>> authorsMap2 = new HashMap<String, ArrayList<String>> (); java authorsMap2.put("FR", frenchAuthors);

Answer: A,C,D

Explanation:
* Option A (Map<String, ArrayList<String>> authorsMap1 = new HashMap<>();)
* #Compilation Fails
* frenchAuthors is declared as List<String>,notArrayList<String>.
* The correct way to declare a Map that allows storing List<String> is to use List<String> as the generic type,notArrayList<String>.
* Fix:
java
Map<String, List<String>> authorsMap1 = new HashMap<>();
authorsMap1.put("FR", frenchAuthors);
* Reason:The type ArrayList<String> is more specific than List<String>, and this would cause a type mismatcherror.
* Option B (Map<String, ? extends List<String>> authorsMap2 = new HashMap<String, ArrayList<String>>();)
* #Compilation Fails
* ? extends List<String>makes the map read-onlyfor adding new elements.
* The line authorsMap2.put("FR", frenchAuthors); causes acompilation errorbecause wildcard (?
extends List<String>) prevents modifying the map.
* Fix:Remove the wildcard:
java
Map<String, List<String>> authorsMap2 = new HashMap<>();
authorsMap2.put("FR", frenchAuthors);
* Option C (var authorsMap3 = new HashMap<>();)
* Compiles Successfully
* The var keyword allows the compiler to infer the type.
* However,the inferred type is HashMap<Object, Object>, which may cause issues when retrieving values.
* Option D (Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>
>();)
* Compiles Successfully
* Valid declaration:HashMap<K, V> can be assigned to Map<K, V>.
* Using new HashMap<String, ArrayList<String>>() with Map<String, List<String>> isallowed due to polymorphism.
* Correct syntax:
java
Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>>(); authorsMap4.put("FR", frenchAuthors);
* Option E (Map<String, List<String>> authorsMap5 = new HashMap<String, List<String>>();)
* Compiles Successfully
* HashMap<String, List<String>> isa valid instantiation.
* Correct usage:
java
Map<String, List<String>> authorsMap5 = new HashMap<>();
authorsMap5.put("FR", frenchAuthors);
Thus, the correct answers are:C, D, E
References:
* Java SE 21 - Generics and Type Inference
* Java SE 21 - var Keyword


NEW QUESTION # 48
......

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