ORACLE 1Z1-830 EXAMCOLLECTION DUMPS, 1Z1-830 EXAM SIMULATOR FEE

Oracle 1z1-830 Examcollection Dumps, 1z1-830 Exam Simulator Fee

Oracle 1z1-830 Examcollection Dumps, 1z1-830 Exam Simulator Fee

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Oracle Java SE 21 Developer Professional Sample Questions (Q31-Q36):

NEW QUESTION # 31
Given:
java
List<String> abc = List.of("a", "b", "c");
abc.stream()
.forEach(x -> {
x = x.toUpperCase();
});
abc.stream()
.forEach(System.out::print);
What is the output?

  • A. An exception is thrown.
  • B. Compilation fails.
  • C. abc
  • D. ABC

Answer: C

Explanation:
In the provided code, a list abc is created containing the strings "a", "b", and "c". The first forEach operation attempts to convert each element to uppercase by assigning x = x.toUpperCase();. However, this assignment only changes the local variable x within the lambda expression and does not modify the elements in the original list abc. Strings in Java are immutable, meaning their values cannot be changed once created.
Therefore, the original list remains unchanged.
The second forEach operation iterates over the original list and prints each element. Since the list was not modified, the output will be the concatenation of the original elements: abc.
To achieve the output ABC, you would need to collect the transformed elements into a new list, as shown below:
java
List<String> abc = List.of("a", "b", "c");
List<String> upperCaseAbc = abc.stream()
map(String::toUpperCase)
collect(Collectors.toList());
upperCaseAbc.forEach(System.out::print);
In this corrected version, the map operation creates a new stream with the uppercase versions of the original elements, which are then collected into a new list upperCaseAbc. The forEach operation then prints ABC.


NEW QUESTION # 32
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. NotSerializableException
  • C. Compilation fails
  • D. 0
  • E. 1

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 # 33
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. [Lille, Lyon]
  • B. [Lyon, Lille, Toulouse]
  • C. Compilation fails
  • D. [Paris]
  • E. [Paris, Toulouse]

Answer: A

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 # 34
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. true
  • B. 3.3
  • C. false
  • D. An exception is thrown at runtime
  • E. Compilation fails

Answer: E

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 # 35
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,h i]
  • B. [ , hello ,hi ]
  • C. [,hello,hi]
  • D. [,hello,h i]

Answer: D

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 # 36
......

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