1Z1-830 DUMP | 1Z1-830 VALUABLE FEEDBACK

1z1-830 Dump | 1z1-830 Valuable Feedback

1z1-830 Dump | 1z1-830 Valuable Feedback

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Oracle Java SE 21 Developer Professional Sample Questions (Q27-Q32):

NEW QUESTION # 27
Which two of the following aren't the correct ways to create a Stream?

  • A. Stream stream = Stream.generate(() -> "a");
  • B. Stream stream = Stream.of("a");
  • C. Stream stream = new Stream();
  • D. Stream<String> stream = Stream.builder().add("a").build();
  • E. Stream stream = Stream.of();
  • F. Stream stream = Stream.ofNullable("a");
  • G. Stream stream = Stream.empty();

Answer: C,D

Explanation:
In Java, the Stream API provides several methods to create streams. However, not all approaches are valid.


NEW QUESTION # 28
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. {b=1, a=3, c=2}
  • B. Compilation fails
  • C. {c=2, a=3, b=1}
  • D. {b=1, c=2, a=3}
  • E. {a=1, b=2, c=3}
  • F. {c=1, b=2, a=3}
  • G. {a=3, b=1, c=2}

Answer: G

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 # 29
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. 0 1 2 3
  • D. An exception is thrown.
  • E. 1 2 3 4
  • F. Compilation fails.

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 # 30
Given:
java
Deque<Integer> deque = new ArrayDeque<>();
deque.offer(1);
deque.offer(2);
var i1 = deque.peek();
var i2 = deque.poll();
var i3 = deque.peek();
System.out.println(i1 + " " + i2 + " " + i3);
What is the output of the given code fragment?

  • A. 1 2 2
  • B. 2 1 2
  • C. 1 1 2
  • D. 1 1 1
  • E. An exception is thrown.
  • F. 2 2 1
  • G. 2 1 1
  • H. 1 2 1
  • I. 2 2 2

Answer: A

Explanation:
In this code, an ArrayDeque named deque is created, and the integers 1 and 2 are added to it using the offer method. The offer method inserts the specified element at the end of the deque.
* State of deque after offers:[1, 2]
The peek method retrieves, but does not remove, the head of the deque, returning 1. Therefore, i1 is assigned the value 1.
* State of deque after peek:[1, 2]
* Value of i1:1
The poll method retrieves and removes the head of the deque, returning 1. Therefore, i2 is assigned the value
1.
* State of deque after poll:[2]
* Value of i2:1
Another peek operation retrieves the current head of the deque, which is now 2, without removing it.
Therefore, i3 is assigned the value 2.
* State of deque after second peek:[2]
* Value of i3:2
The System.out.println statement then outputs the values of i1, i2, and i3, resulting in 1 1 2.


NEW QUESTION # 31
Given:
java
package com.vv;
import java.time.LocalDate;
public class FetchService {
public static void main(String[] args) throws Exception {
FetchService service = new FetchService();
String ack = service.fetch();
LocalDate date = service.fetch();
System.out.println(ack + " the " + date.toString());
}
public String fetch() {
return "ok";
}
public LocalDate fetch() {
return LocalDate.now();
}
}
What will be the output?

  • A. ok the 2024-07-10T07:17:45.523939600
  • B. An exception is thrown
  • C. ok the 2024-07-10
  • D. Compilation fails

Answer: D

Explanation:
In Java, method overloading allows multiple methods with the same name to exist in a class, provided they have different parameter lists (i.e., different number or types of parameters). However, having two methods with the exact same parameter list and only differing in return type is not permitted.
In the provided code, the FetchService class contains two fetch methods:
* public String fetch()
* public LocalDate fetch()
Both methods have identical parameter lists (none) but differ in their return types (String and LocalDate, respectively). This leads to a compilation error because the Java compiler cannot distinguish between the two methods based solely on return type.
The Java Language Specification (JLS) states:
"It is a compile-time error to declare two methods with override-equivalent signatures in a class." In this context, "override-equivalent" means that the methods have the same name and parameter types, regardless of their return types.
Therefore, the code will fail to compile due to the duplicate method signatures, and the correct answer is B:
Compilation fails.


NEW QUESTION # 32
......

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