RELIABLE 1Z1-830 TEST PREP - LATEST 1Z1-830 EXAM FORMAT

Reliable 1z1-830 Test Prep - Latest 1z1-830 Exam Format

Reliable 1z1-830 Test Prep - Latest 1z1-830 Exam Format

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Oracle Java SE 21 Developer Professional Sample Questions (Q28-Q33):

NEW QUESTION # 28
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 a double with value: 42
  • B. Compilation fails.
  • C. It's an integer with value: 42
  • D. It throws an exception at runtime.
  • E. It's a string with value: 42
  • F. null

Answer: B

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 # 29
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. {a=1, b=2, c=3}
  • B. {b=1, c=2, a=3}
  • C. {a=3, b=1, c=2}
  • D. {c=2, a=3, b=1}
  • E. Compilation fails
  • F. {b=1, a=3, c=2}
  • G. {c=1, b=2, a=3}

Answer: C

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 # 30
Which two of the following aren't the correct ways to create a Stream?

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

Answer: A,C

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


NEW QUESTION # 31
Given:
java
var _ = 3;
var $ = 7;
System.out.println(_ + $);
What is printed?

  • A. It throws an exception.
  • B. 0
  • C. Compilation fails.
  • D. _$

Answer: C

Explanation:
* The var keyword and identifier rules:
* The var keyword is used for local variable type inference introduced inJava 10.
* However,Java does not allow _ (underscore) as an identifiersinceJava 9.
* If we try to use _ as a variable name, the compiler will throw an error:
pgsql
error: as of release 9, '_' is a keyword, and may not be used as an identifier
* The $ symbol as an identifier:
* The $ characteris a valid identifierin Java.
* However, since _ is not allowed, the codefails to compile before even reaching $.
Thus,the correct answer is "Compilation fails."
References:
* Java SE 21 - var Local Variable Type Inference
* Java SE 9 - Restrictions on _ Identifier


NEW QUESTION # 32
Given:
java
var deque = new ArrayDeque<>();
deque.add(1);
deque.add(2);
deque.add(3);
deque.add(4);
deque.add(5);
System.out.print(deque.peek() + " ");
System.out.print(deque.poll() + " ");
System.out.print(deque.pop() + " ");
System.out.print(deque.element() + " ");
What is printed?

  • A. 1 5 5 1
  • B. 5 5 2 3
  • C. 1 1 1 1
  • D. 1 1 2 2
  • E. 1 1 2 3

Answer: E

Explanation:
* Understanding ArrayDeque Behavior
* ArrayDeque<E>is a double-ended queue (deque), working as aFIFO (queue) and LIFO (stack).
* Thedefault behaviorisqueue-like (FIFO)unless explicitly used as a stack.
* Step-by-Step Execution
java
var deque = new ArrayDeque<>();
deque.add(1);
deque.add(2);
deque.add(3);
deque.add(4);
deque.add(5);
* Deque after additions# [1, 2, 3, 4, 5]
* Operations Breakdown
* deque.peek()# Returns thehead(first element)without removal.
makefile
Output: 1
* deque.poll()# Removes and returns thehead.
go
Output: 1, Deque after poll # `[2, 3, 4, 5]`
* deque.pop()#Same as removeFirst(); removes and returns thehead.
perl
Output: 2, Deque after pop # `[3, 4, 5]`
* deque.element()# Returns thehead(same as peek(), but throws an exception if empty).
makefile
Output: 3
* Final Output
1 1 2 3
Thus, the correct answer is:1 1 2 3
References:
* Java SE 21 - ArrayDeque
* Java SE 21 - Queue Operations


NEW QUESTION # 33
......

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