ADVANCED 1Z1-830 TESTING ENGINE - OFFICIAL 1Z1-830 PRACTICE TEST

Advanced 1z1-830 Testing Engine - Official 1z1-830 Practice Test

Advanced 1z1-830 Testing Engine - Official 1z1-830 Practice Test

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Oracle Java SE 21 Developer Professional Sample Questions (Q57-Q62):

NEW QUESTION # 57
Given:
java
Optional<String> optionalName = Optional.ofNullable(null);
String bread = optionalName.orElse("Baguette");
System.out.print("bread:" + bread);
String dish = optionalName.orElseGet(() -> "Frog legs");
System.out.print(", dish:" + dish);
try {
String cheese = optionalName.orElseThrow(() -> new Exception());
System.out.println(", cheese:" + cheese);
} catch (Exception exc) {
System.out.println(", no cheese.");
}
What is printed?

  • A. bread:Baguette, dish:Frog legs, cheese.
  • B. bread:bread, dish:dish, cheese.
  • C. bread:Baguette, dish:Frog legs, no cheese.
  • D. Compilation fails.

Answer: C

Explanation:
Understanding Optional.ofNullable(null)
* Optional.ofNullable(null); creates an empty Optional (i.e., it contains no value).
* Optional.of(null); would throw a NullPointerException, but ofNullable(null); safely creates an empty Optional.
Execution of orElse, orElseGet, and orElseThrow
* orElse("Baguette")
* Since optionalName is empty, "Baguette" is returned.
* bread = "Baguette"
* Output:"bread:Baguette"
* orElseGet(() -> "Frog legs")
* Since optionalName is empty, "Frog legs" is returned from the lambda expression.
* dish = "Frog legs"
* Output:", dish:Frog legs"
* orElseThrow(() -> new Exception())
* Since optionalName is empty, an exception is thrown.
* The catch block catches this exception and prints ", no cheese.".
Thus, the final output is:
makefile
bread:Baguette, dish:Frog legs, no cheese.
References:
* Java SE 21 & JDK 21 - Optional
* Java SE 21 - Functional Interfaces


NEW QUESTION # 58
Given:
java
Optional o1 = Optional.empty();
Optional o2 = Optional.of(1);
Optional o3 = Stream.of(o1, o2)
.filter(Optional::isPresent)
.findAny()
.flatMap(o -> o);
System.out.println(o3.orElse(2));
What is the given code fragment's output?

  • A. Optional.empty
  • B. An exception is thrown
  • C. 0
  • D. 1
  • E. 2
  • F. Optional[1]
  • G. Compilation fails

Answer: D

Explanation:
In this code, two Optional objects are created:
* o1 is an empty Optional.
* o2 is an Optional containing the integer 1.
A stream is created from o1 and o2. The filter method retains only the Optional instances that are present (i.e., non-empty). This results in a stream containing only o2.
The findAny method returns an Optional describing some element of the stream, or an empty Optional if the stream is empty. Since the stream contains o2, findAny returns Optional[Optional[1]].
The flatMap method is then used to flatten this nested Optional. It applies the provided mapping function (o -
> o) to the value, resulting in Optional[1].
Finally, o3.orElse(2) returns the value contained in o3 if it is present; otherwise, it returns 2. Since o3 contains
1, the output is 1.


NEW QUESTION # 59
Which StringBuilder variable fails to compile?
java
public class StringBuilderInstantiations {
public static void main(String[] args) {
var stringBuilder1 = new StringBuilder();
var stringBuilder2 = new StringBuilder(10);
var stringBuilder3 = new StringBuilder("Java");
var stringBuilder4 = new StringBuilder(new char[]{'J', 'a', 'v', 'a'});
}
}

  • A. stringBuilder2
  • B. None of them
  • C. stringBuilder3
  • D. stringBuilder1
  • E. stringBuilder4

Answer: E

Explanation:
In the provided code, four StringBuilder instances are being created using different constructors:
* stringBuilder1: new StringBuilder()
* This constructor creates an empty StringBuilder with an initial capacity of 16 characters.
* stringBuilder2: new StringBuilder(10)
* This constructor creates an empty StringBuilder with a specified initial capacity of 10 characters.
* stringBuilder3: new StringBuilder("Java")
* This constructor creates a StringBuilder initialized to the contents of the specified string "Java".
* stringBuilder4: new StringBuilder(new char[]{'J', 'a', 'v', 'a'})
* This line attempts to create a StringBuilder using a char array. However, the StringBuilder class does not have a constructor that accepts a char array directly. The available constructors are:
* StringBuilder()
* StringBuilder(int capacity)
* StringBuilder(String str)
* StringBuilder(CharSequence seq)
Since a char array does not implement the CharSequence interface, and there is no constructor that directly accepts a char array, this line will cause a compilation error.
To initialize a StringBuilder with a char array, you can convert the char array to a String first:
java
var stringBuilder4 = new StringBuilder(new String(new char[]{'J', 'a', 'v', 'a'})); This approach utilizes the String constructor that accepts a char array, and then passes the resulting String to the StringBuilder constructor.


NEW QUESTION # 60
Given:
java
var array1 = new String[]{ "foo", "bar", "buz" };
var array2[] = { "foo", "bar", "buz" };
var array3 = new String[3] { "foo", "bar", "buz" };
var array4 = { "foo", "bar", "buz" };
String array5[] = new String[]{ "foo", "bar", "buz" };
Which arrays compile? (Select 2)

  • A. array2
  • B. array4
  • C. array5
  • D. array1
  • E. array3

Answer: C,D

Explanation:
In Java, array initialization can be performed in several ways, but certain syntaxes are invalid and will cause compilation errors. Let's analyze each declaration:
* var array1 = new String[]{ "foo", "bar", "buz" };
This is a valid declaration. The var keyword allows the compiler to infer the type from the initializer. Here, new String[]{ "foo", "bar", "buz" } creates an anonymous array of String with three elements. The compiler infers array1 as String[]. This syntax is correct and compiles successfully.
* var array2[] = { "foo", "bar", "buz" };
This declaration is invalid. While var can be used for type inference, appending [] after var is not allowed.
The correct syntax would be either String[] array2 = { "foo", "bar", "buz" }; or var array2 = new String[]{
"foo", "bar", "buz" };. Therefore, this line will cause a compilation error.
* var array3 = new String[3] { "foo", "bar", "buz" };
This declaration is invalid. In Java, when specifying the size of the array (new String[3]), you cannot simultaneously provide an initializer. The correct approach is either to provide the size without an initializer (new String[3]) or to provide the initializer without specifying the size (new String[]{ "foo", "bar", "buz" }).
Therefore, this line will cause a compilation error.
* var array4 = { "foo", "bar", "buz" };
This declaration is invalid. The array initializer { "foo", "bar", "buz" } can only be used in an array declaration when the type is explicitly provided. Since var relies on type inference and there's no explicit type provided here, this will cause a compilation error. The correct syntax would be String[] array4 = { "foo",
"bar", "buz" };.
* String array5[] = new String[]{ "foo", "bar", "buz" };
This is a valid declaration. Here, String array5[] declares array5 as an array of String. The initializer new String[]{ "foo", "bar", "buz" } creates an array with three elements. This syntax is correct and compiles successfully.
Therefore, the declarations that compile successfully are array1 and array5.
References:
* Java SE 21 & JDK 21 - Local Variable Type Inference
* Java SE 21 & JDK 21 - Arrays


NEW QUESTION # 61
Given:
java
int post = 5;
int pre = 5;
int postResult = post++ + 10;
int preResult = ++pre + 10;
System.out.println("postResult: " + postResult +
", preResult: " + preResult +
", Final value of post: " + post +
", Final value of pre: " + pre);
What is printed?

  • A. postResult: 16, preResult: 16, Final value of post: 6, Final value of pre: 6
  • B. postResult: 15, preResult: 16, Final value of post: 6, Final value of pre: 6
  • C. postResult: 16, preResult: 15, Final value of post: 6, Final value of pre: 5
  • D. postResult: 15, preResult: 16, Final value of post: 5, Final value of pre: 6

Answer: B

Explanation:
* Understanding post++ (Post-increment)
* post++uses the value first, then increments it.
* postResult = post++ + 10;
* post starts as 5.
* post++ returns 5, then post is incremented to 6.
* postResult = 5 + 10 = 15.
* Final value of post after this line is 6.
* Understanding ++pre (Pre-increment)
* ++preincrements the value first, then uses it.
* preResult = ++pre + 10;
* pre starts as 5.
* ++pre increments pre to 6, then returns 6.
* preResult = 6 + 10 = 16.
* Final value of pre after this line is 6.
Thus, the final output is:
yaml
postResult: 15, preResult: 16, Final value of post: 6, Final value of pre: 6 References:
* Java SE 21 - Operators and Expressions
* Java SE 21 - Arithmetic Operators


NEW QUESTION # 62
......

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