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Oracle Java SE 21 Developer Professional Sample Questions (Q82-Q87):

NEW QUESTION # 82
Given:
java
var now = LocalDate.now();
var format1 = new DateTimeFormatter(ISO_WEEK_DATE);
var format2 = DateTimeFormatter.ISO_WEEK_DATE;
var format3 = new DateFormat(WEEK_OF_YEAR_FIELD);
var format4 = DateFormat.getDateInstance(WEEK_OF_YEAR_FIELD);
System.out.println(now.format(REPLACE_HERE));
Which variable prints 2025-W01-2 (present-day is 12/31/2024)?

  • A. format3
  • B. format4
  • C. format1
  • D. format2

Answer: D

Explanation:
In this code, now is assigned the current date using LocalDate.now(). The goal is to format this date to the ISO week date format, which represents dates in the YYYY-'W'WW-E pattern, where:
* YYYY: Week-based year
* 'W': Literal 'W' character
* WW: Week number
* E: Day of the week
Given that the present day is December 31, 2024, this date falls in the first week of the week-based year 2025.
Therefore, the ISO week date representation would be 2025-W01-2, where '2' denotes Tuesday.
Among the provided formatters:
* format1: This line attempts to create a DateTimeFormatter using a constructor, which is incorrect because DateTimeFormatter does not have a public constructor that accepts a pattern directly. This would result in a compilation error.
* format2: This is correctly assigned the predefined DateTimeFormatter.ISO_WEEK_DATE, which formats dates in the ISO week date format.
* format3: This line attempts to create a DateFormat instance using a field, which is incorrect because DateFormat does not have such a constructor. This would result in a compilation error.
* format4: This line attempts to get a DateFormat instance using an integer field, which is incorrect because DateFormat.getDateInstance() does not accept such parameters. This would result in a compilation error.
Therefore, the only correct and applicable formatter is format2. Using format2 in the now.format() method will produce the desired output: 2025-W01-2.


NEW QUESTION # 83
Given:
java
public static void main(String[] args) {
try {
throw new IOException();
} catch (IOException e) {
throw new RuntimeException();
} finally {
throw new ArithmeticException();
}
}
What is the output?

  • A. Compilation fails
  • B. IOException
  • C. RuntimeException
  • D. ArithmeticException

Answer: D

Explanation:
In this code, the try block throws an IOException. The catch block catches this exception and throws a new RuntimeException. Regardless of exceptions thrown in the try or catch blocks, the finally block is always executed. In this case, the finally block throws an ArithmeticException.
When an exception is thrown in a finally block, it overrides any previous exceptions that were thrown in the try or catch blocks. Therefore, the ArithmeticException thrown in the finally block is the exception that propagates out of the method. As a result, the program terminates with an ArithmeticException.


NEW QUESTION # 84
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 ,hi ]
  • B. [ ,hello,h i]
  • C. [,hello,h i]
  • D. [,hello,hi]

Answer: C

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 # 85
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. false
  • B. true
  • C. Compilation fails
  • D. 3.3
  • E. An exception is thrown at runtime

Answer: C

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 # 86
Which of the following statements oflocal variables declared with varareinvalid?(Choose 4)

  • A. var e;
  • B. var h = (g = 7);
  • C. var f = { 6 };
  • D. var b = 2, c = 3.0;
  • E. var a = 1;(Valid: var correctly infers int)
  • F. var d[] = new int[4];

Answer: A,C,D,F

Explanation:
1. Valid Use Cases of var
* var is alocal variable type inferencefeature.
* The compilerinfers the type from the assigned value.
* Example of valid use:
java
var a = 10; // Type inferred as int
var str = "Hello"; // Type inferred as String
2. Analyzing the Given Statements
Statement
Valid/Invalid
Reason
var a = 1;
Valid
Type inferred as int.
var b = 2, c = 3.0;
#Invalid
var doesnot allow multiple declarationsin one statement.
var d[] = new int[4];
#Invalid
Array brackets []are not allowedwith var.
var e;
#Invalid
varrequires an initializer(cannot be declared without assignment).
var f = { 6 };
#Invalid
{ 6 } is anarray initializer, which must have an explicit type.
var h = (g = 7);
Valid
g is assigned 7, and h gets its value.
Thus, the correct answers are:B, C, D, E
References:
* Java SE 21 - Local Variable Type Inference (var)
* Java SE 21 - var Restrictions


NEW QUESTION # 87
......

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