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Oracle Java SE 21 Developer Professional Sample Questions (Q76-Q81):

NEW QUESTION # 76
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. abc
B. Compilation fails.
C. An exception is thrown.
D. ABC

Answer: A

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 # 77
Given:
java
var counter = 0;
do {
System.out.print(counter + " ");
} while (++counter < 3);
What is printed?

A. Compilation fails.
B. 0 1 2 3
C. 1 2 3
D. An exception is thrown.
E. 0 1 2
F. 1 2 3 4

Answer: E

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 # 78
Which of the following statements oflocal variables declared with varareinvalid?(Choose 4)

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

Answer: B,C,E,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 # 79
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 throws an exception at runtime.
B. null
C. It's a string with value: 42
D. Compilation fails.
E. It's a double with value: 42
F. It's an integer with value: 42

Answer: D

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 # 80
Given:
java
Period p = Period.between(
LocalDate.of(2023, Month.MAY, 4),
LocalDate.of(2024, Month.MAY, 4));
System.out.println(p);
Duration d = Duration.between(
LocalDate.of(2023, Month.MAY, 4),
LocalDate.of(2024, Month.MAY, 4));
System.out.println(d);
What is the output?

A. P1Y
PT8784H
B. UnsupportedTemporalTypeException
C. P1Y
UnsupportedTemporalTypeException
D. PT8784H
P1Y

Answer: C

Explanation:
In this code, two LocalDate instances are created representing May 4, 2023, and May 4, 2024. The Period.
between() method is used to calculate the period between these two dates, and the Duration.between() method is used to calculate the duration between them.
Period Calculation:
The Period.between() method calculates the amount of time between two LocalDate objects in terms of years, months, and days. In this case, the period between May 4, 2023, and May 4, 2024, is exactly one year.
Therefore, p is P1Y, which stands for a period of one year. Printing p will output P1Y.
Duration Calculation:
The Duration.between() method is intended to calculate the duration between two temporal objects that have time components, such as LocalDateTime or Instant. However, LocalDate represents a date without a time component. Attempting to use Duration.between() with LocalDate instances will result in an UnsupportedTemporalTypeException because Duration requires time-based units, which LocalDate does not support.
Exception Details:
The UnsupportedTemporalTypeException is thrown when an unsupported unit is used. In this case, Duration.
between() internally attempts to access time-based fields (like seconds), which are not supported by LocalDate. This behavior is documented in the Java Bug System underJDK-8170275.
Correct Usage:
To calculate the duration between two dates, including time components, you should use LocalDateTime or Instant. For example:
java
LocalDateTime start = LocalDateTime.of(2023, Month.MAY, 4, 0, 0);
LocalDateTime end = LocalDateTime.of(2024, Month.MAY, 4, 0, 0);
Duration d = Duration.between(start, end);
System.out.println(d); // Outputs: PT8784H
This will correctly calculate the duration as PT8784H, representing 8,784 hours (which is 366 days, accounting for a leap year).
Conclusion:
The output of the given code will be:
pgsql
P1Y
Exception in thread "main" java.time.temporal.UnsupportedTemporalTypeException: Unsupported unit:
Seconds
Therefore, the correct answer is D:
nginx
P1Y
UnsupportedTemporalTypeException

NEW QUESTION # 81
......

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Actual4Labs Oracle 1z0-830 Exam Dumps and Practice Test Software

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Oracle Java SE 21 Developer Professional Sample Questions (Q76-Q81):

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