1z0-830 Exam Introduction | Test 1z0-830 Simulator Free

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

NEW QUESTION # 27
Given:
java
CopyOnWriteArrayList<String> list = new CopyOnWriteArrayList<>();
list.add("A");
list.add("B");
list.add("C");
// Writing in one thread
new Thread(() -> {
list.add("D");
System.out.println("Element added: D");
}).start();
// Reading in another thread
new Thread(() -> {
for (String element : list) {
System.out.println("Read element: " + element);
}
}).start();
What is printed?

A. It prints all elements, but changes made during iteration may not be visible.
B. It throws an exception.
C. It prints all elements, including changes made during iteration.
D. Compilation fails.

Answer: A

Explanation:
* Understanding CopyOnWriteArrayList
* CopyOnWriteArrayList is a thread-safe variant of ArrayList whereall mutative operations (add, set, remove, etc.) create a new copy of the underlying array.
* This meansiterations will not reflect modifications made after the iterator was created.
* Instead of modifying the existing array, a new copy is created for modifications, ensuring that readers always see a consistent snapshot.
* Thread Execution Behavior
* Thread 1 (Writer Thread)adds "D" to the list.
* Thread 2 (Reader Thread)iterates over the list.
* The reader thread gets a snapshot of the listbefore"D" is added.
* The output may look like:
mathematica
Read element: A
Read element: B
Read element: C
Element added: D
* "D" may not appear in the output of the reader threadbecause the iteration occurs on a snapshot before the modification.
* Why doesn't it print all elements including changes?
* Since CopyOnWriteArrayList doesnot allow changes to be visible during iteration, the reader threadwill not see "D"if it started iterating before "D" was added.
Thus, the correct answer is:"It prints all elements, but changes made during iteration may not be visible." References:
* Java SE 21 - CopyOnWriteArrayList

NEW QUESTION # 28
Given:
java
String colors = "redn" +
"greenn" +
"bluen";
Which text block can replace the above code?

A. java
String colors = """
red t
greent
blue t
""";
B. java
String colors = """
red
green
blue
""";
C. java
String colors = """
red
green
blue
""";
D. java
String colors = """
red s
greens
blue s
""";
E. None of the propositions

Answer: C

Explanation:
* Understanding Multi-line Strings in Java (""" Text Blocks)
* Java 13 introducedtext blocks ("""), allowing multi-line stringswithout needing explicit n for new lines.
* In a text block,each line is preserved as it appears in the source code.
* Analyzing the Options
* Option A: (Backslash Continuation)
* The backslash () at the end of a lineprevents a new line from being added, meaning:
nginx
red green blue
* Incorrect.
* Option B: s (Whitespace Escape)
* s represents asingle space,not a new line.
* The output would be:
nginx
red green blue
* Incorrect.
* Option C: t (Tab Escape)
* t inserts atab, not a new line.
* The output would be:
nginx
red green blue
* Incorrect.
* Option D: Correct Text Block
java
String colors = """
red
green
blue
""";
* Thispreserves the new lines, producing:
nginx
red
green
blue
* Correct.
Thus, the correct answer is:"String colors = """ red green blue """."
References:
* Java SE 21 - Text Blocks
* Java SE 21 - String Formatting

NEW QUESTION # 29
Given:
java
Stream<String> strings = Stream.of("United", "States");
BinaryOperator<String> operator = (s1, s2) -> s1.concat(s2.toUpperCase()); String result = strings.reduce("-", operator); System.out.println(result); What is the output of this code fragment?

A. -UNITEDSTATES
B. United-STATES
C. United-States
D. UNITED-STATES
E. -UnitedStates
F. UnitedStates
G. -UnitedSTATES

Answer: G

Explanation:
In this code, a Stream of String elements is created containing "United" and "States". A BinaryOperator<String> named operator is defined to concatenate the first string (s1) with the uppercase version of the second string (s2). The reduce method is then used with "-" as the identity value and operator as the accumulator.
The reduce method processes the elements of the stream as follows:
* Initial Identity Value: "-"
* First Iteration:
* Accumulator Operation: "-".concat("United".toUpperCase())
* Result: "-UNITED"
* Second Iteration:
* Accumulator Operation: "-UNITED".concat("States".toUpperCase())
* Result: "-UNITEDSTATES"
Therefore, the final result stored in result is "-UNITEDSTATES", and the output of theSystem.out.println (result); statement is -UNITEDSTATES.

NEW QUESTION # 30
Given:
java
ExecutorService service = Executors.newFixedThreadPool(2);
Runnable task = () -> System.out.println("Task is complete");
service.submit(task);
service.shutdown();
service.submit(task);
What happens when executing the given code fragment?

A. It prints "Task is complete" twice, then exits normally.
B. It exits normally without printing anything to the console.
C. It prints "Task is complete" twice and throws an exception.
D. It prints "Task is complete" once, then exits normally.
E. It prints "Task is complete" once and throws an exception.

Answer: E

Explanation:
In this code, an ExecutorService is created with a fixed thread pool of size 2 using Executors.
newFixedThreadPool(2). A Runnable task is defined to print "Task is complete" to the console.
The sequence of operations is as follows:
* service.submit(task);
This submits the task to the executor service for execution. Since the thread pool has a size of 2 and no other tasks are running, this task will be executed promptly, printing "Task is complete" to the console.
* service.shutdown();
This initiates an orderly shutdown of the executor service. In this state, the service stops accepting new tasks

NEW QUESTION # 31
Given:
java
public class ExceptionPropagation {
public static void main(String[] args) {
try {
thrower();
System.out.print("Dom Perignon, ");
} catch (Exception e) {
System.out.print("Chablis, ");
} finally {
System.out.print("Saint-Emilion");
}
}
static int thrower() {
try {
int i = 0;
return i / i;
} catch (NumberFormatException e) {
System.out.print("Rose");
return -1;
} finally {
System.out.print("Beaujolais Nouveau, ");
}
}
}
What is printed?

A. Beaujolais Nouveau, Chablis, Saint-Emilion
B. Saint-Emilion
C. Rose
D. Beaujolais Nouveau, Chablis, Dom Perignon, Saint-Emilion

Answer: A

Explanation:
* Analyzing the thrower() Method Execution
java
int i = 0;
return i / i;
* i / i evaluates to 0 / 0, whichthrows ArithmeticException (/ by zero).
* Since catch (NumberFormatException e) doesnot matchArithmeticException, it is skipped.
* The finally block always executes, printing:
nginx
Beaujolais Nouveau,
* The exceptionpropagates backto main().
* Handling the Exception in main()
java
try {
thrower();
System.out.print("Dom Perignon, ");
} catch (Exception e) {
System.out.print("Chablis, ");
} finally {
System.out.print("Saint-Emilion");
}
* Since thrower() throws ArithmeticException, it is caught by catch (Exception e).
* "Chablis, "is printed.
* Thefinally block always executes, printing "Saint-Emilion".
* Final Output
nginx
Beaujolais Nouveau, Chablis, Saint-Emilion
Thus, the correct answer is:Beaujolais Nouveau, Chablis, Saint-Emilion
References:
* Java SE 21 - Exception Handling
* Java SE 21 - finally Block Execution

NEW QUESTION # 32
......

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