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Oracle Java SE 21 Developer Professional Sample Questions (Q35-Q40):
NEW QUESTION # 35
Which of the followingisn'ta correct way to write a string to a file?
- A. java
try (FileWriter writer = new FileWriter("file.txt")) {
writer.write("Hello");
} - B. java
try (FileOutputStream outputStream = new FileOutputStream("file.txt")) { byte[] strBytes = "Hello".getBytes(); outputStream.write(strBytes);
} - C. None of the suggestions
- D. java
Path path = Paths.get("file.txt");
byte[] strBytes = "Hello".getBytes();
Files.write(path, strBytes); - E. java
try (PrintWriter printWriter = new PrintWriter("file.txt")) {
printWriter.printf("Hello %s", "James");
} - F. java
try (BufferedWriter writer = new BufferedWriter("file.txt")) {
writer.write("Hello");
}
Answer: F
Explanation:
(BufferedWriter writer = new BufferedWriter("file.txt") is incorrect.)
Theincorrect statementisoption Bbecause BufferedWriterdoes nothave a constructor that accepts a String (file name) directly. The correct way to use BufferedWriter is to wrap it around a FileWriter, like this:
java
try (BufferedWriter writer = new BufferedWriter(new FileWriter("file.txt"))) { writer.write("Hello");
}
Evaluation of Other Options:
Option A (Files.write)# Correct
* Uses Files.write() to write bytes to a file.
* Efficient and concise method for writing small text files.
Option C (FileOutputStream)# Correct
* Uses a FileOutputStream to write raw bytes to a file.
* Works for both text and binary data.
Option D (PrintWriter)# Correct
* Uses PrintWriter for formatted text output.
Option F (FileWriter)# Correct
* Uses FileWriter to write text data.
Option E (None of the suggestions)# Incorrect becauseoption Bis incorrect.
NEW QUESTION # 36
Given:
java
Integer frenchRevolution = 1789;
Object o1 = new String("1789");
Object o2 = frenchRevolution;
frenchRevolution = null;
Object o3 = o2.toString();
System.out.println(o1.equals(o3));
What is printed?
- A. A ClassCastException is thrown.
- B. Compilation fails.
- C. true
- D. A NullPointerException is thrown.
- E. false
Answer: C
Explanation:
* Understanding Variable Assignments
java
Integer frenchRevolution = 1789;
Object o1 = new String("1789");
Object o2 = frenchRevolution;
frenchRevolution = null;
* frenchRevolution is an Integer with value1789.
* o1 is aString with value "1789".
* o2 storesa reference to frenchRevolution, which is an Integer (1789).
* frenchRevolution = null;only nullifies the reference, but o2 still holds the Integer 1789.
* Calling toString() on o2
java
Object o3 = o2.toString();
* o2 refers to an Integer (1789).
* Integer.toString() returns theString representation "1789".
* o3 is assigned "1789" (String).
* Evaluating o1.equals(o3)
java
System.out.println(o1.equals(o3));
* o1.equals(o3) isequivalent to:
java
"1789".equals("1789")
* Since both areequal strings, the output is:
arduino
true
Thus, the correct answer is:true
References:
* Java SE 21 - Integer.toString()
* Java SE 21 - String.equals()
NEW QUESTION # 37
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 exits normally without printing anything to the console.
- B. It prints "Task is complete" once, then exits normally.
- C. It prints "Task is complete" twice and throws an exception.
- D. It prints "Task is complete" once and throws an exception.
- E. It prints "Task is complete" twice, then exits normally.
Answer: D
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 # 38
Given:
java
Runnable task1 = () -> System.out.println("Executing Task-1");
Callable<String> task2 = () -> {
System.out.println("Executing Task-2");
return "Task-2 Finish.";
};
ExecutorService execService = Executors.newCachedThreadPool();
// INSERT CODE HERE
execService.awaitTermination(3, TimeUnit.SECONDS);
execService.shutdownNow();
Which of the following statements, inserted in the code above, printsboth:
"Executing Task-2" and "Executing Task-1"?
- A. execService.call(task1);
- B. execService.execute(task2);
- C. execService.call(task2);
- D. execService.submit(task1);
- E. execService.execute(task1);
- F. execService.run(task2);
- G. execService.submit(task2);
- H. execService.run(task1);
Answer: D,G
Explanation:
* Understanding ExecutorService Methods
* execute(Runnable command)
* Runs the task but only supports Runnable (not Callable).
* #execService.execute(task2); fails because task2 is Callable<String>.
* submit(Runnable task)
* Submits a Runnable task for execution.
* execService.submit(task1); executes "Executing Task-1".
* submit(Callable<T> task)
* Submits a Callable<T> task for execution.
* execService.submit(task2); executes "Executing Task-2".
* call() Does Not Exist in ExecutorService
* #execService.call(task1); and execService.call(task2); are invalid.
* run() Does Not Exist in ExecutorService
* #execService.run(task1); and execService.run(task2); are invalid.
* Correct Code to Print Both Messages:
java
execService.submit(task1);
execService.submit(task2);
Thus, the correct answer is:execService.submit(task1); execService.submit(task2); References:
* Java SE 21 - ExecutorService
* Java SE 21 - Callable and Runnable
NEW QUESTION # 39
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. ArithmeticException
- B. IOException
- C. Compilation fails
- D. RuntimeException
Answer: A
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 # 40
......
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