1z0-830考試資訊 & 1z0-830題庫下載

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最新的 Java SE 1z0-830 免費考試真題 (Q52-Q57):

問題 #52
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.run(task1);
B. execService.run(task2);
C. execService.submit(task1);
D. execService.submit(task2);
E. execService.call(task1);
F. execService.execute(task1);
G. execService.call(task2);
H. execService.execute(task2);

答案：C,D

解題說明：
* 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

問題 #53
Given:
java
var _ = 3;
var $ = 7;
System.out.println(_ + $);
What is printed?

A. _$
B. It throws an exception.
C. Compilation fails.
D. 0

答案：C

解題說明：
* The var keyword and identifier rules:
* The var keyword is used for local variable type inference introduced inJava 10.
* However,Java does not allow _ (underscore) as an identifiersinceJava 9.
* If we try to use _ as a variable name, the compiler will throw an error:
pgsql
error: as of release 9, '_' is a keyword, and may not be used as an identifier
* The $ symbol as an identifier:
* The $ characteris a valid identifierin Java.
* However, since _ is not allowed, the codefails to compile before even reaching $.
Thus,the correct answer is "Compilation fails."
References:
* Java SE 21 - var Local Variable Type Inference
* Java SE 9 - Restrictions on _ Identifier

問題 #54
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. 0
B. Optional.empty
C. Optional[1]
D. 1
E. 2
F. Compilation fails
G. An exception is thrown

答案：D

解題說明：
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.

問題 #55
Given:
java
public class Test {
static int count;
synchronized Test() {
count++;
}
public static void main(String[] args) throws InterruptedException {
Runnable task = Test::new;
Thread t1 = new Thread(task);
Thread t2 = new Thread(task);
t1.start();
t2.start();
t1.join();
t2.join();
System.out.println(count);
}
}
What is the given program's output?

A. It's either 1 or 2
B. It's always 1
C. Compilation fails
D. It's always 2
E. It's either 0 or 1

答案：C

解題說明：
In this code, the Test class has a static integer field count and a constructor that is declared with the synchronized modifier. In Java, the synchronized modifier can be applied to methods to control access to critical sections, but it cannot be applied directly to constructors. Attempting to declare a constructor as synchronized will result in a compilation error.
Compilation Error Details:
The Java Language Specification does not permit the use of the synchronized modifier on constructors.
Therefore, the compiler will produce an error indicating that the synchronized modifier is not allowed in this context.
Correct Usage:
If you need to synchronize the initialization of instances, you can use a synchronized block within the constructor:
java
public class Test {
static int count;
Test() {
synchronized (Test.class) {
count++;
}
}
public static void main(String[] args) throws InterruptedException {
Runnable task = Test::new;
Thread t1 = new Thread(task);
Thread t2 = new Thread(task);
t1.start();
t2.start();
t1.join();
t2.join();
System.out.println(count);
}
}
In this corrected version, the synchronized block within the constructor ensures that the increment operation on count is thread-safe.
Conclusion:
The original program will fail to compile due to the illegal use of the synchronized modifier on the constructor. Therefore, the correct answer is E: Compilation fails.

問題 #56
Which of the following statements is correct about a final class?

A. It must contain at least a final method.
B. It cannot be extended by any other class.
C. It cannot extend another class.
D. The final keyword in its declaration must go right before the class keyword.
E. It cannot implement any interface.

答案：B

解題說明：
In Java, the final keyword can be applied to classes, methods, and variables to impose certain restrictions.
Final Classes:
* Definition:A class declared with the final keyword is known as a final class.
* Purpose:Declaring a class as final prevents it from being subclassed. This is useful when you want to ensure that the class's implementation remains unchanged and cannot be extended or modified through inheritance.
Option Evaluations:
* A. The final keyword in its declaration must go right before the class keyword.
* This is correct. The syntax for declaring a final class is:
java
public final class ClassName {
// class body
}
* However, this statement is about syntax rather than the core characteristic of a final class.
* B. It must contain at least a final method.
* Incorrect. A final class can have zero or more methods, and none of them are required to be declared as final. The final keyword at the class level prevents inheritance, regardless of the methods' finality.
* C. It cannot be extended by any other class.
* Correct. The primary characteristic of a final class is that it cannot be subclassed. Attempting to do so will result in a compilation error.
* D. It cannot implement any interface.
* Incorrect. A final class can implement interfaces. Declaring a class as final restricts inheritance but does not prevent the class from implementing interfaces.
* E. It cannot extend another class.
* Incorrect. A final class can extend another class. The final keyword prevents the class from being subclassed but does not prevent it from being a subclass itself.
Therefore, the correct statement about a final class is option C: "It cannot be extended by any other class."

問題 #57
......

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1z0-830考試資訊 & 1z0-830題庫下載

1z0-830考試資訊 - 你通過Java SE 21 Developer Professional的強大武器

最新的 Java SE 1z0-830 免費考試真題 (Q52-Q57):

Oracle 1z0-830考試資訊：Java SE 21 Developer Professional＆認證成功保證，簡單的培訓方式

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