Consider this example (typical in OOP books): I have an Animal class, where each Animal can have many friends. And subclasses like Dog , Duck , Mouse etc which add specific behavior like bark() , quack() etc. Here's the Animal class: public class Animal { private Map<String,Animal> friends = new HashMap<>(); public void addFriend(String name, Animal animal){
考虑这个例子（典型的OOP书籍）： 我有一个Animal课，每个Animal可以有很多朋友。 像Dog ， Duck ， Mouse等子类可以添加特定的行为，如bark() ， quack()等。 这是Animal类： public class Animal { private Map<String,Animal> friends = new HashMap<>(); public void addFriend(String name, Animal animal){ friends.put(name,animal); } public Animal callFriend(String nam
Here is a sample code package org.example; import java.lang.reflect.Method; class TestRef { public void testA(String ... a) { for (String i : a) { System.out.println(i); } } public static void main(String[] args){ Class testRefClass = TestRef.class; for (Method m: testRefClass.ge
这是一个示例代码 package org.example; import java.lang.reflect.Method; class TestRef { public void testA(String ... a) { for (String i : a) { System.out.println(i); } } public static void main(String[] args){ Class testRefClass = TestRef.class; for (Method m: testRefClass.getMeth
If we check the invokeAll method in AbstractExecutorService class, then we can see that following things are happening: Submission of the tasks in a for loop Running of another for loop. And in case the future is not completed , then do a get on it (to make sure that all the futures are finished before returning from the invokeAll method) If we check the get method in CompletableFuture, th
如果我们检查AbstractExecutorService类中的invokeAll方法，那么我们可以看到以下事情正在发生： 在for循环中提交任务 运行另一个for循环。 如果未来还没有完成 ，那就去做吧（为了确保所有的期货在从invokeAll方法返回之前完成） 如果我们检查CompletableFuture中的get方法，那么我们看到get方法将在计算完成时立即返回结果，否则将等待结果可用。 我的问题是 - 检查点（2） 是否使用未来多余的isDone（）方法检查完成
I am struggling with an issue where, after working smoothly for most of the the day, a callable task is put into a Java Single Thread Executor and apparently never get's executed. Subsequent calls to submit a new task fail and the ExecutorService seems to be dead. At this point the client producing the tasks is out of service until the process can be restarted which is not possible during b
我正在努力解决一个问题，在一天中的大部分时间工作顺利之后，一个可调用的任务被放入一个Java单线程执行程序中，并且显然永远不会执行。 随后调用提交新任务失败， ExecutorService似乎已经死亡。 此时，生产任务的客户端将停止服务，直到流程可以重新启动，这在工作时间内是不可能的。 一些背景：多个高吞吐量生产者线程将他们的任务放到他们自己专用的Single Thread ExecutorService并立即返回。 低延迟对生产者线程非常
I have around 900 tasks. I have set threadpool count to be 50. I am running a loop and submitting every task to the executorService. As soon as control comes out I call shutdown as below for (Entry<String, String> CurrentJob : Tasks.entrySet()) { m_service.submit(new MyTask(CurrentJob.getValue(), CurrentJob.getKey())); } m_service.shutdown(); Each task tak
我有大约900个任务。 我已将threadpool count设置为50.我正在运行一个循环并将每个任务提交给executorService。 一旦控制出现，我会按照如下所述调用shutdown for (Entry<String, String> CurrentJob : Tasks.entrySet()) { m_service.submit(new MyTask(CurrentJob.getValue(), CurrentJob.getKey())); } m_service.shutdown(); 每个任务的平均时间大约为1秒。现在我有两个问题 a）
So I have some Callable tasks, sensitive to interruptions , which I submit to the ExecutorService using invokeAll. After 5 seconds from another method I call executorService.shutdownNow after which I call the awaitTermination, which returns true, so all seems good. The problem is the executor never terminates. Due to logging I know that each one of my tasks finished. nevertheless the invokeA
所以我有一些Callable任务， 对中断很敏感 ，我使用invokeAll提交给ExecutorService。 从另一个方法5秒钟后，我调用executorService.shutdownNow，之后我调用awaitTermination，它返回true，所以一切似乎都很好。 问题是执行者永远不会终止。 由于记录，我知道我的每个任务都完成了。 不过，当我等于执行程序的线程数时，invokeAll仍会阻塞f.get： 下面的代码是从AbstractExecutorService +一些日志记录中获得的。
I have created one callable which prints couple of lines and sleeps for 2 seconds. I have created a main method which creates 10 instances of this callable and pass to invokeALL method of ExecutorService. service.invokeAll(callableList, 3, SECONDS); when I am iterating returned List of Future objects. I am getting CancellationException. I am testing whether all future.get() call would caus
我创建了一个可打印几行并可睡2秒的可调用函数。 我创建了一个主方法，该方法创建10个可调用实例并传递给ExecutorService的invokeALL方法。 service.invokeAll(callableList, 3, SECONDS); 当我迭代未来对象的返回列表。 我得到CancellationException。 我正在测试所有future.get（）调用是否会导致CancellationException或只有那些可能无法完成并被取消的任务。 无论invokeALL中的时间设置如何，我都会得到所有结果或
When are writes that are performed by one thread visible to a different thread when there are no synchronized blocks and no volatile variables? Here is a simplified quicksort example: int middle = partitionForTheFirstTime(array); Thread t = new Thread(new Quicksorter(array, 0, middle - 1)); Thread u = new Thread(new Quicksorter(array, middle + 1, array.size - 1)); t.start() u.start(); t.join
何时是由一个线程执行的写入在不存在同步块且没有易失性变量的情况下对不同线程可见？ 这是一个简化的快速排序示例： int middle = partitionForTheFirstTime(array); Thread t = new Thread(new Quicksorter(array, 0, middle - 1)); Thread u = new Thread(new Quicksorter(array, middle + 1, array.size - 1)); t.start() u.start(); t.join(); u.join(); （为了简单起见，假定两个“工作线程”不会产生任何额外的线程。
Say I have a queue full of tasks which I need to submit to an executor service. I want them processed one at a time. The simplest way I can think of is to: Take a task from the queue Submit it to the executor Call .get on the returned Future and block until a result is available Take another task from the queue... However, I am trying to avoid blocking completely. If I have 10,000 su
假设我有一个队列充满了我需要提交给执行者服务的任务。 我希望他们一次处理一个。 我能想到的最简单的方法是： 从队列中取一个任务 提交给执行者 在返回的Future上调用.get，并阻止，直到结果可用 从队列中取另一个任务... 不过，我试图避免完全阻止。 如果我有10,000个这样的队列，它们需要一次处理它们的任务，那么我将耗尽堆栈空间，因为它们中的大多数将持续阻塞线程。 我想提交一个任务并提供一个回调，当
In Java, when we try to do pattern matching using a regular expression. eg take a input string and use regular expression to find out if it is numeric. If not, throw an exception. In this case, I understand, using regex makes the code less verbose than if we were to take each character of the string, check if it is a number and if not throw an exception. But I was under the assumption that r
在Java中，当我们尝试使用正则表达式进行模式匹配时。 例如，输入一个字符串并使用正则表达式来确定它是否是数字。 如果不是，则抛出异常。 在这种情况下，我明白，使用正则表达式可以减少代码冗长，比如果我们要获取字符串的每个字符，检查它是否是数字，是否抛出异常。 但我认为正则表达式也使得这个过程更有效率。 这是真的？ 在这一点上我找不到任何证据。 正则表达式如何在幕后进行比赛？ 它是不是也遍历字符串并