semaphore

I have a script in python which uses a resource which can not be used by more than a certain amount of concurrent scripts running. Classically, this would be solved by a named semaphores but I can not find those in the documentation of the multiprocessing module or threading . Am I missing something or are named semaphores not implemented / exposed by Python? and more importantly, if the answer is no, what is the best way to emulate one? Thanks, Boaz PS. For reasons which are not so relevant to this question, I can not aggregate the task to a continuously running process/daemon or work with

CountDownLatch CountDownLatch是一个计数器闭锁，通过它可以完成类似于阻塞当前线程的功能，即：一个线程或多个线程一直等待，直到其他线程执行的操作完成。CountDownLatch用一个给定的计数器来初始化，该计数器的操作是原子操作，即同时只能有一个线程去操作该计数器。调用该类await方法的线程会一直处于阻塞状态，直到其他线程调用countDown方法使当前计数器的值变为零，每次调用countDown计数器的值减1。当计数器值减至零时，所有因调用await()方法而处于等待状态的线程就会继续往下执行。这种现象只会出现一次，因为计数器不能被重置，如果业务上需要一个可以重置计数次数的版本，可以考虑使用CycliBarrier。 在某些业务场景中，程序执行需要等待某个条件完成后才能继续执行后续的操作；下面举个栗子，比如秦国需要灭六国才能一统华夏。 代码示例： public class CountDownLatchDemo { public static void main(String[] args) throws InterruptedException { CountDownLatch countDownLatch = new CountDownLatch(6); for (int i = 1; i <= 6; i++) { new Thread(() ->

版权声明：本文为博主转载文章 原文链接： https://blog.csdn.net/lixiaobuaa/article/details/78995572 线程之间相互合作时，需要用到同步器来完成同步，下面介绍几种常用同步器： 1.Semaphore（信号量） 信号量数量限制了访问资源的线程总数，线程请求会消耗一个信号量，当信号量为0时，新的线程会阻塞，直到有线程释放了一个信号量 线程类： package Semaphore; import java.util.concurrent.Semaphore; public class SemaphoreThread extends Thread { Semaphore semaphore; public SemaphoreThread (Semaphore semaphore){ this .semaphore = semaphore; } @Override public void run () { try { semaphore.acquire(); System.out.println( "一个线程正在执行" ); sleep( 3000 ); System.out.println( "一个线程结束运行" ); } catch (InterruptedException e) { e.printStackTrace(); }

when a C# program holds a named semaphore, it does not seem to be released when the application is terminated early (for example by pressing Ctrl+C or closing the console window). At least not until all instances of the process have terminated. With a named mutex an AbandonedMutexException is raised in this case but not with a semaphore. How do you prevent one program instance from stalling when another program instance has been terminated early? class Program { // Same with count > 1 private static Semaphore mySemaphore = new Semaphore(1, 1, "SemaphoreTest"); static void Main(string[] args) {

I'm trying to understand the similarities and differences between named and unnamed semaphore so my google searches yielded me this . I had a question about the wording on the page though, it says: Unnamed semaphores might be usable by more than one process Named semaphores are sharable by several processes Do those two words create any important distinction between those two types of semaphores or are they irrelevant? So so far here's what I have: Similarities -Several processes can do something with the semaphore Difference -Named are referenced with pathname and unnamed are referenced by

Why am I deadlocking? - (void)foo { static dispatch_once_t onceToken; dispatch_once(&onceToken, ^{ [self foo]; }); // whatever... } I expect foo to be executed twice on first call. Neither of the existing answers are quite accurate (one is dead wrong, the other is a bit misleading and misses some critical details). First, let's go right to the source : void dispatch_once_f(dispatch_once_t *val, void *ctxt, dispatch_function_t func) { struct _dispatch_once_waiter_s * volatile *vval = (struct _dispatch_once_waiter_s**)val; struct _dispatch_once_waiter_s dow = { NULL, 0 }; struct _dispatch_once