最简单的并发模式
请您帮我提醒我最简单的并行编程技术之一。
我如何在 C# 中执行以下操作:
初始状态:
semaphore counter = 0
线程 1:
// Block until semaphore is signalled
semaphore.Wait(); // wait until semaphore counter is 1
线程 2:
// Allow thread 1 to run:
semaphore.Signal(); // increments from 0 to 1
它不是互斥锁,因为没有临界区,或者更确切地说,您可以说存在无限临界区。那么它是什么?
Please, would you help me in reminding me of one of the simplest parallel programming techniques.
How do I do the following in C#:
Initial state:
semaphore counter = 0
Thread 1:
// Block until semaphore is signalled
semaphore.Wait(); // wait until semaphore counter is 1
Thread 2:
// Allow thread 1 to run:
semaphore.Signal(); // increments from 0 to 1
It's not a mutex because there is no critical section, or rather you can say there is an infinite critical section. So what is it?
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关于“如何”...好吧,您可以使用 信号量< /a>、互斥体 或重置事件 (< a href="http://msdn.microsoft.com/en-us/library/system.threading.manualresetevent.aspx" rel="nofollow">ManualResetEvent,AutoResetEvent),但我个人使用
Monitor
来实现;线程 1:
线程 2:
Re "how"... well, you could use a Semaphore, Mutex or reset-event (ManualResetEvent, AutoResetEvent), but personally I use
Monitor
for that;thread 1:
thread 2:
你写的只是一个互斥锁;互斥体与值为 1 的信号量基本相同。并且有一个“临界区”——它是任一线程持有互斥体时执行的代码序列。
What you wrote is simply a mutex; a mutex is basically the same as a semaphore with value 1. And there is a 'critical section' - it's the code sequence which is executed while either thread holds the mutex.
信号量初始化为 1,可以被视为[读取:使用]作为互斥锁。
您描述的具体场景是信号量在两个线程之间使用信号系统。
第一个线程尝试锁定资源[获取信号量],但被阻塞[因为信号量已初始化为零],直到第二个线程向线程 1 发出信号,表明他可以输入代码的其余部分。
关于该主题的一篇优秀文章是:
http://geekswithblogs.net/shahed /archive/2006/06/09/81268.aspx
A semaphore initialized to 1, can be seen [read: used] as a mutex.
The specific scenario you are describing is a semaphore used a signaling system between two threads.
The first thread tries to lock the resourse [acquire the semaphore] but gets blocked [because the semaphore has been initialized to zero], until the second thread signal thread 1 that he can enter the remaining part of the code.
An excellent article on the topic is:
http://geekswithblogs.net/shahed/archive/2006/06/09/81268.aspx