c# - 最简单的并发模式

Question

请您帮我提醒一下最简单的并行编程技术之一。

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

它不是互斥体，因为没有临界区，或者更确切地说，你可以说有一个无限的临界区。那么它是什么？

Answer 1

关于“如何”...好吧，你可以使用 Semaphore , Mutex或重置事件( ManualResetEvent 、 AutoResetEvent )，但我个人使用 Monitor ；

object sync = new object();

线程 1:

lock(sync) { // blocks until it successfully takes (re-entrant) lock on sync
    // here: launch thread 2, to ensure thread 1 has the lock first
    Monitor.Wait(sync); // releases lock(s) and waits for pulse,
                        // then retakes locks(s) when free
} // releases lock (or decrement by 1)

线程 2:

lock(sync) { // blocks until it successfully takes (re-entrant) lock on sync
    Monitor.Pulse(sync); // moves a waiting thread into the ready-queue
                         // (but the awoken thread still can't continue until
                         // it gets the lock itself - and we still hold it)
} // releases lock (or decrement by 1)