信号量会阻止指令重新排序吗? [英] Do semaphores prevent instruction reordering?

问题描述

我一直在寻找C#中可以等待的等效锁语句.有人建议通过以下方式使用二进制SemaphoreSlim:

I was looking for an awaitable equivalent of lock statements in C#. Some people suggest using a binary SemaphoreSlim in the following way:

await semaphore.WaitAsync().ConfigureAwait(false);
try
{
    //inner instructions
}
finally
{
    semaphore.Release();
}

我知道它存在一些问题(例如，它不是可重入的)，但是我最大的担心是指令的重新排序.

I know it has some issues (e.g. it's not reentrant), but my biggest concern is with the instruction reeordering.

在普通的旧锁语句中，我们保证不会将锁的内部指令移出锁语句之前(或之后).这种信号量解决方案是否也一样?据我所见，文档没有提到此问题.

In plain old lock statements we have a guarantee that no inner instruction from the lock will be moved outside (before or after) the lock statement. Does the same hold for this semaphore solution? As far as I can see, the documentation doesn't mention this problem.

推荐答案

SemaphoreSlim和几乎所有其他同步构造都是使用Monitor(或其他基于Monitor)内部，这正是lock的实现方式，为您提供相同的保证.

SemaphoreSlim, and pretty much all of the other synchronization constructs, are built using a Monitor (or other types that are built on top of a Monitor) internally, which is exactly how a lock is implemented, giving you the same guarantees.

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