@Craig:

信号量是一种锁定 资源，这样才有保证 当执行一段代码时， 只有这段代码可以访问 该资源。这保留了两个线程 从并发访问资源， 这可能会导致问题。

这不仅仅局限于一个线程。信号量可以配置为允许固定数量的线程访问资源。

互斥:对资源的独占成员访问

信号量:对资源的n个成员访问

也就是说，互斥可以用来同步对计数器、文件、数据库等的访问。

信号量可以做同样的事情，但支持固定数量的同时调用者。例如，我可以将我的数据库调用包装在一个信号量(3)中，这样我的多线程应用程序将最多3个同时连接到数据库。所有的尝试都将被阻塞，直到三个插槽中的一个打开。它们让简单的节流变得非常简单。

硬件或软件标志。在多任务系统中，信号量是一个变量，其值表示公共资源的状态。需要资源的进程检查信号量以确定资源状态，然后决定如何继续。

互斥量只是一个布尔值，而信号量是一个计数器。

两者都用于锁定部分代码，这样就不会有太多线程访问它。

例子

lock.set()
a += 1
lock.unset()

现在，如果lock是一个互斥锁，这意味着无论有多少线程尝试访问受保护的代码片段，它将始终处于锁定或解锁状态(表面下是一个布尔值)。当被锁定时，任何其他线程都会等待它被前一个线程解锁/取消设置。

现在想象一下，如果lock在引擎盖下是一个具有预定义MAX值的计数器(在我们的例子中是2)。然后，如果有两个线程试图访问该资源，那么lock的值将增加到2。如果第三个线程试图访问它，它就会等待计数器低于2，以此类推。

如果lock作为一个信号量的最大值为1，那么它将完全作为一个互斥量。

Michael Barr的文章《互斥量和信号量揭秘》很好地介绍了互斥量和信号量的不同之处，以及什么时候应该和不应该使用它们。我在这里摘录了几个关键段落。

关键在于应该使用互斥来保护共享资源，而应该使用信号量来发送信号。通常不应该使用信号量来保护共享资源，也不应该使用互斥量来发送信号。例如，在使用信号量来保护共享资源方面，使用bouncer类比是有问题的——您可以这样使用它们，但这可能会导致难以诊断错误。

While mutexes and semaphores have some similarities in their implementation, they should always be used differently. The most common (but nonetheless incorrect) answer to the question posed at the top is that mutexes and semaphores are very similar, with the only significant difference being that semaphores can count higher than one. Nearly all engineers seem to properly understand that a mutex is a binary flag used to protect a shared resource by ensuring mutual exclusion inside critical sections of code. But when asked to expand on how to use a "counting semaphore," most engineers—varying only in their degree of confidence—express some flavor of the textbook opinion that these are used to protect several equivalent resources.

...

在这一点上，一个有趣的类比是使用浴室钥匙的想法来保护共享资源-浴室。如果一家商店只有一间浴室，那么一把钥匙就足以保护这一资源，防止多人同时使用。

如果有多个浴室，人们可能会试图用相同的键来设置多个键——这类似于误用信号量。一旦你有了一个键，你实际上不知道哪个浴室是可用的，如果你沿着这条路走下去，你可能最终会使用互斥锁来提供该信息，并确保你没有使用已经被占用的浴室。

A semaphore is the wrong tool to protect several of the essentially same resource, but this is how many people think of it and use it. The bouncer analogy is distinctly different - there aren't several of the same type of resource, instead there is one resource which can accept multiple simultaneous users. I suppose a semaphore can be used in such situations, but rarely are there real-world situations where the analogy actually holds - it's more often that there are several of the same type, but still individual resources, like the bathrooms, which cannot be used this way.

...

The correct use of a semaphore is for signaling from one task to another. A mutex is meant to be taken and released, always in that order, by each task that uses the shared resource it protects. By contrast, tasks that use semaphores either signal or wait—not both. For example, Task 1 may contain code to post (i.e., signal or increment) a particular semaphore when the "power" button is pressed and Task 2, which wakes the display, pends on that same semaphore. In this scenario, one task is the producer of the event signal; the other the consumer.

...

Here an important point is made that mutexes interfere with real time operating systems in a bad way, causing priority inversion where a less important task may be executed before a more important task because of resource sharing. In short, this happens when a lower priority task uses a mutex to grab a resource, A, then tries to grab B, but is paused because B is unavailable. While it's waiting, a higher priority task comes along and needs A, but it's already tied up, and by a process that isn't even running because it's waiting for B. There are many ways to resolve this, but it most often is fixed by altering the mutex and task manager. The mutex is much more complex in these cases than a binary semaphore, and using a semaphore in such an instance will cause priority inversions because the task manager is unaware of the priority inversion and cannot act to correct it.

...

The cause of the widespread modern confusion between mutexes and semaphores is historical, as it dates all the way back to the 1974 invention of the Semaphore (capital "S", in this article) by Djikstra. Prior to that date, none of the interrupt-safe task synchronization and signaling mechanisms known to computer scientists was efficiently scalable for use by more than two tasks. Dijkstra's revolutionary, safe-and-scalable Semaphore was applied in both critical section protection and signaling. And thus the confusion began. However, it later became obvious to operating system developers, after the appearance of the priority-based preemptive RTOS (e.g., VRTX, ca. 1980), publication of academic papers establishing RMA and the problems caused by priority inversion, and a paper on priority inheritance protocols in 1990, 3 it became apparent that mutexes must be more than just semaphores with a binary counter.

互斥:资源共享

信号:信号

在没有仔细考虑副作用的情况下，不要用一种药物来替代另一种药物。

把信号量想象成夜总会的保镖。同时允许进入俱乐部的人数是固定的。如果俱乐部满了，任何人都不允许进入，但只要一个人离开，另一个人就可以进入。

它只是一种限制特定资源的消费者数量的方法。例如，限制应用程序中对数据库的同时调用数量。

下面是一个非常适合教学的c#例子:-)

using System;
using System.Collections.Generic;
using System.Text;
using System.Threading;

namespace TheNightclub
{
    public class Program
    {
        public static Semaphore Bouncer { get; set; }

        public static void Main(string[] args)
        {
            // Create the semaphore with 3 slots, where 3 are available.
            Bouncer = new Semaphore(3, 3);

            // Open the nightclub.
            OpenNightclub();
        }

        public static void OpenNightclub()
        {
            for (int i = 1; i <= 50; i++)
            {
                // Let each guest enter on an own thread.
                Thread thread = new Thread(new ParameterizedThreadStart(Guest));
                thread.Start(i);
            }
        }

        public static void Guest(object args)
        {
            // Wait to enter the nightclub (a semaphore to be released).
            Console.WriteLine("Guest {0} is waiting to entering nightclub.", args);
            Bouncer.WaitOne();          

            // Do some dancing.
            Console.WriteLine("Guest {0} is doing some dancing.", args);
            Thread.Sleep(500);

            // Let one guest out (release one semaphore).
            Console.WriteLine("Guest {0} is leaving the nightclub.", args);
            Bouncer.Release(1);
        }
    }
}