互斥锁控制对单个共享资源的访问。它提供了获取()对资源的访问并在完成后释放()资源的操作。

信号量控制对共享资源池的访问。它提供Wait()操作，直到池中的一个资源可用，并提供Signal()操作，当它返回池时。

当一个信号量保护的资源数量大于1时，它被称为计数信号量。当它控制一个资源时，它被称为布尔信号量。布尔信号量相当于互斥量。

因此，信号量是比互斥量更高级别的抽象。互斥锁可以用信号量来实现，但不能用信号量来实现。

它们的同步语义非常不同:

互斥对象允许对给定资源的序列化访问，即多个线程等待一个锁，一次一个，正如前面所说，线程拥有锁，直到锁完成:只有这个特定的线程可以解锁它。 二进制信号量是一个值为0和1的计数器:任务阻塞在它上，直到任何任务执行sem_post。信号量宣布资源可用，并提供等待机制，直到发出可用信号。

因此，可以将互斥锁视为在任务之间传递的令牌，将信号量视为交通红灯(它向某人发出信号，表示可以继续进行)。

厕所的例子是一个有趣的类比:

Mutex: Is a key to a toilet. One person can have the key - occupy the toilet - at the time. When finished, the person gives (frees) the key to the next person in the queue. Officially: "Mutexes are typically used to serialise access to a section of re-entrant code that cannot be executed concurrently by more than one thread. A mutex object only allows one thread into a controlled section, forcing other threads which attempt to gain access to that section to wait until the first thread has exited from that section." Ref: Symbian Developer Library (A mutex is really a semaphore with value 1.) Semaphore: Is the number of free identical toilet keys. Example, say we have four toilets with identical locks and keys. The semaphore count - the count of keys - is set to 4 at beginning (all four toilets are free), then the count value is decremented as people are coming in. If all toilets are full, ie. there are no free keys left, the semaphore count is 0. Now, when eq. one person leaves the toilet, semaphore is increased to 1 (one free key), and given to the next person in the queue. Officially: "A semaphore restricts the number of simultaneous users of a shared resource up to a maximum number. Threads can request access to the resource (decrementing the semaphore), and can signal that they have finished using the resource (incrementing the semaphore)." Ref: Symbian Developer Library

You obviously use mutex to lock a data in one thread getting accessed by another thread at the same time. Assume that you have just called lock() and in the process of accessing data. This means that you don’t expect any other thread (or another instance of the same thread-code) to access the same data locked by the same mutex. That is, if it is the same thread-code getting executed on a different thread instance, hits the lock, then the lock() should block the control flow there. This applies to a thread that uses a different thread-code, which is also accessing the same data and which is also locked by the same mutex. In this case, you are still in the process of accessing the data and you may take, say, another 15 secs to reach the mutex unlock (so that the other thread that is getting blocked in mutex lock would unblock and would allow the control to access the data). Do you at any cost allow yet another thread to just unlock the same mutex, and in turn, allow the thread that is already waiting (blocking) in the mutex lock to unblock and access the data? Hope you got what I am saying here? As per, agreed upon universal definition!,

使用“互斥”就不会发生这种情况。没有其他线程可以解锁锁 在你的帖子里 使用“二进制信号量”可以实现这种情况。任何其他线程都可以解锁 线程中的锁

因此，如果您非常注重使用二进制信号量而不是互斥量，那么在锁定和解锁的“作用域”时应该非常小心。我的意思是，每个触及每个锁的控制流都应该触及一个解锁调用，也不应该有任何“第一次解锁”，而应该总是“第一次锁定”。

神话:

一些文章说“二进制信号量和互斥量是相同的”或“值为1的信号量是互斥量”，但基本的区别是互斥量只能由获得它的线程释放，而你可以从任何其他线程发出信号量

重点:

一个线程可以获得多个锁(互斥锁)。

只有递归互斥锁才能被锁多次，这里的锁和锁应该是一样的

•如果一个线程已经锁定了一个互斥锁，试图再次锁定互斥锁，它将进入该互斥锁的等待列表，这将导致死锁。

二进制信号量和互斥量相似但不相同。

互斥是昂贵的操作，因为与它相关的保护协议。

互斥的主要目的是实现对资源的原子访问或锁定

互斥锁控制对单个共享资源的访问。它提供了获取()对资源的访问并在完成后释放()资源的操作。

信号量控制对共享资源池的访问。它提供Wait()操作，直到池中的一个资源可用，并提供Signal()操作，当它返回池时。

当一个信号量保护的资源数量大于1时，它被称为计数信号量。当它控制一个资源时，它被称为布尔信号量。布尔信号量相当于互斥量。

因此，信号量是比互斥量更高级别的抽象。互斥锁可以用信号量来实现，但不能用信号量来实现。