互斥锁

Until recently, the only sleeping lock in the kernel was the semaphore. Most users of semaphores instantiated a semaphore with a count of one and treated them as a mutual exclusion lock—a sleeping version of the spin-lock. Unfortunately, semaphores are rather generic and do not impose any usage constraints. This makes them useful for managing exclusive access in obscure situations, such as complicated dances between the kernel and userspace. But it also means that simpler locking is harder to do, and the lack of enforced rules makes any sort of automated debugging or constraint enforcement impossible. Seeking a simpler sleeping lock, the kernel developers introduced the mutex.Yes, as you are now accustomed to, that is a confusing name. Let’s clarify.The term “mutex” is a generic name to refer to any sleeping lock that enforces mutual exclusion, such as a semaphore with a usage count of one. In recent Linux kernels, the proper noun “mutex” is now also a specific type of sleeping lock that implements mutual exclusion.That is, a mutex is a mutex.

互斥锁的简单性和效率来自于它在信号量要求之外强加给用户的附加约束。信号量是按照Dijkstra的原始设计来实现最基本的行为，而互斥锁则不同，它的用例更严格、更窄: n一次只能有一个任务持有互斥锁。也就是说，互斥锁的使用计数总是1。

Whoever locked a mutex must unlock it. That is, you cannot lock a mutex in one context and then unlock it in another. This means that the mutex isn’t suitable for more complicated synchronizations between kernel and user-space. Most use cases, however, cleanly lock and unlock from the same context. Recursive locks and unlocks are not allowed. That is, you cannot recursively acquire the same mutex, and you cannot unlock an unlocked mutex. A process cannot exit while holding a mutex. A mutex cannot be acquired by an interrupt handler or bottom half, even with mutex_trylock(). A mutex can be managed only via the official API: It must be initialized via the methods described in this section and cannot be copied, hand initialized, or reinitialized.

[1] Linux内核开发，第三版Robert Love

互斥量和二进制信号量是相同的用法，但实际上，它们是不同的。

对于互斥锁，只有锁定了它的线程才能解锁它。如果有其他线程来锁定它，它将等待。

对于信号电话来说，情况就不是这样了。信号量没有与特定的线程ID绑定。

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

I think most of the answers here were confusing especially those saying that mutex can be released only by the process that holds it but semaphore can be signaled by ay process. The above line is kind of vague in terms of semaphore. To understand we should know that there are two kinds of semaphore one is called counting semaphore and the other is called a binary semaphore. In counting semaphore handles access to n number of resources where n can be defined before the use. Each semaphore has a count variable, which keeps the count of the number of resources in use, initially, it is set to n. Each process that wishes to uses a resource performs a wait() operation on the semaphore (thereby decrementing the count). When a process releases a resource, it performs a release() operation (incrementing the count). When the count becomes 0, all the resources are being used. After that, the process waits until the count becomes more than 0. Now here is the catch only the process that holds the resource can increase the count no other process can increase the count only the processes holding a resource can increase the count and the process waiting for the semaphore again checks and when it sees the resource available it decreases the count again. So in terms of binary semaphore, only the process holding the semaphore can increase the count, and count remains zero until it stops using the semaphore and increases the count and other process gets the chance to access the semaphore.

二进制信号量和互斥量之间的主要区别在于，信号量是一种信号机制，而互斥量是一种锁定机制，但二进制信号量的功能似乎与互斥量类似，这造成了混乱，但两者是适用于不同类型工作的不同概念。

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)。

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”。