以上几乎所有人都说对了。如果有人还有疑问，让我来澄清一下。

互斥->用于序列化 信号- >同步。

两者的目的是不同的，但是，通过精心的编程，可以实现相同的功能。

标准示例->生产者消费者问题。

initial value of SemaVar=0

Producer                           Consumer
---                                SemaWait()->decrement SemaVar   
produce data
---
SemaSignal SemaVar or SemaVar++  --->consumer unblocks as SemVar is 1 now.

希望我能澄清。

在Windows上，互斥量和二进制信号量之间有两个区别:

互斥锁只能由拥有所有权的线程释放，即之前调用Wait函数的线程(或在创建互斥锁时获得所有权的线程)。任何线程都可以释放信号量。 线程可以在互斥锁上重复调用等待函数而不会阻塞。但是，如果你在一个二进制信号量上调用了两次等待函数，而中间没有释放信号量，线程就会阻塞。

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”。

以上几乎所有人都说对了。如果有人还有疑问，让我来澄清一下。

互斥->用于序列化 信号- >同步。

两者的目的是不同的，但是，通过精心的编程，可以实现相同的功能。

标准示例->生产者消费者问题。

initial value of SemaVar=0

Producer                           Consumer
---                                SemaWait()->decrement SemaVar   
produce data
---
SemaSignal SemaVar or SemaVar++  --->consumer unblocks as SemVar is 1 now.

希望我能澄清。

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)。

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