互斥量是任何想要解决临界区问题的算法都必须遵循的标准，而二进制信号量本身是一个可以取0和1值的变量。

答案可能取决于目标操作系统。例如，我所熟悉的至少一个RTOS实现允许对单个OS互斥量进行多个连续的“get”操作，只要它们都来自同一个线程上下文中。在允许另一个线程获得互斥量之前，多个get必须被相等数量的put替换。这与二进制信号量不同，对于二进制信号量，无论线程上下文如何，一次只允许一个get。

这种互斥锁背后的思想是，通过一次只允许一个上下文修改数据来保护对象。即使线程获得了互斥量，然后调用进一步修改对象的函数(并在自己的操作周围获得/放置保护互斥量)，这些操作仍然应该是安全的，因为它们都发生在单个线程下。

{
    mutexGet();  // Other threads can no longer get the mutex.

    // Make changes to the protected object.
    // ...

    objectModify();  // Also gets/puts the mutex.  Only allowed from this thread context.

    // Make more changes to the protected object.
    // ...

    mutexPut();  // Finally allows other threads to get the mutex.
}

当然，在使用此特性时，必须确保单个线程中的所有访问都是安全的!

我不确定这种方法有多普遍，或者它是否适用于我所熟悉的系统之外。有关这种互斥锁的示例，请参阅ThreadX RTOS。

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”。

Mutex uses a locking mechanism i.e. if a process wants to use a resource then it locks the resource, uses it and then release it. But on the other hand, semaphore uses a signalling mechanism where wait() and signal() methods are used to show if a process is releasing a resource or taking a resource. A mutex is an object but semaphore is an integer variable. In semaphore, we have wait() and signal() functions. But in mutex, there is no such function. A mutex object allows multiple process threads to access a single shared resource but only one at a time. On the other hand, semaphore allows multiple process threads to access the finite instance of the resource until available. In mutex, the lock can be acquired and released by the same process at a time. But the value of the semaphore variable can be modified by any process that needs some resource but only one process can change the value at a time.

一本有用的书，我从这里学习和复制

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

除了互斥对象有一个所有者之外，这两个对象还可以针对不同的用途进行优化。互斥锁被设计为只保留很短的时间;违反这一点会导致糟糕的性能和不公平的调度。例如，一个正在运行的线程可能被允许获取一个互斥量，即使另一个线程已经被阻塞在这个线程上。信号量可以提供更多的公平性，或者可以使用几个条件变量强制实现公平性。