它们不是一回事。它们有不同的用途! 虽然这两种类型的信号量都有一个满/空状态，并且使用相同的API，但它们的用法非常不同。

互斥信号量 互斥信号量用于保护共享资源(数据结构、文件等)。

互斥信号量由接收它的任务“拥有”。如果Task B尝试semGive一个当前由Task a持有的互斥锁，Task B的调用将返回一个错误并失败。

互斥对象总是使用以下顺序:

  - SemTake
  - Critical Section
  - SemGive

这里有一个简单的例子:

  Thread A                     Thread B
   Take Mutex
     access data
     ...                        Take Mutex  <== Will block
     ...
   Give Mutex                     access data  <== Unblocks
                                  ...
                                Give Mutex

二进制信号量 二进制信号量解决了一个完全不同的问题:

任务B被挂起等待某些事情发生(例如传感器被绊倒)。 传感器跳闸和中断服务程序运行。它需要通知任务的行程。 任务B应运行并对传感器跳闸采取适当的操作。然后继续等待。

   Task A                      Task B
   ...                         Take BinSemaphore   <== wait for something
   Do Something Noteworthy
   Give BinSemaphore           do something    <== unblocks

注意，对于二进制信号量，B获取信号量，a给出信号量是可以的。 同样，二进制信号量不能保护资源不被访问。信号量的给予和获取从根本上是分离的。 对于同一个任务来说，对同一个二进制信号量的给予和获取通常没有什么意义。

“二进制信号量”是一种编程语言规避使用«信号量»，如«互斥量»。显然有两个非常大的区别:

你称呼他们的方式。 标识符的最大长度。

关于这个主题的好文章:

互斥量与信号量——第1部分:信号量 互斥量与信号量——第2部分:互斥量 互斥量与信号量——第3部分(最后一部分):互斥问题

来自第二部分:

The mutex is similar to the principles of the binary semaphore with one significant difference: the principle of ownership. Ownership is the simple concept that when a task locks (acquires) a mutex only it can unlock (release) it. If a task tries to unlock a mutex it hasn’t locked (thus doesn’t own) then an error condition is encountered and, most importantly, the mutex is not unlocked. If the mutual exclusion object doesn't have ownership then, irrelevant of what it is called, it is not a mutex.

你可以通过以下方法清楚地记住不同之处:

互斥锁:用于保护关键区域， 互斥锁不能跨进程使用，只能在单个进程中使用 信号量:用于信号资源的可用性。 信号量既可以跨进程使用，也可以跨进程使用。

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。

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”。