既然上面的答案都不能消除困惑，这里有一个答案可以消除我的困惑。

Strictly speaking, a mutex is a locking mechanism used to synchronize access to a resource. Only one task (can be a thread or process based on OS abstraction) can acquire the mutex. It means there will be ownership associated with mutex, and only the owner can release the lock (mutex). Semaphore is signaling mechanism (“I am done, you can carry on” kind of signal). For example, if you are listening songs (assume it as one task) on your mobile and at the same time your friend called you, an interrupt will be triggered upon which an interrupt service routine (ISR) will signal the call processing task to wakeup.

来源:http://www.geeksforgeeks.org/mutex-vs-semaphore/

最好的解决方案

唯一的区别是

1.互斥锁-> lock和unlock属于锁定互斥锁的线程。

2.信号量->没有所有权，即;如果一个线程调用semwait(s)，任何其他线程都可以调用sempost(s)来移除锁。

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。

虽然二进制信号量可以用作互斥量，但互斥量是一个更具体的用例，因为只有锁定了互斥量的进程才应该解锁它。这种所有权限制使我们有可能对以下情况提供保护:

意外释放 递归死锁 任务死亡死锁

这些限制并不总是存在，因为它们降低了速度。在代码开发期间，您可以暂时启用这些检查。

例如，你可以在互斥锁中启用错误检查属性。错误检查互斥量返回EDEADLK，如果你试图锁定同一个互斥量两次，如果你解锁了一个不是你的互斥量，返回EPERM。

pthread_mutex_t mutex;
pthread_mutexattr_t attr;
pthread_mutexattr_init (&attr);
pthread_mutexattr_settype (&attr, PTHREAD_MUTEX_ERRORCHECK_NP);
pthread_mutex_init (&mutex, &attr);

一旦初始化，我们可以将这些检查放在我们的代码中，就像这样:

if(pthread_mutex_unlock(&mutex)==EPERM)
 printf("Unlock failed:Mutex not owned by this thread\n");

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

在看了上面的帖子后，这个概念对我来说很清楚。但仍有一些挥之不去的问题。所以，我写了一小段代码。

当我们试图给出一个信号量而不接收它时，它就会通过。但是，当你试图给出一个互斥量而不获取它时，它会失败。我在Windows平台上进行了测试。启用USE_MUTEX使用MUTEX运行相同的代码。

#include <stdio.h>
#include <windows.h>
#define xUSE_MUTEX 1
#define MAX_SEM_COUNT 1

DWORD WINAPI Thread_no_1( LPVOID lpParam );
DWORD WINAPI Thread_no_2( LPVOID lpParam );

HANDLE Handle_Of_Thread_1 = 0;
HANDLE Handle_Of_Thread_2 = 0;
int Data_Of_Thread_1 = 1;
int Data_Of_Thread_2 = 2;
HANDLE ghMutex = NULL;
HANDLE ghSemaphore = NULL;


int main(void)
{

#ifdef USE_MUTEX
    ghMutex = CreateMutex( NULL, FALSE, NULL);
    if (ghMutex  == NULL) 
    {
        printf("CreateMutex error: %d\n", GetLastError());
        return 1;
    }
#else
    // Create a semaphore with initial and max counts of MAX_SEM_COUNT
    ghSemaphore = CreateSemaphore(NULL,MAX_SEM_COUNT,MAX_SEM_COUNT,NULL);
    if (ghSemaphore == NULL) 
    {
        printf("CreateSemaphore error: %d\n", GetLastError());
        return 1;
    }
#endif
    // Create thread 1.
    Handle_Of_Thread_1 = CreateThread( NULL, 0,Thread_no_1, &Data_Of_Thread_1, 0, NULL);  
    if ( Handle_Of_Thread_1 == NULL)
    {
        printf("Create first thread problem \n");
        return 1;
    }

    /* sleep for 5 seconds **/
    Sleep(5 * 1000);

    /*Create thread 2 */
    Handle_Of_Thread_2 = CreateThread( NULL, 0,Thread_no_2, &Data_Of_Thread_2, 0, NULL);  
    if ( Handle_Of_Thread_2 == NULL)
    {
        printf("Create second thread problem \n");
        return 1;
    }

    // Sleep for 20 seconds
    Sleep(20 * 1000);

    printf("Out of the program \n");
    return 0;
}


int my_critical_section_code(HANDLE thread_handle)
{

#ifdef USE_MUTEX
    if(thread_handle == Handle_Of_Thread_1)
    {
        /* get the lock */
        WaitForSingleObject(ghMutex, INFINITE);
        printf("Thread 1 holding the mutex \n");
    }
#else
    /* get the semaphore */
    if(thread_handle == Handle_Of_Thread_1)
    {
        WaitForSingleObject(ghSemaphore, INFINITE);
        printf("Thread 1 holding semaphore \n");
    }
#endif

    if(thread_handle == Handle_Of_Thread_1)
    {
        /* sleep for 10 seconds */
        Sleep(10 * 1000);
#ifdef USE_MUTEX
        printf("Thread 1 about to release mutex \n");
#else
        printf("Thread 1 about to release semaphore \n");
#endif
    }
    else
    {
        /* sleep for 3 secconds */
        Sleep(3 * 1000);
    }

#ifdef USE_MUTEX
    /* release the lock*/
    if(!ReleaseMutex(ghMutex))
    {
        printf("Release Mutex error in thread %d: error # %d\n", (thread_handle == Handle_Of_Thread_1 ? 1:2),GetLastError());
    }
#else
    if (!ReleaseSemaphore(ghSemaphore,1,NULL) )      
    {
        printf("ReleaseSemaphore error in thread %d: error # %d\n",(thread_handle == Handle_Of_Thread_1 ? 1:2), GetLastError());
    }
#endif

    return 0;
}

DWORD WINAPI Thread_no_1( LPVOID lpParam ) 
{ 
    my_critical_section_code(Handle_Of_Thread_1);
    return 0;
}


DWORD WINAPI Thread_no_2( LPVOID lpParam ) 
{
    my_critical_section_code(Handle_Of_Thread_2);
    return 0;
}

信号量允许您发出“使用资源完成”的信号，即使它从未拥有该资源，这一事实使我认为在信号量的情况下，拥有和发出信号之间存在非常松散的耦合。