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

互斥锁只能由获得它的线程释放。 二进制信号量可以由任何线程(或进程)发出信号。

因此，信号量更适合于一些同步问题，如生产者-消费者。

在Windows上，二进制信号量更像事件对象而不是互斥对象。

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。

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。

二进制信号量和互斥量的区别: 所有权: 信号量甚至可以从非当前所有者发出信号(发布)。这意味着您可以简单地从任何其他线程发布，尽管您不是所有者。

信号量是进程中的公共属性，它可以简单地由非所有者线程发布。 请用粗体字标出这个区别，这意味着很多。

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”。