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。

互斥锁用于阻塞关键区域，而信号量用于计数。

互斥量和二进制信号量是相同的用法，但实际上，它们是不同的。

对于互斥锁，只有锁定了它的线程才能解锁它。如果有其他线程来锁定它，它将等待。

对于信号电话来说，情况就不是这样了。信号量没有与特定的线程ID绑定。

在理论层面上，它们在语义上并无不同。您可以使用信号量实现互斥量，反之亦然(参见这里的示例)。在实践中，实现是不同的，它们提供的服务也略有不同。

实际的区别(就围绕它们的系统服务而言)在于互斥锁的实现旨在成为一种更轻量级的同步机制。在oracle语言中，互斥锁被称为锁存器，而信号量被称为等待。

在最低级别，他们使用某种原子测试和设置机制。它读取内存位置的当前值，计算某种条件，并在一条不能中断的指令中写入该位置的值。这意味着您可以获得一个互斥锁，并测试是否有人在您之前拥有它。

典型的互斥量实现有一个进程或线程执行test-and-set指令，并评估是否有其他东西设置了互斥量。这里的关键点是与调度程序没有交互，因此我们不知道(也不关心)谁设置了锁。然后，我们要么放弃我们的时间片，并在任务重新调度时再次尝试它，要么执行自旋锁。自旋锁是这样一种算法:

Count down from 5000:
     i. Execute the test-and-set instruction
    ii. If the mutex is clear, we have acquired it in the previous instruction 
        so we can exit the loop
   iii. When we get to zero, give up our time slice.

当我们完成执行受保护的代码(称为临界区)时，我们只需将互斥量的值设置为零或其他表示“清除”的值。如果有多个任务试图获取互斥量，那么下一个计划在互斥量释放后的任务将获得对资源的访问权。通常情况下，您可以使用互斥来控制同步资源，在这种资源中，只需要在很短的时间内对其进行独占访问，通常是对共享数据结构进行更新。

A semaphore is a synchronised data structure (typically using a mutex) that has a count and some system call wrappers that interact with the scheduler in a bit more depth than the mutex libraries would. Semaphores are incremented and decremented and used to block tasks until something else is ready. See Producer/Consumer Problem for a simple example of this. Semaphores are initialised to some value - a binary semaphore is just a special case where the semaphore is initialised to 1. Posting to a semaphore has the effect of waking up a waiting process.

一个基本的信号量算法如下所示:

(somewhere in the program startup)
Initialise the semaphore to its start-up value.

Acquiring a semaphore
   i. (synchronised) Attempt to decrement the semaphore value
  ii. If the value would be less than zero, put the task on the tail of the list of tasks waiting on the semaphore and give up the time slice.

Posting a semaphore
   i. (synchronised) Increment the semaphore value
  ii. If the value is greater or equal to the amount requested in the post at the front of the queue, take that task off the queue and make it runnable.  
 iii. Repeat (ii) for all tasks until the posted value is exhausted or there are no more tasks waiting.

在二进制信号量的情况下，两者之间的主要实际区别是围绕实际数据结构的系统服务的性质。

编辑:正如evan正确地指出的那样，自旋锁会降低单个处理器的速度。你只能在多处理器上使用自旋锁，因为在单处理器上，持有互斥锁的进程永远不会在另一个任务运行时重置它。自旋锁只在多处理器架构上有用。

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。

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。