二进制信号量和互斥量之间有区别吗?或者它们本质上是相同的?


当前回答

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。

其他回答

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。

http://www.geeksforgeeks.org/archives/9102将详细讨论。

互斥是一种锁机制,用于同步对资源的访问。 信号量是一种信号机制。

如果他/她想使用二进制信号量来代替互斥量,这取决于程序员。

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。

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/

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。