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

The basic issue is concurrency. There is more than one flow of control. Think about two processes using a shared memory. Now only one process can access the shared memory at a time. If more than one process accesses the shared memory at a time, the contents of shared memory would get corrupted. It is like a railroad track. Only one train can run on it, else there would be an accident.So there is a signalling mechanism, which a driver checks. If the signal is green, the train can go and if it is red it has to wait to use the track. Similarly in case of shared memory, there is a binary semaphore. If the semaphore is 1, a process acquires it (makes it 0) and goes ahead and accesses it. If the semaphore is 0, the process waits. The functionality the binary semaphore has to provide is mutual exclusion (or mutex, in short) so that only one of the many concurrent entities (process or thread) mutually excludes others. It is a plus that we have counting semaphores, which help in synchronizing multiple instances of a resource.

互斥是信号量提供的基本功能。现在在线程上下文中,我们可能有不同的名称和语法。但基本概念是相同的:如何在并发编程中保持代码和数据的完整性。在我看来,像所有权和相关检查这样的东西是由实现提供的改进。

其他回答

正如这里许多人提到的,互斥锁用于保护关键代码段(又名临界段)。你将在同一个线程中获得互斥锁(lock),进入临界区,释放互斥锁(unlock)。

在使用信号量时,您可以让一个线程(例如线程a)等待一个信号量,直到另一个线程(例如线程B)完成任何任务,然后为线程a设置信号量以停止等待,并继续其任务。

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。

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。

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。

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");