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

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。

其他回答

互斥锁:假设我们有临界区线程T1想要访问它,然后按照以下步骤进行。 T1:

锁 使用临界区 解锁

二进制信号量:它基于信号等待和信号工作。 等待将“s”的值减少1,通常“s”的值初始化为值“1”, 信号(s)使“s”值加1。如果“s”值为1表示没有人在使用临界区,当“s”值为0时表示临界区正在使用。 假设线程T2正在使用临界区,那么它遵循以下步骤。 T2:

Wait (s)//最初的s值是1,调用Wait后,它的值减少了1,即0 利用临界区 信号(s) //现在s值增加,变成1

Main difference between Mutex and Binary semaphore is in Mutext if thread lock the critical section then it has to unlock critical section no other thread can unlock it, but in case of Binary semaphore if one thread locks critical section using wait(s) function then value of s become "0" and no one can access it until value of "s" become 1 but suppose some other thread calls signal(s) then value of "s" become 1 and it allows other function to use critical section. hence in Binary semaphore thread doesn't have ownership.

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”。

在Windows上,互斥量和二进制信号量之间有两个区别:

互斥锁只能由拥有所有权的线程释放,即之前调用Wait函数的线程(或在创建互斥锁时获得所有权的线程)。任何线程都可以释放信号量。 线程可以在互斥锁上重复调用等待函数而不会阻塞。但是,如果你在一个二进制信号量上调用了两次等待函数,而中间没有释放信号量,线程就会阻塞。

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");

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。