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
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");
其他回答
在窗口,差异如下所示。 MUTEX:成功执行等待的进程必须执行一个信号,反之亦然。二进制信号量:不同的进程可以在一个信号量上执行等待或信号操作。
厕所的例子是一个有趣的类比:
Mutex: Is a key to a toilet. One person can have the key - occupy the toilet - at the time. When finished, the person gives (frees) the key to the next person in the queue. Officially: "Mutexes are typically used to serialise access to a section of re-entrant code that cannot be executed concurrently by more than one thread. A mutex object only allows one thread into a controlled section, forcing other threads which attempt to gain access to that section to wait until the first thread has exited from that section." Ref: Symbian Developer Library (A mutex is really a semaphore with value 1.) Semaphore: Is the number of free identical toilet keys. Example, say we have four toilets with identical locks and keys. The semaphore count - the count of keys - is set to 4 at beginning (all four toilets are free), then the count value is decremented as people are coming in. If all toilets are full, ie. there are no free keys left, the semaphore count is 0. Now, when eq. one person leaves the toilet, semaphore is increased to 1 (one free key), and given to the next person in the queue. Officially: "A semaphore restricts the number of simultaneous users of a shared resource up to a maximum number. Threads can request access to the resource (decrementing the semaphore), and can signal that they have finished using the resource (incrementing the semaphore)." Ref: Symbian Developer Library
关于这个主题的好文章:
互斥量与信号量——第1部分:信号量 互斥量与信号量——第2部分:互斥量 互斥量与信号量——第3部分(最后一部分):互斥问题
来自第二部分:
The mutex is similar to the principles of the binary semaphore with one significant difference: the principle of ownership. Ownership is the simple concept that when a task locks (acquires) a mutex only it can unlock (release) it. If a task tries to unlock a mutex it hasn’t locked (thus doesn’t own) then an error condition is encountered and, most importantly, the mutex is not unlocked. If the mutual exclusion object doesn't have ownership then, irrelevant of what it is called, it is not a mutex.
它们不是一回事。它们有不同的用途! 虽然这两种类型的信号量都有一个满/空状态,并且使用相同的API,但它们的用法非常不同。
互斥信号量 互斥信号量用于保护共享资源(数据结构、文件等)。
互斥信号量由接收它的任务“拥有”。如果Task B尝试semGive一个当前由Task a持有的互斥锁,Task B的调用将返回一个错误并失败。
互斥对象总是使用以下顺序:
- SemTake - Critical Section - SemGive
这里有一个简单的例子:
Thread A Thread B Take Mutex access data ... Take Mutex <== Will block ... Give Mutex access data <== Unblocks ... Give Mutex
二进制信号量 二进制信号量解决了一个完全不同的问题:
任务B被挂起等待某些事情发生(例如传感器被绊倒)。 传感器跳闸和中断服务程序运行。它需要通知任务的行程。 任务B应运行并对传感器跳闸采取适当的操作。然后继续等待。
Task A Task B
... Take BinSemaphore <== wait for something
Do Something Noteworthy
Give BinSemaphore do something <== unblocks
注意,对于二进制信号量,B获取信号量,a给出信号量是可以的。 同样,二进制信号量不能保护资源不被访问。信号量的给予和获取从根本上是分离的。 对于同一个任务来说,对同一个二进制信号量的给予和获取通常没有什么意义。
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。
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