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

“二进制信号量”是一种编程语言规避使用«信号量»,如«互斥量»。显然有两个非常大的区别:

你称呼他们的方式。 标识符的最大长度。

其他回答

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。

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。

虽然互斥量和信号量被用作同步原语,但它们之间有很大的区别。 在互斥锁的情况下,只有锁定或获得互斥锁的线程才能解锁它。 在信号量的情况下,等待信号量的线程可以由另一个线程发出信号。 一些操作系统支持在进程之间使用互斥量和信号量。通常使用是在共享内存中创建的。

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。

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