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
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
其他回答
Mutex uses a locking mechanism i.e. if a process wants to use a resource then it locks the resource, uses it and then release it. But on the other hand, semaphore uses a signalling mechanism where wait() and signal() methods are used to show if a process is releasing a resource or taking a resource. A mutex is an object but semaphore is an integer variable. In semaphore, we have wait() and signal() functions. But in mutex, there is no such function. A mutex object allows multiple process threads to access a single shared resource but only one at a time. On the other hand, semaphore allows multiple process threads to access the finite instance of the resource until available. In mutex, the lock can be acquired and released by the same process at a time. But the value of the semaphore variable can be modified by any process that needs some resource but only one process can change the value at a time.
一本有用的书,我从这里学习和复制
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.
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。
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。
以上几乎所有人都说对了。如果有人还有疑问,让我来澄清一下。
互斥->用于序列化 信号- >同步。
两者的目的是不同的,但是,通过精心的编程,可以实现相同的功能。
标准示例->生产者消费者问题。
initial value of SemaVar=0
Producer Consumer
--- SemaWait()->decrement SemaVar
produce data
---
SemaSignal SemaVar or SemaVar++ --->consumer unblocks as SemVar is 1 now.
希望我能澄清。
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