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。
其他回答
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.
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
互斥锁用于阻塞关键区域,而信号量用于计数。
在理论层面上,它们在语义上并无不同。您可以使用信号量实现互斥量,反之亦然(参见这里的示例)。在实践中,实现是不同的,它们提供的服务也略有不同。
实际的区别(就围绕它们的系统服务而言)在于互斥锁的实现旨在成为一种更轻量级的同步机制。在oracle语言中,互斥锁被称为锁存器,而信号量被称为等待。
在最低级别,他们使用某种原子测试和设置机制。它读取内存位置的当前值,计算某种条件,并在一条不能中断的指令中写入该位置的值。这意味着您可以获得一个互斥锁,并测试是否有人在您之前拥有它。
典型的互斥量实现有一个进程或线程执行test-and-set指令,并评估是否有其他东西设置了互斥量。这里的关键点是与调度程序没有交互,因此我们不知道(也不关心)谁设置了锁。然后,我们要么放弃我们的时间片,并在任务重新调度时再次尝试它,要么执行自旋锁。自旋锁是这样一种算法:
Count down from 5000:
i. Execute the test-and-set instruction
ii. If the mutex is clear, we have acquired it in the previous instruction
so we can exit the loop
iii. When we get to zero, give up our time slice.
当我们完成执行受保护的代码(称为临界区)时,我们只需将互斥量的值设置为零或其他表示“清除”的值。如果有多个任务试图获取互斥量,那么下一个计划在互斥量释放后的任务将获得对资源的访问权。通常情况下,您可以使用互斥来控制同步资源,在这种资源中,只需要在很短的时间内对其进行独占访问,通常是对共享数据结构进行更新。
A semaphore is a synchronised data structure (typically using a mutex) that has a count and some system call wrappers that interact with the scheduler in a bit more depth than the mutex libraries would. Semaphores are incremented and decremented and used to block tasks until something else is ready. See Producer/Consumer Problem for a simple example of this. Semaphores are initialised to some value - a binary semaphore is just a special case where the semaphore is initialised to 1. Posting to a semaphore has the effect of waking up a waiting process.
一个基本的信号量算法如下所示:
(somewhere in the program startup)
Initialise the semaphore to its start-up value.
Acquiring a semaphore
i. (synchronised) Attempt to decrement the semaphore value
ii. If the value would be less than zero, put the task on the tail of the list of tasks waiting on the semaphore and give up the time slice.
Posting a semaphore
i. (synchronised) Increment the semaphore value
ii. If the value is greater or equal to the amount requested in the post at the front of the queue, take that task off the queue and make it runnable.
iii. Repeat (ii) for all tasks until the posted value is exhausted or there are no more tasks waiting.
在二进制信号量的情况下,两者之间的主要实际区别是围绕实际数据结构的系统服务的性质。
编辑:正如evan正确地指出的那样,自旋锁会降低单个处理器的速度。你只能在多处理器上使用自旋锁,因为在单处理器上,持有互斥锁的进程永远不会在另一个任务运行时重置它。自旋锁只在多处理器架构上有用。
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.
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