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

互斥对象有所有权,不像信号量。尽管在互斥量范围内的任何线程都可以获得一个未锁定的互斥量,并锁定对同一关键代码段的访问,但只有锁定了互斥量的线程才应该解锁它。

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。

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。

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。

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