信号量小书
下面的代码中，每个线程必须等待其他线程完成集合部分，然后等待直到每个线程都完成关键部分。
/* rendezvous code */
mutex.wait()
count++;
mutex_signal()
if(count==n)
sem.signal()
sem.wait()
sem.signal()
mutex.wait()
count--;
mutex.signal()
if(count==0)
sem.wait()
我知道两个进程可能会遇到相同的 count 值（可能是 0 或 n）的情况。因此，可以同时发送两个或多个信号。上次测试怎么会出现僵局呢。我好像没明白这个。
这是一种十字转门类型的信号量安排，作者实际上认为它是一个十字转门，但它是一个信号量，它应该可以无死锁地工作。 请告诉我这段代码怎么会出现死锁！
Below is code in which each thread must wait for each other thread to complete the rendezvous part and then wait until everyone has completed the critical section.
/* rendezvous code */
mutex.wait()
count++;
mutex_signal()
if(count==n)
sem.signal()
sem.wait()
sem.signal()
mutex.wait()
count--;
mutex.signal()
if(count==0)
sem.wait()
I know that two processes can reach the case where both see the same value of count (0 or n may be). Due to this two or more signals may be sent at the same time. How can there be a deadlock in the last test. I don't seem to get this.
This is a turnsile kindof semaphore arrangement and author is actually thinking it is a turnstile, but it's a semaphore and it should work without a deadlock.
Please tell me how is there a deadlock in this code!
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我将尝试解释我的看法。
除最后一个线程外的所有线程都会在第一个 sem.wait() 处等待。一旦最后一个线程到达，它将 sem.signal() （因为 count==n）允许等待线程之一（比如 T1）继续。然后T1将依次执行sem.signal()，这将允许另一个线程继续。这就像连锁反应。请注意，最后一个传递的线程也会发出一个信号，这将使信号量值为 1。
现在，如果两个线程看到 count==0 然后将尝试执行 sem.wait()。但由于信号量值为1，一个线程将无法通过，造成死锁。
I'll try to explain the way I see it.
All threads but the last will come and wait at the first sem.wait(). Once the last thread arrives it will sem.signal() (because count==n) allowing one of the waiting threads(say T1) to continue. Then T1 will in turn do a sem.signal() which will allow another thread to continue. It is something like a chain reaction. Note that the last thread to pass will also do a signal which will make the Semaphore value 1.
Now if two threads come and see that the count==0 then will try to do sem.wait(). But since the semaphore value is 1, one thread will not be able to pass, causing deadlock.
“if”语句也应该位于“mutex”信号量指定的关键部分内，否则竞争条件可能导致死锁。
即，正确的代码是
The "if" statements should also be inside the critical sections designated by the "mutex" semaphore, otherwise race conditions can lead to deadlock.
I.e., the correct code is