使用单个信号量的多进程同步
我们正在参加的课程中介绍多线程编程。教授提出了一个额外的问题，我一直在尝试解决这个问题，但没有成功：
每个进程 P0、P1、P2 和 P3 都必须等待其他三个进程交叉或到达其代码中的特定同步点，并且只有这样该进程才能跨越它自己的同步点。
我已经知道如何用四个信号量来回答这个问题，困难的部分是只用一个信号量来回答这个问题。
关于如何进行的任何建议或提示？
We're covering multithreaded programming in a class I'm taking. The professor offered a bonus question that I have been trying, to no avail, to figure out:
Each of processes P0, P1, P2 and P3 have to wait for the other three to cross or reach a particular synchronization point in their code, and only then may that process cross its own synchronization point.
I already know how to answer the question with four semaphores, the hard part is doing it with only one semaphore.
Any suggestions or hints as to how to proceed?
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您对解决方案所施加的限制不太了解，但请参阅 The Little Book of Semaphores 并阅读通过障碍部分。这应该会给你一些想法。
You are a little light on the constraints imposed on your solution but see The Little Book of Semaphores and read through the sections on barriers. That should give you some ideas.
如果信号量不是二进制信号量，只需将其初始化为 -4 即可。
Just initialize your semaphore at -4 if it's not a binary one.
原来教授的意思是说你可以使用两个信号量而不是一个。正如我在思考这个问题一段时间后所做的那样，他认为用单个信号量是不可能做到的。
Turns out the professor had meant to say that you could use two semaphores instead of one. He believes, as I do after having thought about the problem for a while, that it is impossible to do with a single semaphore.