睡眠理发师问题：如何避免理发师在顾客就绪前开始理发？

我近期在研究经典的睡眠理发师问题，从维基百科上找到一个可行的解决方案：

# The first two are mutexes (only 0 or 1 possible)
Semaphore barberReady = 0
Semaphore accessWRSeats = 1     # if 1, the number of seats in the waiting room can be incremented or decremented
Semaphore custReady = 0         # the number of customers currently in the waiting room, ready to be served
int numberOfFreeWRSeats = N     # total number of seats in the waiting room

def Barber():
  while True:                   # Run in an infinite loop.
    wait(custReady)             # Try to acquire a customer - if none is available, go to sleep.
    wait(accessWRSeats)         # Awake - try to get access to modify # of available seats, otherwise sleep.
    numberOfFreeWRSeats += 1    # One waiting room chair becomes free.
    signal(barberReady)         # I am ready to cut.
    signal(accessWRSeats)       # Don't need the lock on the chairs anymore.
    # (Cut hair here.)

def Customer():
  while True:                   # Run in an infinite loop to simulate multiple customers.
    wait(accessWRSeats)         # Try to get access to the waiting room chairs.
    if numberOfFreeWRSeats > 0: # If there are any free seats:
      numberOfFreeWRSeats -= 1  #   sit down in a chair
      signal(custReady)         #   notify the barber, who's waiting until there is a customer
      signal(accessWRSeats)     #   don't need to lock the chairs anymore
      wait(barberReady)         #   wait until the barber is ready
      # (Have hair cut here.)
    else:                       # otherwise, there are no free seats; tough luck --
      signal(accessWRSeats)     #   but don't forget to release the lock on the seats!
      # (Leave without a haircut.)

起初该方案看似合理，但我发现一个潜在漏洞：根据我对信号量的有限理解，signal()可以在对应的wait()之前执行。假设理发师执行signal(barberReady)后立即开始理发，而此时顾客还未执行wait(barberReady)，这在逻辑上不合理，因为顾客尚未就绪。请问操作系统（或信号量语义）如何防止这种不匹配情况？

其实你担心的这种“时序不匹配”不会导致逻辑问题，核心在于信号量的计数特性：

• 信号量本质是带等待队列的计数器。当signal()先于wait()调用时，信号量的计数会先加1；后续wait()调用时，会直接消耗这个已有的计数，不会阻塞。
• 回到这个场景：理发师只有在wait(custReady)成功后（确认有顾客等待），才会执行signal(barberReady)。这意味着这个signal()必然对应某个已经坐下的顾客——哪怕顾客还没走到wait(barberReady)这一步，信号量的计数会把“理发师就绪”的通知暂存起来。等顾客后续调用wait(barberReady)时，会直接取用这个计数，进入理发环节。
• 从数量匹配上看：每一个signal(custReady)对应一个wait(custReady)，每一个signal(barberReady)对应一个wait(barberReady)，两者的调用是严格一一对应的，不会出现“理发师白理发”或者“顾客永远等不到”的情况。

简单来说，信号量的计数机制天然解决了“通知提前到达”的问题，它相当于把通知暂存起来，等需要的线程来取，既不会丢失也不会乱序。

内容的提问来源于stack exchange，提问作者CN.hitori