Implementing an N process barrier using semaphores

问题:

I'm currently training for an OS exam with previous iterations and I came across this:

Implement a "N Process Barrier", that is, making sure that each process out of a group of them waits, at some point in its respective execution, for the other processes to reach their given point.

You have the following ops available:

init(sem,value), wait(sem) and signal(sem)

N is an arbitrary number. I can make it so that it works for a given number of processes, but not for any number.

Any ideas? It's OK to reply with the pseudo-code, this is not an assignment, just personal study.

回答1:

This is well presented in The Little Book of Semaphores.

n = the number of threads
count = 0
mutex = Semaphore(1)
barrier = Semaphore(0)


mutex.wait()
count = count + 1
mutex.signal()

if count == n: barrier.signal() # unblock ONE thread

barrier.wait()
barrier.signal() # once we are unblocked, it's our duty to unblock the next thread

回答2:

Using N semaphores. Not very sure...

semaphore barr[N]
semaphore excl=1
int count=0

int i=1
while (i<=N)
   barr[i]=0 #initialization
   i=i+1

# use, each thread (tid)
wait(excl)
count=count+1
if (count==N)
   int j=1
   while (j<=N)
       signal(barr[j])
       j=j+1
   count=0
signal(excl)
wait(barr[tid])

回答3:

Only 2 barrier semaphores, but not sure...

semaphore barr[0..1] # two semaphores: barr[0] and barr[1]
semaphore excl=1
int count=0
int whichOne=0 # select semaphore to avoid race conditions

barr[0]=0 #initialization
barr[1]=0

# sample use
int current   #local for each thread
wait(excl)
current=whichOne
count=count+1
if (count==N)
   int j=1
   while (j<=N)
       signal(barr[current])
       j=j+1
   count=0
   whichOne=1-whichOne # swap barrier to avoid race conditions
signal(excl)
wait(barr[current])