Implementing an N process barrier using semaphores
I'm currently training for an OS exam with previous iterations and I came across this:
Implement a "N Process Barrier", that is, makin开发者_如何学JAVAg 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.
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
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])
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])
I think this should also work, using only one semaphore (at least if the barrier is not required to be reusable)?
n = number of threads
barrier = Semaphore(1 - n)
barrier.signal()
barrier.wait()
barrier.signal()
精彩评论