在此网站http://www.cs.cornell.edu/courses/cs4410/2010fa/synchreview.pdf上,它说必须以下列方式实现多次迭代的屏障:
class Barrier:
def __init__(self, N):
self.incount = 0
self.outcount = 0
self.N = N
self.lock = Lock()
self.everyoneatbarrier = Condition(self.lock)
def barrier(self, processid):
with self.lock:
self.incount += 1
if self.incount == self.N
self.everoneatbarrier.notifyAll()
while self.incount < self.N or
(self.incount >= self.N and self.outcount < self.N):
self.everyoneatbarrier.wait()
self.outcount += 1
if self.outcount == self.N:
self.outcount = 0
self.incount = self.N
如果我们只使用一个简单的 if-else 语句,会不会更简单:
def barrier(self, processid):
with self.lock:
self.count += 1
if self.count == self.N :
self.everyoneatbarrier.notifyAll()
else :
self.everyoneatbarrier.wait()
self.count = 0
我不明白所有额外的努力。感谢大家。