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对于生产者 - 消费者问题，我想出了这个解决方案：生产者 - 消费者在python
import threading
import random
import time
class Bucket:
def __init__(self, size):
self.size = size
self.current_size = 0
self.cond_var = threading.Condition()
def available_for_put(self):
return self.current_size < self.size
def available_for_get(self):
return self.current_size > 0
def put(self):
self.current_size = self.current_size + 1
print(self)
self.cond_var.notify_all()
def get(self):
self.current_size = self.current_size - 1
print(self)
self.cond_var.notify_all()
def acquire(self):
self.cond_var.acquire()
def release(self):
self.cond_var.release()
def wait(self):
self.cond_var.wait()
def __str__(self):
return "Size is {0}".format(self.current_size)
class Worker(threading.Thread):
PRODUCER = 1
CONSUMER = 0
def __init__(self, bucket, kind):
threading.Thread.__init__(self)
self.kind = kind
self.bucket = bucket
def run(self):
while(1):
self.bucket.acquire()
while(((self.kind == Worker.PRODUCER) and (not self.bucket.available_for_put())) or \
((self.kind == Worker.CONSUMER) and (not self.bucket.available_for_get()))):
self.bucket.wait()
### self.bucket.acquire()
if self.kind == Worker.PRODUCER:
self.bucket.put()
else:
self.bucket.get()
time.sleep(0.1)
self.bucket.release()
bucket = Bucket(10)
workers = []
for i in range(10):
workers.append(Worker(bucket, i % 2))
for w in workers:
w.start()
print("Thread started")
for w in workers:
w.join()
显然，如果删除while(1)循环，使每个线程只运行一次达到循环内的块僵局，我不明白为什么。