◎知识点
经典生产者消费者问题(多线程Condition版)
经典生产者消费者问题(多进程Condition版)
◎脚本练习
▽ 经典生产者消费者问题(多线程Condition版)
"""
经典生产者消费者问题：
假设有一群生产者(Producer)和一群消费者(Consumer)通过一个市场来交换产品。
生产者的策略是：如果市场上剩余的产品少于20个，那么就生产4个产品放到市场上；
消费者的策略是：如果市场上剩余的产品多于10个，那么就从市场上消费3个产品。
"""
from threading import Thread, Condition
import time
cond = Condition()
count = 0
class Producer(Thread):
def run(self):
global count, cond
while True:
cond.acquire()
if count < 20:
count += 4
print('%s：生产者生产了4个产品，当前总共%d个产品' % (self.name, count))
cond.notify()
else:
print('%s：不生产，等待' % self.name)
cond.wait()
cond.release()
time.sleep(2)
class Consumer(Thread):
def run(self):
global count, cond
while True:
cond.acquire()
if count > 10:
count -= 3
print('%s：消费者消费了3个产品，当前总共%d个产品' % (self.name, count))
cond.notify()
else:
print('%s：不消费，等待' % self.name)
cond.wait()
cond.release()
time.sleep(2)
for i in range(3):
Producer().start()
for i in range(3):
Consumer().start()▽ 经典生产者消费者问题(多进程Condition版)
from multiprocessing import Process, Value, Condition
import time
cond = Condition()
count = Value('i', 0)
class Producer(Process):
def run(self):
global count, cond
while True:
cond.acquire()
if count.value < 20:
count.value += 4
print('%s：生产者生产了4个产品，当前总共%d个产品' % (self.name, count.value))
cond.notify()
else:
print('%s：不生产，等待' % self.name)
cond.wait()
cond.release()
time.sleep(2)
class Consumer(Process):
def run(self):
global count, cond
while True:
cond.acquire()
if count.value > 10:
count.value -= 3
print('%s：消费者消费了3个产品，当前总共%d个产品' % (self.name, count.value))
cond.notify()
else:
print('%s：不消费，等待' % self.name)
cond.wait()
cond.release()
time.sleep(2)
if __name__ == '__main__':
for i in range(3):
Producer().start()
for i in range(3):
Consumer().start()
