Python 使用线程从数据块中弹出整数

我已经四天累了，但仍然坚持下去 请帮我解决这个问题

下一张幻灯片上的程序有两个线程，分别运行一个producer和一个producer 消费者。他们通过电话交流

•生产者写入整数0，1，2。。。9点到德克

•消费者尝试从数据块中弹出整数以构建列表。但是 列表中的数字随机缺失。对于每个省略的数字，都有 是否打印pop错误消息

修改程序，这样就不会再有弹出错误了 整数出现在输出列表中

import time
import threading
import random
from collections import deque

def int_producer(n, d):
    for i in range(n):
        time.sleep(random.random()/10)
        d.append(i)

def int_consumer(n, d):
    my_list = []
    for i in range(n):
        time.sleep(random.random()/10)
        try:
            my_list.append(d.popleft())
        except IndexError:
            print("pop error")
    print(my_list)

def main():
    d = deque(maxlen=3)
    threads = [threading.Thread(daemon=True, target=int_producer,
                                args=(10,d)) for i in range(3)]
    threads.extend([threading.Thread(daemon=True, target=int_consumer,
                                    args=(10,d)) for i in range(3)])
    [thread.start() for thread in threads]
    [thread.join() for thread in threads]

if __name__ == '__main__':
    main()

这是我的密码

import time
import threading
import random
from collections import deque

def int_producer(n, d):
    global has_space, has_full 

    for i in range(n):
        time.sleep(random.random()/10)
        with has_full:
            while len(d) == 3 :
                print ( ' wait here ' )
                has_full.wait()
        with has_space:
            d.append(i)
            has_space.notify()


def int_consumer(n, d):
    global has_space, has_full 

    my_list = []
    for i in range(n):
        time.sleep(random.random()/10)
        try:
            with has_space:
                while len(d) == 0 :
                    has_space.wait()


            with has_full:
                while len(d) == 3 :
                    has_full.notify()


            my_list.append(d.popleft())    


        except IndexError:
            print("pop error")
    print(my_list)

def main():
    global cv, has_space, has_full 
    has_full = threading.Condition()
    has_space = threading.Condition()

    d = deque(maxlen=3)
    threads = [threading.Thread(daemon=True, target=int_producer,
                                args=(10,d)) for i in range(3)]
    threads.extend([threading.Thread(daemon=True, target=int_consumer,
                                    args=(10,d)) for i in range(3)])
    [thread.start() for thread in threads]
    [thread.join() for thread in threads]

if __name__ == '__main__':
    main()

结果是这样的，列表应该是[0,1,2,3,4,5,6,7,8,9]

[0, 0, 1, 3, 2, 3, 5, 6, 7, 7]
[2, 2, 3, 4, 5, 5, 6, 7, 8, 8]
[0, 1, 1, 4, 4, 6, 9, 8, 9, 9]

---

那么，您已经给了我们您在家庭作业中给出的代码，但是您试图解决这个问题的方法是什么？您是否可以将

deque

替换为固有的线程

queue.queue（）

？您希望输出3行还是一行？因为您正在启动3个生产者和3个消费者，所以您的输出仍将是混合的。如果希望3行符合每个线程的要求，则可以输出正确的结果->[0,1,2,3,4,5,6,7,8,9[0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,6,7,8，9@AKX它必须使用deque