python使用threading.Condition交替打印两个字符

Python中使用threading.Condition交替打印两个字符的程序。

这个程序涉及到两个线程的的协调问题,两个线程为了能够相互协调运行,必须持有一个共同的状态,通过这个状态来维护两个线程的执行,通过使用threading.Condition对象就能够完成两个线程之间的这种协调工作。

threading.Condition默认情况下会通过持有一个ReentrantLock来协调线程之间的工作,所谓可重入锁,是只一个可以由一个线程递归获取的锁,此锁对象会维护当前锁的所有者(线程)和当前所有者递归获取锁的次数(本文在逻辑上和可重入锁没有任何关系,完全可以用一个普通锁替代)。

Python文档中给出的描述是:它是一个与某个锁相联系的变量。同时它实现了上下文管理协议。其对象中除了acquire和release方法之外,其它方法的调用的前提是,当前线程必须是这个锁的所有者。

通过代码和其中的注释,能够非常明白地弄清楚Condition的原理是怎样的:

import threading

import time

import functools

def worker(cond, name):

"""worker running in different thread"""

with cond: # 通过__enter__方法,获取cond对象中的锁,默认是一个ReentrantLock对象

print('...{}-{}-{}'.format(name, threading.current_thread().getName(), cond._is_owned()))

cond.wait() # 创建一个新的锁NEWLOCK,调用acquire将NEWLOCK获取,然后将NEWLOCK放入等待列表中,\

# 释放cond._lock锁(_release_save),最后再次调用acquire让NEWLOCK阻塞

print('wait returned in {}'.format(name))

if __name__ == '__main__':

condition = threading.Condition()

t1 = threading.Thread(target=functools.partial(worker, condition, 't1'))

t2 = threading.Thread(target=functools.partial(worker, condition, 't2'))

t2.start() # 启动线程2

t1.start() # 启动线程1

time.sleep(2)

with condition:

condition.notify(1) # 按照FIFO顺序(wait调用顺序),释放一个锁,并将其从等待列表中删除

time.sleep(2)

with condition:

condition.notify(1) # 按照FIFO顺序(wait调用顺序),释放另一个锁,并将其从等待队列中删除

t1.join() # 主线程等待子线程结束

t2.join() # 主线程等待子线程结束

print('All done')

其输出为:

...t2-Thread-2-True

...t1-Thread-1-True

wait returned in t2

wait returned in t1

All done

其中wait方法要求获取到threading.Condition对象中的锁(如果没有提供,默认使用一个可重入锁),然后自己创建一个新的普通锁(NEWLOCK),并获取这个NEWLOCK;之后调用_release_save方法释放threading.Condition对象中的锁,让其它线程能够获取到;最后再次调用NEWLOCK上的acquire方法,由于在创建时已经acquire过,所以此线程会阻塞在此。而wait想要继续执行,必须等待其它线程将产生阻塞的这个NEWLOCK给release掉,当然,这就是notify方法的责任了。

notify方法接收一个数字n,从等待列表中取出相应数量的等待对象(让wait方法阻塞的锁对象),调用其release方法,让对应的wait方法能够返回。而notify_all方法仅仅就是将n设置为等待列表的总长度而已。

在理解了threading.Condition对象中wait和notify的工作原理之后,我们就可以利用它们来实现两个线程交替打印字符的功能了:

import threading

import functools

import time

def print_a(state):

while True:

if state.closed:

print('Close a')

return

print('A')

time.sleep(2)

state.set_current_is_a(True)

state.wait_for_b()

def print_b(state):

while True:

if state.closed:

print('Close b')

return

state.wait_for_a()

print('B')

time.sleep(2)

state.set_current_is_a(False)

if __name__ == '__main__':

class State(object):

"""state used to coordinate multiple(two here) threads"""

def __init__(self):

self.condition = threading.Condition()

self.current_is_a = False

self.closed = False

def wait_for_a(self):

with self.condition:

while not self.current_is_a:

self.condition.wait()

def wait_for_b(self):

with self.condition:

while self.current_is_a:

self.condition.wait()

def set_current_is_a(self, flag):

self.current_is_a = flag

with self.condition:

self.condition.notify_all()

state = State()

t1 = threading.Thread(target=functools.partial(print_a, state))

t2 = threading.Thread(target=functools.partial(print_b, state))

try:

t1.start()

t2.start()

t1.join()

t2.join()

except KeyboardInterrupt:

state.closed = True

print('Closed')

可以看到有两种类型的任务,一个用于打印字符A,一个用于打印字符B,我们的实现种让A先于B打印,所以在print_a中,先打印A,再设置当前字符状态并释放等待列表中的所有锁(set_current_is_a),如果没有这一步,current_is_a将一直是False,wait_for_b能够返回,而wait_for_a却永远不会返回,最终效果就是每隔两秒就打印一个字符A,而B永远不会打印。另一个副作用是如果wait_for_a永远不会返回,那print_b所在线程的关闭逻辑也就无法执行,最终会成为僵尸线程(这里的关闭逻辑只用作示例,生产环境需要更加完善的关闭机制)。

考虑另一种情况,print_a种将set_current_is_a和wait_for_b交换一下位置会怎么样。从观察到的输出我们看到,程序首先输出了一个字符A,以后,每隔2秒钟,就会同时输出A和B,而不是交替输出。原因在于,由于current_is_a还是False,我们先调用的wait_for_b其会立即返回,之后调用set_current_is_a,将current_is_a设置为True,并释放所有的阻塞wait的锁(notify_all),这个过程中没有阻塞,print_a紧接着进入了下一个打印循环;与此同时,print_b中的wait_for_a也返回了,进入到B的打印循环,故最终我们看到A和B总是一起打印。

可见对于threading.Condition的使用需要多加小心,要注意逻辑上的严谨性。

附一个队列版本:

import threading

import functools

import time

from queue import Queue

def print_a(q_a, q_b):

while True:

char_a = q_a.get()

if char_a == 'closed':

return

print(char_a)

time.sleep(2)

q_b.put('B')

def print_b(q_a, q_b):

while True:

char_b = q_b.get()

if char_b == 'closed':

return

print(char_b)

time.sleep(2)

q_a.put('A')

if __name__ == '__main__':

q_a = Queue()

q_b = Queue()

t1 = threading.Thread(target=functools.partial(print_a, q_a, q_b))

t2 = threading.Thread(target=functools.partial(print_b, q_a, q_b))

try:

t1.start()

t2.start()

q_a.put('A')

t1.join()

t2.join()

except KeyboardInterrupt:

q_a.put('closed')

q_b.put('closed')

print('Done')

队列版本逻辑更清晰,更不容易出错,实际应用中应该选用队列。 

附一个协程版本(Python 3.5+):

import time

import asyncio

async def print_a():

while True:

print('a')

time.sleep(2) # simulate the CPU block time

await asyncio.sleep(0) # release control to event loop

async def print_b():

while True:

print('b')

time.sleep(2) # simulate the CPU block time

await asyncio.sleep(0) # release control to event loop

async def main():

await asyncio.wait([print_a(), print_b()])

if __name__ == '__main__':

loop = asyncio.get_event_loop()

loop.run_until_complete(main())

协程的运行需要依附于一个事件循环(select/poll/epoll/kqueue),通过async def将一个函数定义为协程,通过await主动让渡控制权,通过相互让渡控制权完成交替打印字符。整个程序运行于一个线程中,这样就没有线程间协调的工作,仅仅是控制权的让渡逻辑。对于IO密集型操作,而没有明显的CPU阻塞(计算复杂,以致出现明显的延时,比如复杂加解密算法)的情况下非常合适。

附一个Java版本:

PrintMain类,用于管理和协调打印A和打印B的两个线程:

package com.cuttyfox.tests.self.version1;

import java.util.concurrent.ExecutorService;

import java.util.concurrent.Executors;

import java.util.concurrent.TimeUnit;

public class PrintMain {

private boolean currentIsA = false;

public synchronized void waitingForPrintingA() throws InterruptedException {

while (this.currentIsA == false) {

wait();

}

}

public synchronized void waitingForPrintingB() throws InterruptedException {

while (this.currentIsA == true) {

wait();

}

}

public synchronized void setCurrentIsA(boolean flag) {

this.currentIsA = flag;

notifyAll();

}

public static void main(String[] args) throws Exception {

PrintMain state = new PrintMain();

ExecutorService executorService = Executors.newCachedThreadPool();

executorService.execute(new PrintB(state));

executorService.execute(new PrintA(state));

executorService.shutdown();

executorService.awaitTermination(10, TimeUnit.SECONDS);

System.out.println("Done");

System.exit(0);

}

}

打印A的线程(首先打印A):

package com.cuttyfox.tests.self.version1;

import java.util.concurrent.TimeUnit;

public class PrintA implements Runnable{

private PrintMain state;

public PrintA(PrintMain state) {

this.state = state;

}

public void run() {

try {

while (!Thread.interrupted()){

System.out.println("Print A");

TimeUnit.SECONDS.sleep(1);

this.state.setCurrentIsA(true);

this.state.waitingForPrintingB();

}

} catch (InterruptedException e) {

System.out.println("Exit through Interrupting.");

}

}

}

打印B的线程:

package com.cuttyfox.tests.self.version1;

import java.util.concurrent.TimeUnit;

public class PrintB implements Runnable{

private PrintMain state;

public PrintB(PrintMain state) {

this.state = state;

}

public void run() {

try{

while (!Thread.interrupted()) {

this.state.waitingForPrintingA();

System.out.println("Print B");

TimeUnit.SECONDS.sleep(1);

this.state.setCurrentIsA(false);

}

} catch (InterruptedException e) {

System.out.println("Exit through Interrupting.");

}

}

}

Java对象本身有对象锁,故这里没有像Python中那样需要显式通过创建一个Condition对象来得到一把锁。

使用Python实现交替打印abcdef的过程:

import threading

import time

import functools

from collections import deque

LETTERS = [chr(code) for code in range(97, 97+6)]

LENGTH = len(LETTERS)

class State(object):

def __init__(self):

self.condition = threading.Condition()

self.index_value = 0

def set_next_index(self, index):

with self.condition:

self.index_value = index

self.condition.notify_all()

def wait_for(self, index_value):

with self.condition:

while not self.index_value == index_value:

self.condition.wait()

def print_letter(state: State, wait_ident: int):

print('Got: {}!'.format(wait_ident))

while True:

state.wait_for(wait_ident)

time.sleep(2)

print(LETTERS[state.index_value])

print('PRINT: {} AND SET NEXT: {}'.format(state.index_value,

(state.index_value + 1) % LENGTH

))

state.set_next_index((state.index_value + 1) % LENGTH)

state = State()

d = deque()

d.extend(range(LENGTH))

d.rotate(1)

print(d)

threads = []

for wait_ident in d:

t = threading.Thread(target=functools.partial(print_letter, state, wait_ident))

threads.append(t)

for thread in threads:

thread.start()

for thread in threads:

thread.join()

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