python使用threading.Condition交替打印两个字符
作者:tinyid
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()
以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持脚本之家。