time.sleep 与 threading.Lock 的交互

Question

我试图了解 Python 中的锁并编写了以下示例代码。

import threading
import time

LOCK = threading.Lock()

class Printer(threading.Thread):
    def __init__(self, *args, **kwargs):
        super(Printer, self).__init__(*args, **kwargs)
        self.daemon = True
        self.i = 0

    def run(self):
        while True:
            with LOCK:
                print(self.i)
                self.i += 1
            time.sleep(1)

raw_input('press enter to start thread\n')
Printer().start()
raw_input('press enter to pause thread\n')
LOCK.acquire()
raw_input('press enter to resume thread\n')
LOCK.release()
raw_input('press enter to exit program\n')
print('bye!')

这按预期工作并产生 output ，例如：

press enter to start thread

0
 press enter to pause thread
1
2
3
4

press enter to resume thread

press enter to exit program
5
6
7
8

bye!

问题

为什么在with块中缩进time.sleep(1)会破坏程序？

和

def run(self):
    while True:
        with LOCK:
            print(self.i)
            self.i += 1
            time.sleep(1)

打印机不能中断！

Answer 1

将您的代码更改为此，看看发生了什么：

import threading
import time

LOCK = threading.Lock()


class Printer(threading.Thread):
    def __init__(self, *args, **kwargs):
        super(Printer, self).__init__(*args, **kwargs)
        self.daemon = True
        self.i = 0

    def run(self):
        while True:
            with LOCK:
                print(self.i)
                self.i += 1
                time.sleep(1)


raw_input('press enter to start thread\n')
Printer().start()
raw_input('press enter to pause thread\n')
print('acquiring...')
LOCK.acquire()
print('acquired')
raw_input('press enter to resume thread\n')
LOCK.release()
raw_input('press enter to exit program\n')
print('bye!')

当您的代码尝试运行LOCK.acquire()时，我认为比赛从with LOCK和LOCK.acquire()开始，并且线程大部分时间都赢得了游戏并获得 LOCK 1 秒。 但是在第一种方法中，当 Printer() 等待 1 秒时，主线程获取 LOCK 没有任何问题，因为它是空闲的。 我希望你能明白。 对不起我的语言:))

Answer 2

Hameda169想通了！

我编写了一个带有“on”和“off”按钮的小型演示 GUI，以便将来解释/记住这个概念。

如果您在 LOCK.release() 之后LOCK.release() time.sleep(0.01) ，则 GUI 按钮在尝试获取锁时可能会冻结很长时间。

它还展示了为什么需要锁。 如果你在on和off回调with LOCK注释掉，你很快就会在按下按钮时得到一个AttributeError 。

# see what happens when there's no sleep AFTER release
# see what happens if on and off functions do not acquire the lock

import Tkinter as tk
import threading
import time
import ttk

class ON(object):
    pass

class Printer(threading.Thread):
    def __init__(self, *args, **kwargs):
        super(Printer, self).__init__()
        self.daemon = True
        self.on = None
        self.i = 0

    def run(self):
        while True:
            LOCK.acquire()
            if self.on is not None: 
                time.sleep(0.1) # make it very likely that self.on can be set to
                                # None from outside before trying to access
                                # dummy_attribute -> AttributeError w/o lock!
                self.on.dummy_attribute # can error without lock
                print(self.i)
                self.i += 1
            LOCK.release()

            time.sleep(0.01) # must sleep after release, else lock can be
                             # acquired immediately again!

ROOT = tk.Tk()
PRINTER = Printer()
LOCK = threading.Lock()

def on():
    with LOCK:
        PRINTER.on = ON()
        PRINTER.on.dummy_attribute = 'foo'

def off():
    with LOCK:
        PRINTER.on = None

ttk.Button(ROOT, text='on', command=on).grid()
ttk.Button(ROOT, text='off', command=off).grid()

if __name__ == '__main__':
    PRINTER.start()
    ROOT.mainloop()