python - time.sleep 与 threading.Lock 的交互

Question

我试图理解 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如下输出:

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 block 中缩进 time.sleep(1) 会破坏程序？

与

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

打印机不能被中断!

Answer 1

将您的代码更改为此并查看发生了什么:

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!')

当你的代码尝试运行LOCK.acquire()时，我认为with LOCK和LOCK.acquire()之间开始了一场竞争，并且该线程大多数时候赢得了游戏并获取 LOCK 1 秒。但在第一种方法中，当 Printer() 等待 1 秒时，主线程毫无问题地获取 LOCK，因为它是空闲的。我希望你能明白。抱歉我的语言:))