Python上的信号量

Question

几个星期前我开始用Python编程,并试图使用Semaphores同步两个简单的线程,用于学习目的.这是我得到的:

import threading
sem = threading.Semaphore()

def fun1():
    while True:
        sem.acquire()
        print(1)
        sem.release()

def fun2():
    while True:
        sem.acquire()
        print(2)
        sem.release()

t = threading.Thread(target = fun1)
t.start()
t2 = threading.Thread(target = fun2)
t2.start()

但它一直打印只有1.如何对照片进行内部缩放？

Answer 1

此外，您可以使用Lock/mutex方法如下：

import threading
import time

mutex = threading.Lock()  # is equal to threading.Semaphore(1)

def fun1():
    while True:
        mutex.acquire()
        print(1)
        mutex.release()
        time.sleep(.5)

def fun2():
    while True:
        mutex.acquire()
        print(2)
        mutex.release()
        time.sleep(.5)

t1 = threading.Thread(target=fun1).start()
t2 = threading.Thread(target=fun2).start()

使用“ with”的更简单的样式：

import threading
import time

mutex = threading.Lock()  # is equal to threading.Semaphore(1)

def fun1():
    while True:
        with mutex:
            print(1)
        time.sleep(.5)

def fun2():
    while True:
        with mutex:
            print(2)
        time.sleep(.5)

t1 = threading.Thread(target=fun1).start()
t2 = threading.Thread(target=fun2).start()

[注意]：

互斥量、信号量和锁的区别

Answer 2

它工作正常,只是它的打印速度太快,你无法看到.尝试将time.sleep()两个函数(少量)放入线程中休眠那段时间,实际上能够同时看到1和2.

示例 -

import threading
import time
sem = threading.Semaphore()

def fun1():
    while True:
        sem.acquire()
        print(1)
        sem.release()
        time.sleep(0.25)

def fun2():
    while True:
        sem.acquire()
        print(2)
        sem.release()
        time.sleep(0.25)

t = threading.Thread(target = fun1)
t.start()
t2 = threading.Thread(target = fun2)
t2.start()

Answer 3

事实上，我想找到asyncio.Semaphores，而不是threading.Semaphore，而且我相信有人可能也想要它。

所以，我决定分享asyncio。Semaphores，希望你不要介意。

from asyncio import (
    Task,
    Semaphore,
)
import asyncio
from typing import List


async def shopping(sem: Semaphore):
    while True:
        async with sem:
            print(shopping.__name__)
        await asyncio.sleep(0.25)  # Transfer control to the loop, and it will assign another job (is idle) to run.


async def coding(sem: Semaphore):
    while True:
        async with sem:
            print(coding.__name__)
        await asyncio.sleep(0.25)


async def main():
    sem = Semaphore(value=1)
    list_task: List[Task] = [asyncio.create_task(_coroutine(sem)) for _coroutine in (shopping, coding)]
    """ 
    # Normally, we will wait until all the task has done, but that is impossible in your case.
    for task in list_task:
        await task
    """
    await asyncio.sleep(2)  # So, I let the main loop wait for 2 seconds, then close the program.


asyncio.run(main())

输出

shopping
coding
shopping
coding
shopping
coding
shopping
coding
shopping
coding
shopping
coding
shopping
coding
shopping
coding

16*0.25 = 2