Python 上的信号量

Question

几周前我开始使用 Python 进行编程，并尝试使用 Semaphores 同步两个简单线程，以用于学习目的。这是我所拥有的：

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()

但它一直只打印 1。如何插入打印件？

Answer 1

它工作正常，只是它的打印速度太快，你看不到。尝试在两个函数（少量）中添加 time.sleep() 以使线程休眠这么长的时间，以便实际上能够同时看到 1 和 2。

例子-

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 2

我使用此代码演示了如何 1 个线程可以使用信号量，而另一个线程将等待（非阻塞）直到信号量可用。

这是使用 Python3.6 编写的；未在任何其他版本上测试。

这只有在从同一个线程进行同步时才有效，使用这种机制来自不同进程的 IPC 将失败。

import threading
from  time import sleep
sem = threading.Semaphore()

def fun1():
    print("fun1 starting")
    sem.acquire()
    for loop in range(1,5):
        print("Fun1 Working {}".format(loop))
        sleep(1)
    sem.release()
    print("fun1 finished")



def fun2():
    print("fun2 starting")
    while not sem.acquire(blocking=False):
        print("Fun2 No Semaphore available")
        sleep(1)
    else:
        print("Got Semphore")
        for loop in range(1, 5):
            print("Fun2 Working {}".format(loop))
            sleep(1)
    sem.release()

t1 = threading.Thread(target = fun1)
t2 = threading.Thread(target = fun2)
t1.start()
t2.start()
t1.join()
t2.join()
print("All Threads done Exiting")

当我运行它时 - 我得到以下输出。

fun1 starting
Fun1 Working 1
fun2 starting
Fun2 No Semaphore available
Fun1 Working 2
Fun2 No Semaphore available
Fun1 Working 3
Fun2 No Semaphore available
Fun1 Working 4
Fun2 No Semaphore available
fun1 finished
Got Semphore
Fun2 Working 1
Fun2 Working 2
Fun2 Working 3
Fun2 Working 4
All Threads done Exiting

Answer 3

另外，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]：

The difference between mutex, semaphore, and lock

Answer 4

其实我是想找asyncio.Semaphores，不是threading.Semaphore， 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