Python上的信号量

Question

几周前，我开始用Python编程，为了学习的目的，我试图使用信号量来同步两个简单的线程。以下是我所得到的：

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()

但它只印了1个。我怎么能把这些印出来？

Answer 1

它运行良好，只是它的打印速度太快，你看不见。试着在两个函数中都放一个time.sleep()（少量）来让线程睡眠那么长时间，以便能够同时看到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 2

我用这段代码演示了一个线程如何使用一个信号量，而另一个线程将等待（非阻塞）直到Sempahore可用。

这是用Python3.6编写的；没有在任何其他版本上进行测试。

只有同步是从同一个线程完成的，来自不同进程的IPC才会使用此机制失败。

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")

当我运行这个-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

您还可以按如下方式使用Lock/mutex方法：

import threading
import time

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


def fun1():
    while True:
        mutex.acquire()
        print(1)
        mutex.release()
        time.sleep(0.5)


def fun2():
    while True:
        mutex.acquire()
        print(2)
        mutex.release()
        time.sleep(0.5)

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

另一种/更简单的用法类型：

import threading
import time

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


def fun1():
    while True:
        with mutex:
            print(1)
        time.sleep(0.5)


def fun2():
    while True:
        with mutex:
            print(2)
        time.sleep(0.5)

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

---

[注意：

另外，the difference between mutex, semaphore, and lock