Semaphores on Python

Question

I've started programming in Python a few weeks ago and was trying to use Semaphores to synchronize two simple threads, for learning purposes. Here is what I've got:

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()

But it keeps printing just 1's. How can I intercale the prints?

Answer 1

It is working fine, its just that its printing too fast for you to see . Try putting a time.sleep() in both functions (a small amount) to sleep the thread for that much amount of time, to actually be able to see both 1 as well as 2.

Example -

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

You can also use Lock/mutex method as following:

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()

Another/Simpler usage type:

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()

Note: additionaly difference between mutex and semaphore, mutex VS semaphore

Answer 3

I used this code to demonstrate how 1 thread can use a Semaphore and the other thread will wait (non-blocking) until the Sempahore is available.

This was written using Python3.6; Not tested on any other version.

This will only work is the synchronization is being done from the same thread, IPC from separate processes will fail using this mechanism.

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")

When I run this - I get the following output.

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