How to use Queue with threading properly

Question

I am new to queue & threads kindly help with the below code, here I am trying to execute the function hd , I need to run the function multiple times but only after a single run has been completed

import queue
import threading
import time

fifo_queue = queue.Queue()

def hd():
    print("hi")
    time.sleep(1)
    print("done")


for i in range(3):
    cc = threading.Thread(target=hd)
    fifo_queue.put(cc)
    cc.start()

Current Output

hi
hi
hi
donedonedone

Expected Output

hi
done   
hi
done
hi
done​

Answer 1

You can use a Semaphore for your purposes

A semaphore manages an internal counter which is decremented by each acquire() call and incremented by each release() call. The counter can never go below zero; when acquire() finds that it is zero, it blocks, waiting until some other thread calls release().

A default value of Semaphore is 1 ,

class threading.Semaphore(value=1)

so only one thread would be active at once:

import queue
import threading
import time

fifo_queue = queue.Queue()

semaphore = threading.Semaphore()


def hd():
    with semaphore:
        print("hi")
        time.sleep(1)
        print("done")


for i in range(3):
    cc = threading.Thread(target=hd)
    fifo_queue.put(cc)
    cc.start()

hi
done
hi
done
hi
done

---

As @user2357112supportsMonica mentioned in comments RLock would be more safe option

class threading.RLock

This class implements reentrant lock objects. A reentrant lock must be released by the thread that acquired it . Once a thread has acquired a reentrant lock, the same thread may acquire it again without blocking; the thread must release it once for each time it has acquired it.

import queue
import threading
import time

fifo_queue = queue.Queue()

lock = threading.RLock()


def hd():
    with lock:
        print("hi")
        time.sleep(1)
        print("done")


for i in range(3):
    cc = threading.Thread(target=hd)
    fifo_queue.put(cc)
    cc.start()

Answer 2

please put the print("down") before sleep. it will work fine. Reason: your program will do this: thread1
: print
sleep
print but while the thread is sleeping, other threads will be working and printing their first command. in my way the thread will write the first, write the second and then go to sleep and wait for other threads to show up.