Problem: To know when will a launched thread actually starts running.
A key feature of threads is that they execute independently and nondeterministically. This can present a tricky synchronization problem if other threads in the program need to know if a thread has reached a certain point in its execution before carrying out further operations. To solve such problems, use the Event object from the threading library.
Event instances are similar to a “sticky” flag that allows threads to wait for something to happen. Initially, an event is set to 0. If the event is unset and a thread waits on the event, it will block (i.e., go to sleep) until the event gets set. A thread that sets the event will wake up all of the threads that happen to be waiting (if any). If a thread waits on an event that has already been set, it merely moves on, continuing to execute.
Code #1 : Code that uses an Event to coordinate the startup of a thread.
from threading import Thread, Event import time # Code to execute in an independent thread def countdown(n, started_evt): print('countdown starting') started_evt.set() while n > 0: print('T-minus', n) n -= 1 time.sleep(5) # Create the event object that # will be used to signal startup started_evt = Event() # Launch the thread and pass the startup event print('Launching countdown') t = Thread(target = countdown, args =(10, started_evt)) t.start() # Wait for the thread to start started_evt.wait() print('countdown is running') |
On running the code above, the “countdown is running” message will always appear after the “countdown starting” message. This is coordinated by the event that makes the main thread wait until the countdown() function has first printed the startup message.
- Event objects are best used for one-time events. That is, create an event, threads wait for the event to be set, and once set, the Event is discarded.
- Although it is possible to clear an event using its
clear()method, safely clearing an event and waiting for it to be set again is tricky to coordinate, and can lead to missed events, deadlock, or other problems (in particular, it can’t be guaranteed that a request to clear an event after setting it will execute before a released thread cycles back to wait on the event again). - If a thread is going to repeatedly signal an event over and over, it’s probably better off using a Condition object instead.
Code #2 : Implementing a periodic timer that other threads can monitor to see whenever the timer expires.
import threading import time class PeriodicTimer: def __init__(self, interval): self._interval = interval self._flag = 0 self._cv = threading.Condition() def start(self): t = threading.Thread(target = self.run) t.daemon = True t.start() def run(self): ''' Run the timer and notify waiting threads after each interval ''' while True: time.sleep(self._interval) with self._cv: self._flag ^= 1 self._cv.notify_all() def wait_for_tick(self): ''' Wait for the next tick of the timer ''' with self._cv: last_flag = self._flag while last_flag == self._flag: self._cv.wait() |
Code #3 : Use of timer
# Example use of the timer ptimer = PeriodicTimer(5) ptimer.start() # Two threads that synchronize on the timer def countdown(nticks): while nticks > 0: ptimer.wait_for_tick() print('T-minus', nticks) nticks -= 1 def countup(last): n = 0 while n < last: ptimer.wait_for_tick() print('Counting', n) n += 1 threading.Thread(target = countdown, args =(10, )).start() threading.Thread(target = countup, args =(5, )).start() |
A critical feature of Event objects is that they wake all waiting threads. If writing a program where it is only desired to wake up a single waiting thread, it is probably better to use a Semaphore or Condition object instead.
Code #4 : Code involving semaphores
def worker(n, sema): # Wait to be signaled sema.acquire() # Do some work print('Working', n) # Create some threads sema = threading.Semaphore(0) nworkers = 10 for n in range(nworkers): t = threading.Thread(target = worker, args =(n, sema, )) t.start() |
On running this code, a pool of threads will start, but nothing happens because they’re all blocked waiting to acquire the semaphore. Each time the semaphore is released, only one worker will wake up and run as shown in the code below
Code #5 :
sema.release() sema.release() |
Output :
Working 0 Working 1
Writing code that involves a lot of tricky synchronization between threads is likely to make your head explode. A more sane approach is to thread threads as communicating tasks using queues or as actors.
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