Python asyncio task ordering

Question

I have a question about how the event loop in python's asyncio module manages outstanding tasks. Consider the following code:

import asyncio

@asyncio.coroutine
def a():
   for i in range(0, 3):
      print('a.' + str(i))
      yield


@asyncio.coroutine
def b():
   for i in range(0, 3):
      print('b.' + str(i))
      yield


@asyncio.coroutine
def c():
   for i in range(0, 3):
      print('c.' + str(i))
      yield


tasks = [
   asyncio.Task(a()),
   asyncio.Task(b()),
   asyncio.Task(c()),
]

loop = asyncio.get_event_loop()
loop.run_until_complete(asyncio.wait([t1, t2, t3]))

Running this will print:

a.0
b.0
c.0
a.1
b.1
c.1
a.2
b.2
c.2

Notice that it always prints out 'a' then 'b' then 'c'. I'm guessing that no matter how many iterations each coroutine goes through it will always print in that order. So you'd never see something like

b.100
c.100
a.100

Coming from a node.js background, this tells me that the event loop here is maintaining a queue internally that it uses to decide which task to run next. It initially puts a() at the front of the queue, then b(), then c() since that's the order of the tasks in the list passed to asyncio.wait(). Then whenever it hits a yield statement it puts that task at the end of the queue. I guess in a more realistic example, say if you were doing an async http request, it would put a() back on the end of the queue after the http response came back.

Can I get an amen on this?

Answer 1

Currently your example doesn't include any blocking I/O code. Try this to simulate some tasks:

import asyncio


@asyncio.coroutine
def coro(tag, delay):
    for i in range(1, 8):
        print(tag, i)
        yield from asyncio.sleep(delay)


loop = asyncio.get_event_loop()

print("---- await 0 seconds :-) --- ")
tasks = [
    asyncio.Task(coro("A", 0)),
    asyncio.Task(coro("B", 0)),
    asyncio.Task(coro("C", 0)),
]

loop.run_until_complete(asyncio.wait(tasks))

print("---- simulate some blocking I/O --- ")
tasks = [
    asyncio.Task(coro("A", 0.1)),
    asyncio.Task(coro("B", 0.3)),
    asyncio.Task(coro("C", 0.5)),
]

loop.run_until_complete(asyncio.wait(tasks))

loop.close()

As you can see, coroutines are scheduled as needed, and not in order.

Answer 2

DISCLAIMER For at least v3.9 with the default implementation this appears to be true. However, the inner workings of the event loop are not public interface and thus may be changed with new versions. Additionally, asyncio allows for BaseEventLoop implementation to be substituted, which may change its behavior.

When a Task object is created, it calls loop.call_soon to register its _step method as a callback. The _step method actually does the work of calling your coroutine with calls to send() and processing the results.

In BaseEventLoop, loop.call_soon places the _step callback at the end of a _ready list of callbacks. Each run of the event loop, iterates the list of _ready callbacks in a FIFO order and calls them. Thus, for the initial run of tasks, they are executed in the order they are created.

When the task awaits or yields a future, it really depends on the nature of that future when the task's _wakeup method get put into the queue.

Also, note that other callbacks can be registered in between creation of tasks. While it is true that if TaskA is created before TaskB, the initial run of TaskA will happen before TaskB, there could still be other callbacks that get run in between.

Last, the above behavior is also for the default Task class that comes with asyncio. Its possible however to specify a custom task factory and use an alternative task implementation which could also change this behavior.