如何从线程异步获取返回值？

Question

我的问题：启动一个线程函数并异步根据返回的值进行操作

我知道如何：

• 使用threading 启动线程函数。问题：没有简单的方法可以返回结果
• get the return value 来自线程函数。问题：它是同步的

我想要实现的和 JavaScript 类似的

aFunctionThatReturnsAPromise()
  .then(r => {// do something with the returned value when it is available})
// the code here runs synchronously right after aFunctionThatReturnsAPromise is started

在伪 Python 中，我会考虑类似（将示例从 the answer 修改为链接线程）

import time
import concurrent.futures

def foo(bar):
    print('hello {}'.format(bar))
    time.sleep(10)
    return 'foo'

def the_callback(something):
    print(f"the thread returned {something}")

with concurrent.futures.ThreadPoolExecutor() as executor:
    # submit the threaded call ...
    future = executor.submit(foo, 'world!')
# ... and set a callback
future.callback(the_callback, future.result())  # ← this is the made up part
# or, all in one: future = executor.submit(foo, 'world!', callback=the_callback) # in which case the parameters probably would need to be passed the JS way
# the threaded call runs at its pace
# the following line is ran right after the call above
print("after submit")
# after some time (~10 seconds) the callback is finished (and has printed out what was passed to it)
# there should probably be some kind of join() so that the scripts waits until the thread is done

如果可能的话，我想留在线程中（它们按照自己的节奏做事，我不在乎它们何时完成），而不是 asyncio（我必须在单个线程中明确地 await 事情）

Answer 1

您可以使用concurrent.futures.add_done_callback，如下所示。回调必须是一个带有单个参数的可调用对象，Future 实例——它必须从中获取结果，如图所示。该示例还向其中添加了一些附加信息，回调函数用于打印其消息。

请注意，回调函数将被并发调用，因此如果涉及共享资源，则应采取通常的互斥预防措施。这没有在下面的示例中完成，所以有时打印输出会混乱。

from concurrent import futures
import random
import time

def foo(bar, delay):
    print(f'hello {bar} - {delay}')
    time.sleep(delay)
    return bar

def the_callback(fn):
    if fn.cancelled():
        print(f'args {fn.args}: canceled')
    elif fn.done():
        error = fn.exception()
        if error:
            print(f'args {fn.args}: caused error {error}')
        else:
            print(f'args {fn.args}: returned: {fn.result()}')

with futures.ThreadPoolExecutor(max_workers=2) as executor:
    for name in ('foo', 'bar', 'bas'):
        delay = random.randint(1, 5)
        f = executor.submit(foo, name, delay)
        f.args = name, delay
        f.add_done_callback(the_callback)

print('fini')

样本输出：

hello foo - 5
hello bar - 3
args ('bar', 3): returned: bar
hello bas - 4
args ('foo', 5): returned: foo
args ('bas', 4): returned: bas
fini

Answer 2

您可以使用concurrent.futures 库中的add_done_callback，因此您可以像这样修改您的示例：

def the_callback(something):
    print(f"the thread returned {something.result()}")
 
with concurrent.futures.ThreadPoolExecutor() as executor:
    future = executor.submit(foo, 'world!')
    future.add_done_callback(the_callback)