Kindall在Python3中的回答

class ThreadWithReturnValue(Thread):
    def __init__(self, group=None, target=None, name=None,
                 args=(), kwargs={}, *, daemon=None):
        Thread.__init__(self, group, target, name, args, kwargs, daemon)
        self._return = None 

    def run(self):
        try:
            if self._target:
                self._return = self._target(*self._args, **self._kwargs)
        finally:
            del self._target, self._args, self._kwargs 

    def join(self,timeout=None):
        Thread.join(self,timeout)
        return self._return

join总是返回None，我认为你应该子类化Thread来处理返回代码等。

我发现做到这一点的最短和最简单的方法是利用Python类及其动态属性。您可以使用threading.current_thread()从派生线程的上下文中检索当前线程，并将返回值赋给一个属性。

import threading

def some_target_function():
    # Your code here.
    threading.current_thread().return_value = "Some return value."

your_thread = threading.Thread(target=some_target_function)
your_thread.start()
your_thread.join()

return_value = your_thread.return_value
print(return_value)

定义你的目标 1)采取一个论点q 2)用q.put(foo)替换return foo的任何语句;返回

一个函数

def func(a):
    ans = a * a
    return ans

将成为

def func(a, q):
    ans = a * a
    q.put(ans)
    return

然后你就可以这样做了

from Queue import Queue
from threading import Thread

ans_q = Queue()
arg_tups = [(i, ans_q) for i in xrange(10)]

threads = [Thread(target=func, args=arg_tup) for arg_tup in arg_tups]
_ = [t.start() for t in threads]
_ = [t.join() for t in threads]
results = [q.get() for _ in xrange(len(threads))]

你可以使用函数装饰器/包装器来实现它，这样你就可以使用现有的函数作为目标，而不需要修改它们，但要遵循这个基本方案。

GuySoft的想法很棒，但我认为对象不一定要从Thread继承，start()可以从接口中删除:

from threading import Thread
import queue
class ThreadWithReturnValue(object):
    def __init__(self, target=None, args=(), **kwargs):
        self._que = queue.Queue()
        self._t = Thread(target=lambda q,arg1,kwargs1: q.put(target(*arg1, **kwargs1)) ,
                args=(self._que, args, kwargs), )
        self._t.start()

    def join(self):
        self._t.join()
        return self._que.get()


def foo(bar):
    print('hello {0}'.format(bar))
    return "foo"

twrv = ThreadWithReturnValue(target=foo, args=('world!',))

print(twrv.join())   # prints foo

我偷了kindall的答案，稍微整理了一下。

关键部分是为join()添加*args和**kwargs，以便处理超时

class threadWithReturn(Thread):
    def __init__(self, *args, **kwargs):
        super(threadWithReturn, self).__init__(*args, **kwargs)
        
        self._return = None
    
    def run(self):
        if self._Thread__target is not None:
            self._return = self._Thread__target(*self._Thread__args, **self._Thread__kwargs)
    
    def join(self, *args, **kwargs):
        super(threadWithReturn, self).join(*args, **kwargs)
        
        return self._return

更新答案如下

这是我得到最多好评的答案，所以我决定更新可以在py2和py3上运行的代码。

此外，我看到许多对这个问题的回答都显示出对Thread.join()缺乏理解。有些完全不能处理timeout参数。但是当你有(1)一个可以返回None的目标函数并且(2)你也将timeout参数传递给join()时，还有一种极端情况你应该注意。请参阅“TEST 4”以理解这个极端情况。

ThreadWithReturn类，用于py2和py3:

import sys
from threading import Thread
from builtins import super    # https://stackoverflow.com/a/30159479

_thread_target_key, _thread_args_key, _thread_kwargs_key = (
    ('_target', '_args', '_kwargs')
    if sys.version_info >= (3, 0) else
    ('_Thread__target', '_Thread__args', '_Thread__kwargs')
)

class ThreadWithReturn(Thread):
    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)
        self._return = None
    
    def run(self):
        target = getattr(self, _thread_target_key)
        if target is not None:
            self._return = target(
                *getattr(self, _thread_args_key),
                **getattr(self, _thread_kwargs_key)
            )
    
    def join(self, *args, **kwargs):
        super().join(*args, **kwargs)
        return self._return

一些示例测试如下所示:

import time, random

# TEST TARGET FUNCTION
def giveMe(arg, seconds=None):
    if not seconds is None:
        time.sleep(seconds)
    return arg

# TEST 1
my_thread = ThreadWithReturn(target=giveMe, args=('stringy',))
my_thread.start()
returned = my_thread.join()
# (returned == 'stringy')

# TEST 2
my_thread = ThreadWithReturn(target=giveMe, args=(None,))
my_thread.start()
returned = my_thread.join()
# (returned is None)

# TEST 3
my_thread = ThreadWithReturn(target=giveMe, args=('stringy',), kwargs={'seconds': 5})
my_thread.start()
returned = my_thread.join(timeout=2)
# (returned is None) # because join() timed out before giveMe() finished

# TEST 4
my_thread = ThreadWithReturn(target=giveMe, args=(None,), kwargs={'seconds': 5})
my_thread.start()
returned = my_thread.join(timeout=random.randint(1, 10))

你能确定我们在测试4中可能遇到的极端情况吗?

问题是我们期望giveMe()返回None(参见TEST 2)，但我们也期望join()在超时时返回None。

None表示:

(1)这就是giveMe()返回的，或者

(2) join()超时

这个例子很简单，因为我们知道giveMe()总是返回None。但在真实的实例中(目标可能返回None或其他内容)，我们希望显式地检查发生了什么。

下面是如何解决这种极端情况:

# TEST 4
my_thread = ThreadWithReturn(target=giveMe, args=(None,), kwargs={'seconds': 5})
my_thread.start()
returned = my_thread.join(timeout=random.randint(1, 10))

if my_thread.isAlive():
    # returned is None because join() timed out
    # this also means that giveMe() is still running in the background
    pass
    # handle this based on your app's logic
else:
    # join() is finished, and so is giveMe()
    # BUT we could also be in a race condition, so we need to update returned, just in case
    returned = my_thread.join()