下面的函数foo返回一个字符串'foo'。我如何才能获得从线程的目标返回的值'foo' ?
from threading import Thread
def foo(bar):
print('hello {}'.format(bar))
return 'foo'
thread = Thread(target=foo, args=('world!',))
thread.start()
return_value = thread.join()
上面所示的“一种明显的方法”不起作用:thread.join()返回None。
当前回答
我知道这个线程是旧的....但我也遇到了同样的问题…如果你愿意使用thread.join()
import threading
class test:
def __init__(self):
self.msg=""
def hello(self,bar):
print('hello {}'.format(bar))
self.msg="foo"
def main(self):
thread = threading.Thread(target=self.hello, args=('world!',))
thread.start()
thread.join()
print(self.msg)
g=test()
g.main()
其他回答
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
我知道这个线程是旧的....但我也遇到了同样的问题…如果你愿意使用thread.join()
import threading
class test:
def __init__(self):
self.msg=""
def hello(self,bar):
print('hello {}'.format(bar))
self.msg="foo"
def main(self):
thread = threading.Thread(target=self.hello, args=('world!',))
thread.start()
thread.join()
print(self.msg)
g=test()
g.main()
Parris / kindall的answer join/return answer移植到Python 3:
from threading import Thread
def foo(bar):
print('hello {0}'.format(bar))
return "foo"
class ThreadWithReturnValue(Thread):
def __init__(self, group=None, target=None, name=None, args=(), kwargs=None, *, daemon=None):
Thread.__init__(self, group, target, name, args, kwargs, daemon=daemon)
self._return = None
def run(self):
if self._target is not None:
self._return = self._target(*self._args, **self._kwargs)
def join(self):
Thread.join(self)
return self._return
twrv = ThreadWithReturnValue(target=foo, args=('world!',))
twrv.start()
print(twrv.join()) # prints foo
注意，Thread类在Python 3中实现的方式不同。
这是我根据@Kindall的回答创建的版本。
这个版本使得您所要做的就是输入带有参数的命令来创建新线程。
这是用Python 3.8做的:
from threading import Thread
from typing import Any
def test(plug, plug2, plug3):
print(f"hello {plug}")
print(f'I am the second plug : {plug2}')
print(plug3)
return 'I am the return Value!'
def test2(msg):
return f'I am from the second test: {msg}'
def test3():
print('hello world')
def NewThread(com, Returning: bool, *arguments) -> Any:
"""
Will create a new thread for a function/command.
:param com: Command to be Executed
:param arguments: Arguments to be sent to Command
:param Returning: True/False Will this command need to return anything
"""
class NewThreadWorker(Thread):
def __init__(self, group = None, target = None, name = None, args = (), kwargs = None, *,
daemon = None):
Thread.__init__(self, group, target, name, args, kwargs, daemon = daemon)
self._return = None
def run(self):
if self._target is not None:
self._return = self._target(*self._args, **self._kwargs)
def join(self):
Thread.join(self)
return self._return
ntw = NewThreadWorker(target = com, args = (*arguments,))
ntw.start()
if Returning:
return ntw.join()
if __name__ == "__main__":
print(NewThread(test, True, 'hi', 'test', test2('hi')))
NewThread(test3, True)
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
