我使用tornado.ioloop.IOLoop.run_in_executor将同步函数更改为异步函数,但是每次调用该函数时,都会创建一个线程,而不是终止线程。
下面是一个模拟可复制的示例(至少在我的机器上是可复制的):
#!/usr/bin/env python3
import time
import tornado.ioloop
import tornado.web
def slow_func():
time.sleep(1)
print('slow func returned')
return 'succ\n'
class TestHandler(tornado.web.RequestHandler):
async def get(self):
print('GET called')
try:
result = await tornado.ioloop.IOLoop.current().run_in_executor(None, slow_func)
except Exception as e:
print(e)
self.write(result)
print('GET returned')
if __name__ == '__main__':
tornado.web.Application([
(r'/', TestHandler),
]).listen(3000)
print('Serving at 3000')
tornado.ioloop.IOLoop.current().start()对此TestHandler的每个请求都将创建一个新线程来运行slow_func,但是在函数返回后线程仍然保持不变。我可以在ps H中看到它们,它会创建新的线程,直到达到ulimit为止。我这里的环境是:
$ uname -a
Linux xxx 2.6.32-754.6.3.el6.x86_64 #1 SMP Tue Sep 18 10:29:08 EDT 2018 x86_64 x86_64 x86_64 GNU/Linux
$ python3 --version
Python 3.7.4
$ pip3 show tornado
Name: tornado
Version: 6.0.3
Summary: Tornado is a Python web framework and asynchronous networking library, originally developed at FriendFeed.
Home-page: http://www.tornadoweb.org/
Author: Facebook
Author-email: python-tornado@googlegroups.com
License: http://www.apache.org/licenses/LICENSE-2.0
Location: /xxx/lib/python3.7/site-packages
Requires:
Required-by:tornado.ioloop.IOLoop.run_in_executor使用concurrent.futures.Executor并返回一个可访问的Future对象。13
函数返回后,线程在做什么?为什么在未来的目标解决后他们不被杀死?我该怎么避免这件事?
发布于 2019-08-29 10:03:17
run_in_executor接受一个concurrent.futures.Executor对象作为第一个参数。
您可以创建一个执行器并限制线程池的大小:
from concurrent.futures import ThreadPoolExecutor
executor = ThreadPoolExecutor(max_workers=8)
IOLoop.current().run_in_executor(executor, slow_func) https://stackoverflow.com/questions/57707073
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