无法在ARM处理器上运行Python3 HTTPServer

Question

I recently bought Odroid XU4 , a single-board computer with an ARM CPU. I try to run a simple web server using HTTTPServer on Python3.

import http.server
import socketserver

PORT = 8000

Handler = http.server.SimpleHTTPRequestHandler

with socketserver.TCPServer(("", PORT), Handler) as httpd:
    print("serving at port", PORT)
    httpd.serve_forever()

This code runs well on my Mac machine. But when I try to run this on Odroid XU4, I got this error message.

$ python3 webserver.py
Traceback (most recent call last):
  File "test.py", line 8, in <module>
    with socketserver.TCPServer(("", PORT), Handler) as httpd:
AttributeError: __exit__

Can anyone explain why I got this error? For your information, I've attached the information about the OS and Python interpreter.

$ uname -a
Linux odroid 4.9.44-54 #1 SMP PREEMPT Sun Aug 20 20:24:08 UTC 2017 armv7l armv7l armv7l GNU/Linu

$ python 
Python 3.5.2 (default, Aug 18 2017, 17:48:00)
[GCC 5.4.0 20160609] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>>

Answer 1

From the documentation it would seem that the contextmanager protocol ( with ... ) form for TCPServer 's base class (and therefore TCPServer was added in python3.6. This is not available in python3.5

Changed in version 3.6: Support for the context manager protocol was added. Exiting the context manager is equivalent to calling server_close() .

Fortunately, you can use the previous approach. This roughly means taking your with statement and turning it into a plain assignment:

httpd = socketserver.TCPServer(("", PORT), Handler)
print("serving at port", PORT)
httpd.serve_forever()