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Creating non-blocking socket in python

I was trying to understand how non-blocking sockets work ,so I wrote this simple server in python .

import socket

s=socket.socket(socket.AF_INET, socket.SOCK_STREAM)

s.bind(('127.0.0.1',1000))
s.listen(5)
s.setblocking(0)
while True:
      try:
            conn, addr = s.accept()
            print ('connection from',addr)
            data=conn.recv(100)
            print ('recived: ',data,len(data))

      except:
          pass  

Then I tried to connect to this server from multiple instances of this client

import socket

s=socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect(('127.0.0.1',1000))
while True:
    continue

But for some reason putting blocking to 0 or 1 dose not seem to have an effect and server's recv method always block the execution. So, dose creating non-blocking socket in python require more than just setting the blocking flag to 0.

like image 992
Gray Avatar asked Mar 18 '23 16:03

Gray


2 Answers

setblocking only affects the socket you use it on. So you have to add conn.setblocking(0) to see an effect: The recv will then return immediately if there is no data available.

like image 69
Phillip Avatar answered Mar 28 '23 13:03

Phillip


You just need to call setblocking(0) on the connected socket, i.e. conn.

import socket

s = socket.socket()
s.bind(('127.0.0.1', 12345))
s.listen(5)
s.setblocking(0)
>>> conn, addr = s.accept()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib64/python2.7/socket.py", line 202, in accept
    sock, addr = self._sock.accept()
socket.error: [Errno 11] Resource temporarily unavailable

# start your client...
>>> conn, addr = s.accept()
>>> conn.recv()            # this will hang until the client sends some data....
'hi there\n'
>>> conn.setblocking(0)    # set non-blocking on the connected socket "conn"
>>> conn.recv()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
socket.error: [Errno 11] Resource temporarily unavailable
like image 29
mhawke Avatar answered Mar 28 '23 11:03

mhawke