Python Socket accept() Timeout-CodePudding

Using sockets, the below Python code opens a port and waits for a connection. Then, it sends a response when a connection is made.

import socket
ip = 127.0.0.1
port = 80
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
s.bind((ip, port))
s.listen(1)
conn, addr = s.accept()
conn.send(response)
conn.close()

If a connection is not established within 10 seconds, I would like it to move on. Is there a way to define a timeout for s.accept()?

  s.accept(timeout=10)

Maybe something like the above line?

Thanks in advance for your help!

CodePudding user response：

Use socket.settimeout:

s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.settimeout(10)

timeout = False
while not timeout:
  try: 
    (conn, addr) = s.accept() 
  except socket.timeout:
    pass
  else:
    # your code

To make a 10 second timeout the default behavior for any socket, you can use

socket.setdefaulttimeout(10)

CodePudding user response：

to set a timeout for socket s before you ~~connect~~ listen do s.settimeout(10)

edit

I assume it works when listening