Python: Binding Socket: “ Address already in use”

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最佳答案

Try using the SO_REUSEADDR socket option before binding the socket.

comSocket.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)

Edit: I see you're still having trouble with this. There is a case where SO_REUSEADDR won't work. If you try to bind a socket and reconnect to the same destination (with SO_REUSEADDR enabled), then TIME_WAIT will still be in effect. It will however allow you to connect to a different host:port.

A couple of solutions come to mind. You can either continue retrying until you can gain a connection again. Or if the client initiates the closing of the socket (not the server), then it should magically work.

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socket.socket() should run before socket.bind() and use REUSEADDR as said

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As Felipe Cruze mentioned, you must set the SO_REUSEADDR before binding. I found a solution on another site - solution on other site, reproduced below

The problem is that the SO_REUSEADDR socket option must be set before the address is bound to the socket. This can be done by subclassing ThreadingTCPServer and overriding the server_bind method as follows:

import SocketServer, socket


class MyThreadingTCPServer(SocketServer.ThreadingTCPServer):
def server_bind(self):
self.socket.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
self.socket.bind(self.server_address)

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I know you've already accepted an answer but I believe the problem has to do with calling bind() on a client socket. This might be OK but bind() and shutdown() don't seem to play well together. Also, SO_REUSEADDR is generally used with listen sockets. i.e. on the server side.

You should be passing and ip/port to connect(). Like this:

comSocket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
comSocket.connect(('', 5555))

Don't call bind(), don't set SO_REUSEADDR.

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For me the better solution was the following. Since the initiative of closing the connection was done by the server, the setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1) had no effect and the TIME_WAIT was avoiding a new connection on the same port with error:

[Errno 10048]: Address already in use. Only one usage of each socket address (protocol/IP address/port) is normally permitted

I finally used the solution to let the OS choose the port itself, then another port is used if the precedent is still in TIME_WAIT.

I replaced:

self._socket.bind((guest, port))

with:

self._socket.bind((guest, 0))

As it was indicated in the python socket documentation of a tcp address:

If supplied, source_address must be a 2-tuple (host, port) for the socket to bind to as its source address before connecting. If host or port are ‘’ or 0 respectively the OS default behavior will be used.

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Here is the complete code that I've tested and absolutely does NOT give me a "address already in use" error. You can save this in a file and run the file from within the base directory of the HTML files you want to serve. Additionally, you could programmatically change directories prior to starting the server

import socket
import SimpleHTTPServer
import SocketServer
# import os # uncomment if you want to change directories within the program


PORT = 8000


# Absolutely essential!  This ensures that socket resuse is setup BEFORE
# it is bound.  Will avoid the TIME_WAIT issue


class MyTCPServer(SocketServer.TCPServer):
def server_bind(self):
self.socket.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
self.socket.bind(self.server_address)


Handler = SimpleHTTPServer.SimpleHTTPRequestHandler


httpd = MyTCPServer(("", PORT), Handler)


# os.chdir("/My/Webpages/Live/here.html")


httpd.serve_forever()


# httpd.shutdown() # If you want to programmatically shut off the server

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According to this link

Actually, SO_REUSEADDR flag can lead to much greater consequences: SO_REUSADDR permits you to use a port that is stuck in TIME_WAIT, but you still can not use that port to establish a connection to the last place it connected to. What? Suppose I pick local port 1010, and connect to foobar.com port 300, and then close locally, leaving that port in TIME_WAIT. I can reuse local port 1010 right away to connect to anywhere except for foobar.com port 300.

However you can completely avoid TIME_WAIT state by ensuring that the remote end initiates the closure (close event). So the server can avoid problems by letting the client close first. The application protocol must be designed so that the client knows when to close. The server can safely close in response to an EOF from the client, however it will also need to set a timeout when it is expecting an EOF in case the client has left the network ungracefully. In many cases simply waiting a few seconds before the server closes will be adequate.

I also advice you to learn more about networking and network programming. You should now at least how tcp protocol works. The protocol is quite trivial and small and hence, may save you a lot of time in future.

With netstat command you can easily see which programs ( (program_name,pid) tuple) are binded to which ports and what is the socket current state: TIME_WAIT, CLOSING, FIN_WAIT and so on.

A really good explanation of linux network configurations can be found https://serverfault.com/questions/212093/how-to-reduce-number-of-sockets-in-time-wait.

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another solution, in development environment of course, is killing process using it, for example

def serve():
server = HTTPServer(('', PORT_NUMBER), BaseHTTPRequestHandler)
print 'Started httpserver on port ' , PORT_NUMBER
server.serve_forever()
try:
serve()
except Exception, e:
print "probably port is used. killing processes using given port %d, %s"%(PORT_NUMBER,e)
os.system("xterm -e 'sudo fuser -kuv %d/tcp'" % PORT_NUMBER)
serve()
raise e

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You need to set the allow_reuse_address before binding. Instead of the SimpleHTTPServer run this snippet:

Handler = SimpleHTTPServer.SimpleHTTPRequestHandler
httpd = SocketServer.TCPServer(("", PORT), Handler, bind_and_activate=False)
httpd.allow_reuse_address = True
httpd.server_bind()
httpd.server_activate()
httpd.serve_forever()

This prevents the server from binding before we got a chance to set the flags.

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In case you face the problem using TCPServer or SimpleHTTPServer, override SocketServer.TCPServer.allow_reuse_address (python 2.7.x) or socketserver.TCPServer.allow_reuse_address (python 3.x) attribute

class MyServer(SocketServer.TCPServer):
allow_reuse_address = True


server = MyServer((HOST, PORT), MyHandler)
server.serve_forever()

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I think the best way is just to kill the process on that port, by typing in the terminal fuser -k [PORT NUMBER]/tcp, e.g. fuser -k 5001/tcp.

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I found another reason for this exception. When running the application from Spyder IDE (in my case it was Spyder3 on Raspbian) and the program terminated by ^C or an exception, the socket was still active:

sudo netstat -ap | grep 31416
tcp  0  0 0.0.0.0:31416  0.0.0.0:*    LISTEN      13210/python3

Running the program again found the "Address already in use"; the IDE seems to start the new 'run' as a separate process which finds the socket used by the previous 'run'.

socket.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)

did NOT help.

Killing process 13210 helped. Starting the python script from command-line like

python3 <app-name>.py

always worked well when SO_REUSEADDR was set to true. The new Thonny IDE or Idle3 IDE did not have this problem.

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I had the same problem and I couldn't find any other solution (reuse options didn't work) except restarting Raspberry Pi each time. Then I found a workaround;

comSocket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
comSocket.close()
comSocket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
comSocket.connect(('', 5555))

This means, define socket first, close it, then define again, so you can use the same port if it is stuck.