socket.error:[errno 99] cannot assign requested address and namespace in python
Solution 1:
Stripping things down to basics this is what you would want to test with:
import socket
server = socket.socket()
server.bind(("10.0.0.1", 6677))
server.listen(4)
client_socket, client_address = server.accept()
print(client_address, "has connected")
while True:
recvieved_data = client_socket.recv(1024)
print(recvieved_data)
This works assuming a few things:
- Your local IP address (on the server) is 10.0.0.1 (This video shows you how)
- No other software is listening on port 6677
Also note the basic concept of IP addresses:
Try the following, open the start menu, in the "search" field type cmd
and press enter.
Once the black console opens up type ping www.google.com
and this should give you and IP address for google. This address is googles local IP and they bind to that and obviously you can not bind to an IP address owned by google.
With that in mind, you own your own set of IP addresses.
First you have the local IP of the server, but then you have the local IP of your house.
In the below picture 192.168.1.50
is the local IP of the server which you can bind to.
You still own 83.55.102.40
but the problem is that it's owned by the Router and not your server. So even if you visit http://whatsmyip.com and that tells you that your IP is 83.55.102.40
that is not the case because it can only see where you're coming from.. and you're accessing your internet from a router.
In order for your friends to access your server (which is bound to 192.168.1.50
) you need to forward port 6677
to 192.168.1.50
and this is done in your router.
Assuming you are behind one.
If you're in school there's other dilemmas and routers in the way most likely.
Solution 2:
This error will also appear if you try to connect to an exposed port from within a Docker container, when nothing is actively serving the port.
On a host where nothing is listening/bound to that port you'd get a No connection could be made because the target machine actively refused it
error instead when making a request to a local URL that is not served, eg: localhost:5000
. However, if you start a container that binds to the port, but there is no server running inside of it actually serving the port, any requests to that port on localhost will result in:
-
[Errno 99] Cannot assign requested address
(if called from within the container), or -
[Errno 0] Error
(if called from outside of the container).
You can reproduce this error and the behaviour described above as follows:
Start a dummy container (note: this will pull the python image if not found locally):
docker run --name serv1 -p 5000:5000 -dit python
Then for [Errno 0] Error
enter a Python console on host, while for [Errno 99] Cannot assign requested address
access a Python console on the container by calling:
docker exec -it -u 0 serv1 python
And then in either case call:
import urllib.request
urllib.request.urlopen('https://localhost:5000')
I concluded with treating either of these errors as equivalent to No connection could be made because the target machine actively refused it
rather than trying to fix their cause - although please advise if that's a bad idea.
I've spent over a day figuring this one out, given that all resources and answers I could find on the [Errno 99] Cannot assign requested address
point in the direction of binding to an occupied port, connecting to an invalid IP, sysctl
conflicts, docker network issues, TIME_WAIT
being incorrect, and many more things. Therefore I wanted to leave this answer here, despite not being a direct answer to the question at hand, given that it can be a common cause for the error described in this question.
Solution 3:
Try like this:
server.bind(("0.0.0.0", 6677))
Solution 4:
When you bind localhost
or 127.0.0.1
, it means you can only connect to your service from local.
You cannot bind 10.0.0.1
because it not belong to you, you can only bind ip owned by your computer
You can bind 0.0.0.0
because it means all ip on your computer, so any ip can connect to your service if they can connect to any of your ip