Send UDP to local server, but not via loopback interface

Question

I'm ultimately trying to test a UDP client, and want to make sure that it works when sending data not through the loopback interface, to avoid any subtle issues this introduces, such as differences in checksum validation (Bad UDP checksum has no effect: why?).

However, even when sending data to the result of socket.gethostbyname(socket.gethostname()), which is not 127.0.0.1, then according to Wireshark, the data seems to go via the loopback interface.

The below program sends and receives b'somedata' successfully, and has the below capture from Wireshark.

import asyncio
import socket

async def server():
    with socket.socket(socket.AF_INET, socket.SOCK_DGRAM) as sock:
        sock.setblocking(False)
        sock.bind(('', 4567))
        data = await loop.sock_recv(sock, 512)
        print('Received', data)

async def main():
    local_ip = socket.gethostbyname(socket.gethostname())
    print('Local IP', local_ip)  # Outputs 192.168.0.34

    asyncio.ensure_future(server())
    await asyncio.sleep(0)

    with socket.socket(socket.AF_INET, socket.SOCK_DGRAM) as sock:
        sock.setblocking(False)
        sock.connect((local_ip, 4567))
        await loop.sock_sendall(sock, b'somedata')
        await asyncio.sleep(1)

loop = asyncio.get_event_loop()
loop.run_until_complete(main())
loop.close()

How can I send data from a client running locally, to a server running locally, but avoiding the loopback interface and actually sending data out into the network?

Ideally answers would be applicable to both Linux and macOS.

May I ask what you are trying to achieve? Are you testing your code, or are you testing the network interface? If you want to test how your code handles bad packets, then maybe you should create a method that works on a socket, so you could mock the socket object in a unit test. If you are creating a production test / integration test / system test, then the test setup should mimic the real (expected) setup as close as possible. Ideally you would have a real server and client separated from each other. Maybe you can achieve this using virtual machines, or using two USB-to-Ethernet adapters? — wovano
– wovano, Commented Apr 29, 2019 at 14:03
@wovano I'm testing my code, and how it interacts with the rest of the system, avoiding assumptions that mocking introduces, so it's more of an integration-style test... "then the test setup should mimic the real (expected) setup as close as possible." Agreed, and seeing how far I can get towards that without the overhead of multiple machines/adapters. — Michal Charemza
– Michal Charemza, Commented Apr 29, 2019 at 21:42
I certainly understand that you want to avoid that overhead. But I don't think you can test how your application interacts with the rest of the system without having the rest of the system ;-) To realistically test the network interface, you'll need at least two network interfaces to guarantee that the packets are being transmitted over the line. And even then the behavior without switch can be very different than with a switch (e.g. due to packet reordering, IP fragmentation, etc.). And a Linux system might have subtle differences than a Windows system, for example. — wovano
– wovano, Commented Apr 29, 2019 at 22:22
What you can do is test the normal (expected) behavior, and all error conditions you can think of (such as a bad UDP checksum). If the interface boundaries of your application are chosen correctly, you'll probably be able to test most of the situations. Can you describe what you want to test, i.e. which test cases? That might help to provide better solutions. And what assumptions do you think mocking will introduce? — wovano
– wovano, Commented Apr 29, 2019 at 22:23
"Can you describe what you want to test". For example, how a packet arriving that has a bad UDP checksum affects the application. Is it ignored? Processed as usual? An exception raised? "what assumptions do you think mocking will introduce": that the lower level behaviour actually acts like the mock. Understood that different systems have subtle differences, and this is exactly why I want the tests to use the lower-level behaviour of the system to ensure that fewer assumptions are made about it, and that the tests passing means there is a good chance that the application works. — Michal Charemza
– Michal Charemza, Commented Apr 30, 2019 at 10:26

VPfB · Accepted Answer · 2019-04-25 08:15:51Z

1

+200

To 'convince' the networking stack to physically transmit the frame using the Ethernet (or WiFi) card rather than the loopback, use a broadcast address.

I have successfully sent and received an UDP packet this way on my Linux system. I verified it with tcpdump. It shows one packet on Ethernet interface and no activity on loopback.

I have used literal broadcast address. The socket module documentation mentions also the string '<broadcast>' as a special case address. I did not tried it.

with socket.socket(socket.AF_INET, socket.SOCK_DGRAM) as sock:
    sock.setblocking(False)
    sock.setsockopt(socket.SOL_SOCKET, socket.SO_BROADCAST, 1)
    sock.connect(('192.168.0.255', 4567))
    await loop.sock_sendall(sock, b'somedata')
    await asyncio.sleep(1)

Notes:

other hosts on the same network will receive the UDP packet as well.
make sure the firewall/packet filter (e.g. Linux iptables/nftables) will not block the packet.
regarding the setsockopt: Python socket.error: [Errno 13] Permission denied

answered Apr 25, 2019 at 8:15

VPfB

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2 Comments

Michal Charemza Over a year ago

Interesting... never considered this. For my case however, while making the tests more like the real case in terms of going over the network, it’s also less like the real case since broadcast isn’t used

VPfB Over a year ago

I have found a related topic: serverfault.com/q/483209 I tried 2 IP addresses on a single Ethernet card and was able to transmit the datagram from addr1 to addr2, but it was not received back from the switch.

andreihondrari · Accepted Answer · 2019-04-22 09:37:04Z

0

That's probably because your hostname is pointing to the loopback address hence socket.gethostbyname(socket.gethostname()) will yield 127.0.0.1

What you need to do is cancel that pointing from the hostname to the loopback address:

in Linux edit the /etc/hosts and comment out the line 127.0.0.1 YOUR_HOSTNAME
in Windows you should have c:\windows\system32\drivers\etc\hosts which looks similar to the Linux one

After this if you call the socket.gethostbyname(socket.gethostname()) it will yield your DHCP assigned IP.

Though even in this case, calling from yourip to yourip might result in the network driver to route the package through the loopback interface. Alternative would be to use the public IP, outside your network router. You could use an external service as described in this answer

edited Apr 22, 2019 at 9:37

answered Apr 22, 2019 at 9:07

andreihondrari

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3 Comments

Michal Charemza Over a year ago

Ah I don't think that's what's happening here socket.gethostbyname(socket.gethostname()) does not yield 127.0.0.1. I have edited the question to make this more explicit.

user207421 Over a year ago

The Wireshark dump shows conclusively that the packets are not addressed to or from 127.0.0.1.

andreihondrari Over a year ago

@user207421 I think I might have missed that detail the fist time. Nevertheless it would still go through the lo interface even like this.

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