Find IP address of directly connected device

Is there a way to find out the IP address of a device that is directly connected to a specific ethernet interface? I.e. given one host, one wired ethernet connection and one second host connected to this wired connection, which layer or protocol below IP could be used to find this out.

I would also be comfortable with a Windows-only solution using some Windows-API function or callback.

(I know that the real way to do this would probably via DHCP, but this is about discovering a legacy device.)

Answers


To use DHCP, you'd have to run a DHCP server on the primary and a client on the secondary; the primary could then query the server to find out what address it handed out. Probably overkill.

I can't help you with Windows directly. On Unix, the "arp" command will tell you what IP addresses are known to be attached to the local ethernet segment. Windows will have this same information (since it's a core part of the IP/Ethernet interface) but I don't know how you get at it.

Of course, the networking stack will only know about the other host if it has previously seen traffic from it. You may have to first send a broadcast packet on the interface to elicit some sort of response and thus populate the local ARP table.


Mmh ... there are many ways. I answer another network discovery question, and I write a little getting started.

Some tcpip stacks reply to icmp broadcasts. So you can try a PING to your network broadcast address.

For example, you have ip 192.168.1.1 and subnet 255.255.255.0

  1. ping 192.168.1.255
  2. stop the ping after 5 seconds
  3. watch the devices replies : arp -a

Note : on step 3. you get the lists of the MAC-to-IP cached entries, so there are also the hosts in your subnet you exchange data to in the last minutes, even if they don't reply to icmp_get.

Note (2) : now I am on linux. I am not sure, but it can be windows doesn't reply to icm_get via broadcast.

Is it the only one device attached to your pc ? Is it a router or another simple pc ?


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