1

In NFX I'm trying to create - Devices in a Network with 2 Clients, which can communicate with each other. But it does not work, is it because I have not set a Gateway which can tell the Client, where the other Client are?

Do I need the gateway for Communication in the same network?

Thanks for you help.

|improve this question
  • If the clients are in the same subnet, they do not need a gateway to ping each other. – user36472 Sep 26 '18 at 9:52
  • Maybe, you can help me with my Problem? link – Mustafa Sep 26 '18 at 9:54
  • 1
    It might be better to post a new question describing your problem. – Ron Trunk Sep 26 '18 at 11:55
  • Did any answer help you? If so, you should accept the answer so that the question doesn't keep popping up forever, looking for an answer. Alternatively, you can provide and accept your own answer. – Ron Maupin♦ Dec 25 '18 at 9:33
3

You do not need a gateway if both endpoints are in the same network and they have the same network mask. They will send an ARP broadcast to FFFF.FFFF.FFFF and the endpoint with that IP address will reply that ARP with its MAC address, they will fill their ARP tables (associating IP - MAC) and the communication between them will be p2p.

|improve this answer
  • Maybe, you can help me with my Problem? LINK – Mustafa Sep 26 '18 at 10:35
1

Two clients in the same subnet need to be within the same layer-2 segment or broadcast domain - most often, this is a simple Ethernet switch or a VLAN.

In a MAC-based layer 2 like Ethernet, the source will ARP the destination's IP address to discover its MAC address, encapsulate the ping packet in an L2 frame addressed to the discovered MAC and send it over - no gateway required or used.

If this doesn't work there's no common layer-2 segment between both clients or there's a filter in between, disrupting communication.

|improve this answer

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.