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In network programming (C & Linux) with a TCP socket, how can I check the interface that received the packet. I know this can be done very easily with a SOCK_DGRAM and the IP option IP_PKTINFO, but this option does not work on SOCK_STREAM. In this case (for what im programming) I must know the interface that received the packet (the respond that gives back the server depends on that).

One possible solution that I found is to use SOCK_RAW and IPPROTO_TCP (with SOCK_RAW I can set the IP_PKTINFO option) but that implies that i'll be receiving all the TCP traffic that arrives to the machine and i don't want that. Another possible solution that I'm thinking is to check the destination IP against the routing table and get the exit interface from there but what about IPv6? What if the destiny IP its IPv6 link-local?

If anyone has another solution or idea I'll be glad to read it. Thanks in advance

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    In TCP you would gather that information at accept time, not per packet. It is given by getsockname() on the accepted socket. But the weak end system model means that it isn't reliable. Commented Apr 17, 2016 at 4:22
  • You are right, thanks a lot Commented Apr 18, 2016 at 14:54

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When your TCP server accepted a client connection, you will have a socket to represent the connection, you could call getsockname on the socket, it will give you the address associated on the socket, namely the ip addresses on your side. Next step, you can call getifaddrs to get all the interfaces and their information on your system, including name(like eth0) and ip address. At this point, you can search this list with the ip obtained previously via getsockname, you can get the interface you want as a result.

reference:

http://linux.die.net/man/2/getsockname

http://linux.die.net/man/3/getifaddrs

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Yes! That is exactly right and i know the getsockname() funcionality but i did not think about that before. Anyway thanks to you and thanks to the guy who fixed the grammatically errors on the original post :)

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