Date: Wed, 06 Aug 2014 19:08:54 -0400 From: Fernando Gont <fernando@gont.com.ar> To: FreeBSD Net <freebsd-net@freebsd.org> Subject: Routing IPv6 packets towards oneself with routing sockets? Message-ID: <53E2B586.3080700@gont.com.ar>
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Folks, I've found a "tricky" scenario when consulting the IPv6 routing table with routing sockets. Short version of the question: I'm currently consulting the IPv6 routing table with raw sockets. My own host is assigned the address fc00:1::1, and it is directly connected to fc00:1::/64 with em0. The corresponding entries from its routing table are: fc00:1::/64 link#1 U em0 fc00:1::1 link#1 UHS lo0 Essentially, packets sent to fc00:1::1 don't go through em0 but rather go through the loopback interface (if you ping6 fc00:1::1, you'll see the packets on lo0 rather than em0). However, whenever I lookup an entry for fc00:1::1 with routing sockets, the only entry I obtain is fc00:1::/64 (a network route) rather than fc00:1::1/128 (a host route). As a result, I kind of have to figure out that since fc00:1::1 is my own address, I must override whatever I learned via routing sockets, and just send my packets to loopback. I would assume that I must be doing something wrong, since I would expect the host-specific route (i.e. longest-matching route) to be route learned via routing sockets. And that I shouldn't be implementing this "is the dst address my own address?" hack. Any thoughts? P.S.: I can provide a code snippet if that'd be of any help. Thanks! Best regards, -- Fernando Gont e-mail: fernando@gont.com.ar || fgont@si6networks.com PGP Fingerprint: 7809 84F5 322E 45C7 F1C9 3945 96EE A9EF D076 FFF1
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