Make the ip6_tunnel reflect the true mtu.

The ip6_tunnel always assumes it consumes 40 bytes (ip6 hdr) of the mtu of the
underlaying device. So for a normal ethernet bearer, the mtu of the ip6_tunnel is
1460.
However, when creating a tunnel the encap limit option is enabled by default, and it
consumes 8 bytes more, so the true mtu shall be 1452.

I dont really know if this breaks some statement in some RFC, so this is a request for
comments.

Signed-off-by: Anders Franzen <anders.franzen@ericsson.com>
Signed-off-by: David S. Miller <davem@davemloft.net>

1 files changed, 7 insertions, 0 deletions

diff --git a/net/ipv6/ip6_tunnel.c b/net/ipv6/ip6_tunnel.c
index 2a59610c2a58..70e891a20fb9 100644
--- a/net/ipv6/ip6_tunnel.c
+++ b/net/ipv6/ip6_tunnel.c
@@ -1175,6 +1175,8 @@ static void ip6_tnl_link_config(struct ip6_tnl *t)
                                 sizeof (struct ipv6hdr);
 
                         dev->mtu = rt->rt6i_dev->mtu - sizeof (struct ipv6hdr);
+                        if (!(t->parms.flags & IP6_TNL_F_IGN_ENCAP_LIMIT))
+                                dev->mtu-=8;
 
                         if (dev->mtu < IPV6_MIN_MTU)
                                 dev->mtu = IPV6_MIN_MTU;
@@ -1363,12 +1365,17 @@ static const struct net_device_ops ip6_tnl_netdev_ops = {
 
 static void ip6_tnl_dev_setup(struct net_device *dev)
 {
+        struct ip6_tnl *t;
+
         dev->netdev_ops = &ip6_tnl_netdev_ops;
         dev->destructor = ip6_dev_free;
 
         dev->type = ARPHRD_TUNNEL6;
         dev->hard_header_len = LL_MAX_HEADER + sizeof (struct ipv6hdr);
         dev->mtu = ETH_DATA_LEN - sizeof (struct ipv6hdr);
+        t = netdev_priv(dev);
+        if (!(t->parms.flags & IP6_TNL_F_IGN_ENCAP_LIMIT))
+                dev->mtu-=8;
         dev->flags |= IFF_NOARP;
         dev->addr_len = sizeof(struct in6_addr);
         dev->features |= NETIF_F_NETNS_LOCAL;