Re: ARP out the wrong interface

From: Stephen Degler (sdegler@degler.net)
Date: Fri Feb 09 2001 - 00:43:16 EST


Hi,

What you describe below is having the client mis-addressed to have
the same IP as the server. Is this what you meant?

skd

On Thu, Feb 08, 2001 at 09:09:49PM -0800, dean gaudet wrote:
> this appears to occur with both 2.2.16 and 2.4.1.
>
> server:
>
> eth0 is 192.168.250.11 netmask 255.255.255.0
> eth1 is 192.168.251.11 netmask 255.255.255.0
>
> they're both connected to the same switch.
>
> client:
>
> eth0 is 192.168.251.11 netmask 255.255.255.0
>
> connected to the same switch as both of server's eth.
>
> on client i try "ping 192.168.251.11".
>
> responses come back from both eth0 and eth1, listing each of their
> respective MAC addresses... it's essentially a race condition at this
> point as to whether i'll get the right MAC address. ("right" means the
> MAC for server:eth1).
>
> client# tcpdump -n arp
> Kernel filter, protocol ALL, datagram packet socket
> tcpdump: listening on all devices
> 21:03:05.695089 eth0 > arp who-has 192.168.251.11 tell 192.168.251.25 (0:3:47:0:25:80)
> 21:03:05.695405 eth0 < arp reply 192.168.251.11 is-at 0:d0:b7:be:3e:aa (0:3:47:0:25:80)
> 21:03:05.695523 eth0 < arp reply 192.168.251.11 is-at 0:d0:b7:1f:ea:35 (0:3:47:0:25:80)
>
>
> server# cat /proc/sys/net/ipv4/ip_forward
> 0
> server# cat /proc/sys/net/ipv4/conf/*/proxy_arp
> 0
> 0
> 0
> 0
> 0
> 0
> 0
>
> is this expected? it seems broken.
>
> thanks
> -dean
>
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