01-04-2012 03:34 AM - edited 03-04-2019 02:49 PM
Hi everybody
I was reading about how eui- 64 format is used to derive interface id in ipv6.
The book says the eui-format requires 7 bit of mac address must be set to one. But why? I mean if are using burned-in mac address, it is already unique. So why do we need to flip 7th bit in eui-64 format ?
thanks and have a good day.
Solved! Go to Solution.
01-04-2012 06:05 AM
Hi,
Cisco decided to flip the 7th bit all the time whether the MAC address from which it was derived had the U/L bit set or not.
Regards.
Alain
01-04-2012 05:48 AM
Have you seen this reference?
http://wiki.nil.com/IPv6_EUI-64_interface_addressing
From RFC 2373, Section 2.5.1 at http://tools.ietf.org/html/rfc2373#section-2.5.1
"The motivation for inverting the "u" bit when forming the interface identifier is to make it easy for system administrators to hand configure local scope identifiers when hardware tokens are not available. This is expected to be case for serial links, tunnel end- points, etc. The alternative would have been for these to be of the form 0200:0:0:1, 0200:0:0:2, etc., instead of the much simpler ::1, ::2, etc."
Don't forget to rate all posts that are helpful.
Regards
Sean
01-04-2012 06:05 AM
Hi,
Cisco decided to flip the 7th bit all the time whether the MAC address from which it was derived had the U/L bit set or not.
Regards.
Alain
Find answers to your questions by entering keywords or phrases in the Search bar above. New here? Use these resources to familiarize yourself with the community: