05-16-2012 09:54 AM - edited 03-01-2019 05:34 PM
Hi everybody,
someone knows why in the process of building IPV6 unicast address through EUI-64 the seventh bit from left must be complemented ?? I can't understand .....
thanks
05-16-2012 12:14 PM
Paolo,
You are refrring to 7th bit of mac address the U/L bit not the 7th bit of IPv6 address I guess :-)
You need to look at bit side of things. It makes thing easier in notation.
It's described here:
http://en.wikipedia.org/wiki/IPv6_address#Modified_EUI-64
Take a very simple MAC address and convert it to EUI-64 with and without the bit flipped.
You will see that it's "easier" to write it down when it's flipped.
M.
05-17-2012 06:02 AM
You may also want to check on the theory behind the U/L bit in mac addresses. Wikipedia is an excellent source.
Regards,
Leo
05-18-2012 04:53 AM
Hi Paolo,
hope this helps you
Understanding IPv6 EUI-64 Bit Address
Regards,
Sunil Khanna
PS: Please rate the helpful post.
05-23-2012 08:18 AM
This is explained in RFC4291 Section 2.5.1. In short, it is so that the implicit modified EUI-64 portion of manually assigned addresses (such as 2001:db8::1) will be recognized as locally administered.
http://tools.ietf.org/html/rfc4291#section-2.5.1
Relevant text:
The motivation for inverting the "u" bit when forming an interface identifier is to make it easy for system administrators to hand configure non-global identifiers when hardware tokens are not available. This is expected to be the case for serial links and tunnel end-points, for example. 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 0:0:0:1, 0:0:0:2, etc.
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: