U/L bit

major133

hi everone into ieoc

when we using EUI-64 format to allocate interface-id into ipv6 address , we are using mac address

 

for ex:- 

mac= 0x0c0c:dede:1234 and ipv6 address is 2001:0db8:0:1::/64 eui-64

with ipv6 EUI format will be 2001:0db8::0:1:0e0c:deff:fede:1234

now we flip the 7th bit 0000 1100 will be 1110

why we do that?

wy we modify the 7th bil U/L bit ? why we left 7th bit U/L bit as it was?

why the desighners of IPv6 think about this way?

what is the problem it solve

JoeM

Hello Major Muhamed,

Jeremey Stretch of Packetlife.net helps us with this answer....and he takes it from the RFC.

http://packetlife.net/blog/2008/aug/4/eui-64-ipv6/

Again, you're probably wondering why this is done. The answer lies buried in section 2.5.1 of RFC 2373:

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.

major133

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.

JoeM

how are you my friend

i read the article , but i don`t understanding it , forgive me could you explain it to me little easily with easy english and easy way?

iam really sorry my friend but i don`t understand these lines

major133

[6]