From: Victor Cappuccio (cvictor@protokolgroup.com)
Date: Tue Aug 15 2006 - 20:07:49 ART
I do not know if this helps but
http://www.internetworkexpert.com/rfc/rfc2464.txt
The Interface Identifier [AARCH] for an Ethernet interface is based
on the EUI-64 identifier [EUI64] derived from the interface's built-
in 48-bit IEEE 802 address. The EUI-64 is formed as follows.
(Canonical bit order is assumed throughout.)
The OUI of the Ethernet address (the first three octets) becomes the
company_id of the EUI-64 (the first three octets). The fourth and
fifth octets of the EUI are set to the fixed value FFFE hexadecimal.
The last three octets of the Ethernet address become the last three
octets of the EUI-64.
The Interface Identifier is then formed from the EUI-64 by
complementing the "Universal/Local" (U/L) bit, which is the next-to-
lowest order bit of the first octet of the EUI-64. Complementing
this bit will generally change a 0 value to a 1, since an interface's
built-in address is expected to be from a universally administered
address space and hence have a globally unique value. A universally
administered IEEE 802 address or an EUI-64 is signified by a 0 in the
U/L bit position, while a globally unique IPv6 Interface Identifier
is signified by a 1 in the corresponding position. For further
discussion on this point, see [AARCH].
For example, the Interface Identifier for an Ethernet interface whose
built-in address is, in hexadecimal,
34-56-78-9A-BC-DE
would be
36-56-78-FF-FE-9A-BC-DE.
A different MAC address set manually or by software should not be
used to derive the Interface Identifier. If such a MAC address must
be used, its global uniqueness property should be reflected in the
value of the U/L bit.
An IPv6 address prefix used for stateless autoconfiguration [ACONF]
of an Ethernet interface must have a length of 64 bits.
-----Mensaje original-----
De: nobody@groupstudy.com [mailto:nobody@groupstudy.com] En nombre de uyota
oyearone
Enviado el: Martes, 15 de Agosto de 2006 06:45 p.m.
Para: ccielab@groupstudy.com
Asunto: EUI -64 to MAC
hey guys, am trying to understand a scenerio, and would appreciate a
better explanation. task:--------------Conf. IPv6 addre.
2001:192:10:1:1234:56ff:fe:78:9abc on r1 fa0/0. The host portion be
gotten from eui-64.solution:---------------r1#int fa0/0mac-addre
1034.5678.9abcipv6 add 2001:192:10:1::/64 eui-64 their explanation was :
Invert the 7most significant bit, then remove ff:fe, this equals
1034.5678.9abc. Understandable, but the issue i have is: :when they say
invert the 7most significant bit, do this mean the 7 most significant bit
in the first octect?? regards, Uyota
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This archive was generated by hypermail 2.1.4 : Fri Sep 01 2006 - 15:41:57 ART