How is an EUI-64 format interface ID created from a 48-bit MAC address?
A.
by appending 0xFF to the MAC address
B.
by prefixing the MAC address with 0xFFEE
C.
by prefixing the MAC address with 0xFF and appending 0xFF to it
D.
by inserting 0xFFFE between the upper three bytes and the lower three bytes of the MAC
address
E.
by prefixing the MAC address with 0xF and inserting 0xF after each of its first three bytes
Explanation
We convert a 48-bit MAC address (IEEE 802) to a 64-bit value by breaking the MAC address into its two 24-bit halves. The first part is the Organizationally Unique Identifier (OUI) and the next part is the NIC specific part. Then the 16-bit hex value 0xFFFE is inserted between them to create a 64-bit value.
Just for your information, to obtain an IPv6 interface identifier from EUI-64 address, we have to complement the U/L bit (the seventh bit of the first byte and is used to determine whether the address is universally or locally administered). This means if it is a 1, it is set to 0; and if it is a 0, it is set to 1. In the above example, the U/L bit is 0 (from 00 = 0000 0000). Therefore we have to set this bit to 1 to create an IPv6 interface address.
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The inversion of the 7th bit is one of those poorly documented “features”. Half of the sources say yes and others don’t. Cisco’s test question here is another example. Whoever came up with that needs to be shot, I’ve yet to hear a good technical functional reason for doing this in the EUI64 addressing scheme. But alas, it is in fact used as Aldrin states.
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