MAC address layout!
solnsusie
Member Posts: 128
in CCNA & CCENT
hi
as im studying for the CCNA, i see that the low order numbers of the MAC Address are on the left and the high order numbers are on the right, waht does it mean high - low order numbers, is it on every byte, or octed, or bits? i dont understand it,
please check this out from the Odom ICND2 book page # 595:
To be complete, the figure points out one other small detail regarding the EUI-64 interface
ID value. Splitting the MAC address into two halves, and injecting FFFE, is easy. However,
the EUI-64 format requires setting the seventh bit in the first byte of the value to binary 1.
The underlying reason is that Ethernet MAC addresses are listed with the low-order bits of
each byte on the left, and the high-order bits on the right. So, the eighth bit in a byte (reading
from left to right) is the highest-order bit in the address, and the seventh bit (reading from
left to right) is the second highest-order bit. This second highest-order bit in the first byte—
the seventh bit reading from left to right—is called the universal/local (U/L) bit. Set to
binary 0, it means that the MAC address is a burned-in MAC address. Set to 1, it means that
the MAC address has been configured locally. EUI-64 says that the U/L bit should be set to
1, meaning local.
thanks
as im studying for the CCNA, i see that the low order numbers of the MAC Address are on the left and the high order numbers are on the right, waht does it mean high - low order numbers, is it on every byte, or octed, or bits? i dont understand it,
please check this out from the Odom ICND2 book page # 595:
To be complete, the figure points out one other small detail regarding the EUI-64 interface
ID value. Splitting the MAC address into two halves, and injecting FFFE, is easy. However,
the EUI-64 format requires setting the seventh bit in the first byte of the value to binary 1.
The underlying reason is that Ethernet MAC addresses are listed with the low-order bits of
each byte on the left, and the high-order bits on the right. So, the eighth bit in a byte (reading
from left to right) is the highest-order bit in the address, and the seventh bit (reading from
left to right) is the second highest-order bit. This second highest-order bit in the first byte—
the seventh bit reading from left to right—is called the universal/local (U/L) bit. Set to
binary 0, it means that the MAC address is a burned-in MAC address. Set to 1, it means that
the MAC address has been configured locally. EUI-64 says that the U/L bit should be set to
1, meaning local.
thanks
Comments
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SharkDiver
Member Posts: 844
solnsusie,
I read over that several times, and it still doesn't make sense.
I have always heard the highest-order bit referred to as the one with the most value. So, the highest-order bit in a byte would be the first bit. I am not sure what they are saying.
When I read it, I did notice that they say that the seventh bit should always be set to a one. I seemed to remember that the bit should be flipped either way. If it was a one, it needs flipped to a zero and vice versa.
I searched for the errata for the book, and sure enough, the bit flipping is corrected in the errata, but nothing about the highest/lowest order bits.
The errata can be found here.
CCNA Official Exam Certification Library (CCNA Exam 640-802), 2nd Edition
Scroll down to just above the "Product Description" and click "Updates"
As for converting the mac into EUI-64:
If the MAC was 0000.0c07.ac0a, you would insert the fffe directly in the middle, like so:
0000:0cff:fe07:ac0a
Then you would convert the first two hexadecimal characters into binary:
0=0000
0=0000
So, the first two (00) = 00000000
The seventh bit would get flipped:
00000010
Then convert back to hex:
0000=0
0010=2
So you would end up with:
0200:0cff:fe07:ac0a -
solnsusie
Member Posts: 128
hi
thanks a lot for your reply,
I understand everything about the EUI-64, except the concept of the high - low order values!!!! I just don’t get it,