Chris's math is 100% correct but I would offer a slightly different
accompanying
explanation, taking into account that a byte is comprised of only two
hexadecimal
digits.
The MAC address consists of 48 bits which is 6 bytes with 8 bits per byte.
The
difference between Token Ring and Ethernet is the bit order within the
bytes, so you
need to look at 8 bits at a time, that is, you reverse the bit-order in each
of the 6
bytes.
It takes TWO hex digits (of 4 bits per hex digit) to equal a byte.
The best thing to do is to ignore the dots (periods) and take the MAC
address two hex
digits (a byte) at a time, converting as follows:
1. Change the two hex digits (this byte) to binary
2. reverse the entire byte bit-by-bit, starting with the last bit in line,
3. finally convert that result back to hex.
So when Chris states "I'll take the first octet: 0110" he is actually
taking two octets
(2 bytes), although in his example he correctly reversed each byte
individually.
Since 0110 hex is symmetric and looks kind of binary, perhaps a different
example might
help. Let's take the byte D7 (represented as two 4-bit hex numbers) and
reverse it.
1. Convert from hex to binary:
D 7
1101 0111
2. Reverse the entire byte:
1110 1011
3. Convert back to hex
E B
Hope this helps.
-Jonathan
Chris Cell wrote:
> Basically what you do is divide the 8 bit octal number into two four bit
> numbers, swap the first four bits with the last four, then read it
backwards
> in each of the two four bit parts. (confusing, but follow below)
>
> 0110.1234.5678
>
> I'll take the first octet:
> 0110
>
> Seperate it into two four bit parts:
> 01 (split) 10
>
> make it binary:
> 0000 0001 (split) 0001 0000
>
> read it backwards:
> 1000 0000 (split) 0000 1000
>
> bring it back to octal:
> 80 (split) 08
>
> make it one 8 bit octal number again:
> 8008
>
> Do that for all the parts, and you have completed the bit swapping.
Message Posted at:
http://www.groupstudy.com/form/read.php?f=7&i=20227&t=20214
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