|
Home > Archive > CCNP > April 2001 > Simple MAC address question
You are viewing an archived Text-only version of the thread.
To view this thread in it's original format and/or if you want to reply to
this thread please [click here]
| Author |
Simple MAC address question
|
|
| strikeattack 2001-04-29, 4:47 pm |
| Hey guys.
I have a simple MAC address question that is frustrating me because I feel I should know the answer after all my training, but I don't, and it seems very simple.
We all know that a layer 2 MAC address is a 48-bit address expressed as 12 hexadecimal digits. The first six are the OUI and the last six are the serial number or unique identifier for that NIC from that manufactorer. This is a total of 12 digits.
The question is, since it is expressed hexadecimally, which expresses a 16-bit number in one char (0-F), and there are 12 spaces, 16 x 12 = 192 bits and not 48 bits.
Am I missing something very obviuos here? How can a MAC address be 48 bits if 12 hexadecimal digits at 16bits apiece equal 192 bits?
- Michael | |
|
|
| dmaftei 2001-04-29, 5:33 pm |
| quote:
Originally posted by strikeattack
...hexadecimally, which expresses a 16-bit number in one char (0-F)...
One hex digit is four bits:
Hex = Bin
0 = 0000
1 = 0001
2 = 0010
...
E = 1110
F = 1111 | |
| strikeattack 2001-04-29, 5:38 pm |
| Got it, and verified it with Windows Calc in Scientific mode. Thanks guys. | |
| dmaftei 2001-04-29, 5:50 pm |
| . |
|
|
|
|