-1

This is one of my network adapter, the MAC address is: a0:99:9b:17:52:07:

en0: flags=8863<UP,BROADCAST,SMART,RUNNING,SIMPLEX,MULTICAST> mtu 1500
    ether a0:99:9b:17:52:07 
    inet6 fe80::c48:cc97:a24d:e029%en0 prefixlen 64 secured scopeid 0x4 
    inet 192.168.1.106 netmask 0xffffff00 broadcast 192.168.1.255
    nd6 options=201<PERFORMNUD,DAD>
    media: autoselect
    status: active

we know the MAC address is 48 bits = 6 Bytes.

but in the a0:99:9b:17:50:07, we know hexadecimal digits is from 0 to f. the xx:xx:xx:xx:xx:xx each xx can from 00 to ff range, why say xx can only stand 8 bits?


EDIT-01

Because the xx:xx:xx:xx:xx:xx stands for 48 bits, so the xx stand 8 bits. the xx is from 00 to ff range, so why xx can only stand 8 bit?

  • 1
    Please quote the article claiming that two hexadecimal characters stand for 8 bits, without that, this question is not making much sense. In anyway, two hexadecimal characters stand for 16 bits, 8 bits per character. – Teun Vink Oct 2 '18 at 5:01
  • @Teun ... I'm sure the question meant precisely hex digit; not how much space it might take to write in ASCII or whatever. – jonathanjo Oct 2 '18 at 6:23
5

A hexadecimal digit can have sixteen different values, 0..9 and A..F, representing 0..15 in decimal.

For the same value in binary you need four bits, 0000..1111, representing the same 24 values. Two of those four-bit nibbles or hexadecimal digits form a byte, representing 16*16 = 256 values.

48 bits or six bytes require 12 hex digits, often written with - or : in between (byte) pairs.

  • 2
    ... and occasionally written with '.' between groups of four digits, as on Cisco equipment: 649e.f312.3456 which is the same as 64:9e:f3:12:34:56 – jonathanjo Oct 2 '18 at 13:21
-2

because you need 4 bits to calculate an hex degit for example

1101 = 13 in decimal and "d" in hexadecimal so every 4 bit represents an hex degit in a mac address

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