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generate IEEE 802 conform MAC addresses
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README.md

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A IEEE 802 MAC address generator

Purpose

This generator can be used to create IEEE 802 conform local administered MAC addresses

get it

go get github.com/cseeger-epages/mac-gen-go

Usage

package main

import (
  "fmt"
  "net"

  gm "github.com/cseeger-epages/mac-gen-go"
)

func main() {

  // generate a random local administered unicast mac prefix
  prefix := gm.GenerateRandomLocalMacPrefix(true)

  // calculates the NIC Sufix by ip address
  sufix, err := gm.CalculateNICSufix(net.ParseIP("129.168.12.127"))
  if err != nil {
    // your error handling here
  }
  mac := fmt.Sprintf("%s:%s", prefix, sufix)

  fmt.Println(mac)
}

see examples for more information

testing and coverage

mage test
mage coverage

Background

general stuff

  • a /8 net has 16,777,216 addresses
  • a mac address has 6 octets (2^48 addresses)
  • the first 3 octets are the OUI (Organisationally Unique Identifier)
  • and the last 3 octets are the NIC (Network interface controler specific)
  • every octet has 8 bit
  • [b7][b6][b5][b4][b3][b2][b1][b0]
  • the first octet has special meaning for b0 and b1
  • b0 -> 0: unicast, 1: multicast
  • b1 -> 0: globally unique (OUI), 1: locally administered
  • to generate a mac address without having a OUI
  • b0 ->0
  • using only the last 3 octets (16^6 = 2^24 addresses = 16777216 = /8 net)
  • is enough to create mac addresses for all ip addresses
  • per net the first 3 octets need to be generated as a new prefix

prefix block generation

generating a prefix block of 3 octets

  • first octet in binary xxxxxx10 ( or xxxxxx11 if you want a multicast address)

  • second and 3rd octet can be generated freely

  • 8 bit block in binary

  • [0|1] [0|1] [0|1] [0|1] [0|1] [0|1] [0|1] [0|1]

  • left 4 bits does not matter since they do not effect the b0 and b1 bit

  • to have b0 -> 0 and b1 -> 1 (a LA unicast address)

  • [0|1] [0|1] 1 0

  • so we have sth + 2

  • where sth is ether 0 (0 0) or 8 (1 0) or 4 (0 1) or 12 (1 1)

  • so all valid combinations are 2, 6, 10 and 14 which are in hex

  • 2, 6, A, E

  • the first octet has to be sth like

  • [0..F][2|6|A|E]

  • if the address should be multicast

  • b0 -> 1 and b1 -> 1

  • so sth + 3

  • which results in 3, 7, B, F

  • and the first octet has to be sth like

  • [0..F][3|7|B|F]

  • there are 6 free bits in this octet so we have

  • 2^6 permutations for this octet

  • if we include the other 2 octests (sum up to additonal 16 bits)

  • we have 2^22 permutations for free local administratered mac addresses (either uni or multicast)

  • we should exclude some of them because of some registries failed in correctly checking if an OUI has local administraited address (LAA) bit set or not -.-

  • here is a list https://gist.github.com/aallan/b4bb86db86079509e6159810ae9bd3e4

  • and checked against http:-standards-oui.ieee.org/oui/oui.txt

  • to exclude

  • 02 since here are a few registraited ones

  • AA due to DEC LAA

  • so we now have

  • 0[3|6|7|A|B|E|F] - 7

  • [1..9][2|3|6|7|A|B|E|F] - 9*8

  • A[2|3|6|7|B|E|F] - 7

  • and

  • [B..F][2|3|6|7|A|B|E|F] - 5*8

  • we will lose some of our actual permutations

  • without the first octet we have 2^16 permutations (remember 2 octest = 8 bit per octet -> 16 bits)

  • for our first octet we now have 126 possibile permutations left

  • so we get 2^16*126 = 8257536 possible permutations (this includes multicast)

  • this is about 49,21% of the full addresspace (2^24) so we lose about half of the addresses

  • in a perfect world where no registrie failed we would exactly lose 50% (2^16*128 / 2^24) = (2^16 * 2^7 = 2^23) / 2^24 = 0.5

  • so we lose about 0.79% addresses due to registration failure exclusions

  • but yeah we do have over 8 million left - show me someone who has more than 8 millions local networks

  • the registrated private networks are 10.0.0.0/8 (16.77.216) 172.16.0.0/12 (1.048.576) and 192.168.0.0/16 (65.536)

  • so in the worst case you can cut all these networks into /32 networks with only one ip address so you can have

  • up to 17891328 networks so our 2^16*126 permutations fit about 46.15% of this worst case szenario or 23.44% when only using unicast

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