Currently allocated IPv6 network ranges
---------------------------------------


IPv4 compatible address:
(for IPv6 only nodes trying to reach IPv4 [nobody only supports IPv6])

                0:0:0:0:0:0::/96        (displayed as ::192.168.0.1)

IPv4 mapped address:
(for binded IPv6 socket that can receive IPv4 traffic [impossible in OpenBSD])

                0:0:0:0:0:ffff::/96     (displayed as ::ffff:192.168.0.1)


unicast         [2]***::/3
                [3]

                sub-tla 2001::/16

                        IANA    2001:0[0]**::/23
                                      [1]

                        APNIC   2001:0[2]**::/23
                                      [3]

                        ARIN    2001:0[4]**::/23
                                      [5]

                        RIPE    2001:0[6]**::/23
                                      [7]

                        RIPE    2001:0[8]**::/23
                                      [9]

                        APNIC   2001:0[c]**::/23
                                      [d]


                6to4    2002::/16

                        (Every owner of a IPv4 address has a corresponding 6to4
                        IPv6 /48 network of the form: 2002:IPv4-1:IPv4-2::/48
                        and can communicate directly over IPv4 encapsulation to
                        any 6to4 site.) [not supported in OpenBSD]

                        e.g.: IPv4 129.128.5.191 = 6to4 IPv6 2002:8180:5bf::/48

                6bone   3ffe::/16


link-local      fe[8]*::/10
                  [9]
                  [a]
                  [b]

site-local      fe[c]*::/10
                  [d]
                  [e]
                  [f]

multi-cast      ff**::/8


--
Legend (beside regular IPv6 constructs):

        [x]     means that this particular digit (from 0 to F) is allowed
                at this position in the current IPv6 address
        *       any digit (from 0 to F) is allowed at this position


$Id: ipv6.netranges.txt,v 1.2 2003/07/16 01:04:09 harpagon Exp $