[throw0101a]

> I want something:x.x.x.x to get routed to me over v6 if I had x.x.x.x in v4, as the default and recommended way of contacting an ipv6 host, without needing additional config or middleboxes.

Anyone with an IPv4 automatically got a 'free' IPv6 allocation:

> For any 32-bit global IPv4 address that is assigned to a host, a 48-bit 6to4 IPv6 prefix can be constructed for use by that host (and if applicable the network behind it) by appending the IPv4 address to 2002::/16.

> For example, the global IPv4 address 192.0.2.4 has the corresponding 6to4 prefix 2002:c000:0204::/48. This gives a prefix length of 48 bits, which leaves room for a 16-bit subnet field and 64 bit host addresses within the subnets.

* https://en.wikipedia.org/wiki/6to4

"Connection of IPv6 Domains via IPv4 Clouds", etc:

* https://datatracker.ietf.org/doc/html/rfc3056

[zadikian]

Owning 192.0.2.4 in v4 didn't really give you 2002:192.0.2.4 in v6, as in v6 packets can't reach you that way. If someone sent a v6 packet there, some router on their side would intercept it and relay as v4. Aside from this still relying on v4, it was very unreliable in practice because of uncertainty around what v6 route is taken to the relay. There's an RFC somewhere that looked at this in hindsight.

[throw0101a]

AIUI, you encapsulate anything to 2002:192.0.2.4/48 into an IPv4 packet with "41" in the protocol field (as opposed to 6 (TCP) or 17 (UDP)) and send it to 192.0.2.4. 192.0.2.4 would then work as the relay and extract the IPv6 traffic and handle it.

* https://en.wikipedia.org/wiki/IPv4#Protocol

* https://simple.wikipedia.org/wiki/Protocol_41