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0 pointswjholden1mo ago0 comments

The header of an IPv4 packet has the source and destination addresses, both as 32-bit values. These fields are adjacent, and there's other stuff next to them. If you appended more bytes to the source address, routers would think that those new bytes are the destination address. This would not be backward compatible.

Interestingly, what you're describing really is similar to how many languages represent an IPv4 address internally. Go embeds IPv4 addresses inside of IPv6 structs as ::ffff:{IPv4 address}: https://cs.opensource.google/go/go/+/go1.26.2:src/net/ip.go;...

0 comments

urineaut1mo ago

That's not a language-specific thing, but is actually part of the IPv6 RFCs as IPv4-mapped IPv6 addresses: [1], [2]

This is super useful because (at least on Linux) IPv6 sockets per default are dual-stack and bind to both IPv6 and IPv6 (except if you are using the IPV6_V6ONLY sockopt or a sysctl), so you don't need to open and handle IPv4 and IPv6 sockets separately (well, maybe some extra code for logging/checking properly with the actual IPv4 address).

That is also documented in ipv6(7):

  IPv4 connections can be handled with the v6 API by using 
  v4-mapped-on-v6 address type; thus a program needs to support only
  this API type to support both protocols.  This is handled
  transparently by the address handling functions in the C library.
  
  IPv4 and IPv6 share the local port space.  When you get an IPv4
  connection or packet to an IPv6 socket, its source address will be
  mapped to v6.

[0]: https://datatracker.ietf.org/doc/html/rfc5156#section-2.2 [1]: https://datatracker.ietf.org/doc/html/rfc4291#section-2.5.5....

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