|
|
|
|
|
by j1elo
80 days ago
|
|
In IPv4 you only need to transmit IPv4 addresses. If the "cannot be" in parent post is referring to the exact byte disposition in packets, then I go the other way around to claim that I agree. Because the only way that a UTF8 character can pretend to be ASCII is because ASCII didn't use all of the 8 bits in a byte to begin with. Only way to have something similar in this case, would be that IPv4 didn't use all of the allocated bits for addresses... Which is not the case. What I argued was that IPv4 could be embedded into IPv6 address space if they had designed for it. But I agree, that the actual packet header layouts would need to look at least a bit different. |
|
Like:
> Addresses in this group consist of an 80-bit prefix of zeros, the next 16 bits are ones, and the remaining, least-significant 32 bits contain the IPv4 address. For example, ::ffff:192.0.2.128 represents the IPv4 address 192.0.2.128. A previous format, called "IPv4-compatible IPv6 address", was ::192.0.2.128; however, this method is deprecated.[5]
* https://en.wikipedia.org/wiki/IPv6#IPv4-mapped_IPv6_addresse...