this post was submitted on 17 Sep 2025
466 points (98.5% liked)

Programmer Humor

26420 readers
802 users here now

Welcome to Programmer Humor!

This is a place where you can post jokes, memes, humor, etc. related to programming!

For sharing awful code theres also Programming Horror.

Rules

founded 2 years ago
MODERATORS
 
you are viewing a single comment's thread
view the rest of the comments
[–] comfy@lemmy.ml 13 points 2 days ago (1 children)

Just make your IP addresses pronouncable words like feed:deaf:babe:beef:cafe:: problem solved ez (working 2023!)

[–] some_kind_of_guy@lemmy.world 2 points 1 day ago (2 children)

But you're limited to a-f. I wonder if anyone's figured out how many addresses are actually possible with that system.

[–] JargonWagon@lemmy.world 3 points 1 day ago

throw some 1337 speek in there and you're all set!

[–] Ghoelian@lemmy.dbzer0.com 2 points 1 day ago* (last edited 1 day ago) (1 children)

I think that's just 6^32, no? (Amount of options^string length). Which is 7958661109E24.

[–] some_kind_of_guy@lemmy.world 2 points 1 day ago* (last edited 1 day ago) (1 children)

Yeah, calculation of the amount of possible strings containing only a-f is trivial. But the idea is for addresses to be memorable. So I'm wondering how many strings which are valid IPv6 addresses are possible if you are limited to actual English (or, pick a language) 4-letter words containing only a-f. As someone mentioned, this could be expanded with 1337-speak.

[–] Ghoelian@lemmy.dbzer0.com 1 points 16 hours ago

Ahh right, that would be a bit more difficult to calculate.

I guess you could make a script which just bruteforces all combinations of a-f against an English dictionary. I might try to do that tonight.