The Haskell Unfolder Episode 1: unfoldr

[@sebastianselander161]

Suggestion/request for the future: An episode on debugging Haskell code. The lack of a debugger, the purity of the type system, as well as Haskell being lazy has made debugging hard for me in the past. I think it'd be interesting

[@Axman6]

GHCi does actually have a debugger, it’s just not very well known about. GHCi actually has a lot of tools for stepping through the execution/evaluation of your program.

[@MikolajKonarski]

You stopped when it was the coolest and the most fun. Reminds me of childhood. ;)

[@ribosomerocker]

Oh! Could you provide a time-stamp? I'm quite curious

[@MikolajKonarski]

@@ribosomerocker: Haha, the point was totally subjective, it was the programming strategies remark; I'd love to learn about more about these: kzbin.infoS_HSt6jEtWM?feature=share&t=2062

[@ribosomerocker]

​@@MikolajKonarski I completely agree about it being the coolest part! Thank you for the link

[@vadimvinnik]

I found this variation to be useful for explaining the unfold concept to beginners: xUnfold2 :: (a -> Bool) -> (a -> b) -> (a -> a) -> a -> [b] xUnfold2 p g h x | not (p x) = [] | otherwise = (g x) : (xUnfold2 p g h (h x)) Here * p detects whether the generation should continue or finish, * g selects a new element from the current state, * h makes a step to the next state.

[@fixpoint]

Very excited about your video series as I’ve decided to learn Haskell (again, I tried 10 years ago, but got stuck).

[@vadimvinnik]

just one of many possible solutions: filter' xs = join $ unfoldr f xs -- or concat that, for lists, is the same as join where f u = case u of [] -> Nothing y : ys -> Just (if p y then [y] else [], ys)

[@worgenzwithm14z]

Whoa what's your vim setup like?

[@andresloeh]

If you mean the LSP support, I'm using neovim and CoC in this episode, although I've mostly switched to neovim-native lsp in the meantime.

[@logauit]

This is awesome! Thank you so much for making this playlist.

[@logauit]

i guess, just minor mistake in 26:50, the type signature of `foldr` should be `foldr :: (a -> r -> r) -> r -> [a] -> r`

[@well-typed]

@@logauit You're absolutely right, of course. Fortunately, all the rest of the argument still holds.

[@virus3103]

There is a typo. The first argument of foldr has type a->r->r

[@well-typed]

Yes, sorry, that's indeed a typo. It was already mentioned in one of the other comment threads. I've now added it to the video description as well.

[@maxreuv]

Very nice!

[@jameender]

What is the font used? It's so pretty!

[@andresloeh]

Iosevka (which is extremely configurable; I'm using a custom build)

[@marcthatcher]

good stuff, thanks. What editor are you using?

[@andresloeh]

Thanks! I am using neovim.