Lazily consuming a self-referential linked list

VegOwOtenks:

go :: [q] -> s -> (q -> s -> [q] -> ([q], s)) -> ([q], s)
go qs s0 f = case qs of
  [] -> ([], s0)
  q:rest -> let
      (qs1, s1) = f q s0 qs2 -- why is this qs2 and not rest?
      (qs2, s2) = go rest s1 f -- why is this rest and not qs1?
    in (qs1, s2) -- why is this qs1 and not qs2?

I think you could condense go to look something like:

go :: [q] -> s -> (q -> s -> [q] -> ([q], s)) -> ([q], s)
go qs s0 f = case qs of
  [] -> ([], s0)
  q:rest -> let
      (qs1, s1) = f q s0 rest
      newQ = rest ++ qs1
      in go newQ s1 f

Unless I’m missing something.

Your original approach causes an infinite loop because the recursive call to go depends on s1 whose result depends on qs2 which is produced by go.