haskell - 类型族 - 不能派生 Base Int？

Question

这个想法是实现“惰性”长度函数来将列表长度与 Int 进行比较而不计算整个长度。

{-# LANGUAGE DeriveFunctor
           , TypeFamilies
           , FlexibleInstances #-}
import Data.Functor.Foldable

type instance Base Int   = Maybe

现在我们可以拥有可折叠/不可折叠

instance Foldable Int where
  project 0 = Nothing
  project x = Just (x-1)

instance Unfoldable Int where
  embed Nothing  = 0
  embed (Just x) = x+1

我想将 [a] 转换为 Base Int Int:

leng :: [a] -> Base Int Int
leng = ana phi where
  phi :: [a] -> Base Int [a]
  phi []    = Nothing
  phi (_:t) = Just t

但这不起作用。它提示 [a] -> Base (Maybe Int) [a] 应该是 phi 的类型。我不明白为什么。

如果可行，那么我可以比较:

gt = curry $ hylo psi phi where
  phi (Just _, Nothing)  = Left True
  phi (Nothing, _)       = Left False
  phi (Just t, Just n)   = Right (t, n)

  psi (Left t)  = t
  psi (Right t) = t

main = print $ (leng [1..]) `gt` (ana project 4)

len 有什么问题？

Answer 1

ana 的类型是 (a -> Base t a) -> a -> t .请注意，它返回普通的 t而不是 Base t t .所以 leng 的正确类型是

leng :: [a] -> Int