Question

根据this post，我正在尝试通过变形来创建Cofree结构。但是编译器抱怨类型不匹配：

Expected type: Base (Cofree Term E) (E, Fix Term)
  Actual type: CofreeF Term E (E, Fix Term)

但是在recursion-schemes包的源代码中，有一个类型实例定义：

type instance Base (Cofree f a) = CofreeF f a

如何使haskell编译器通过此特定类型实例方程式成功地统一该类型？

代码与链接中的代码几乎相同：

import qualified Control.Comonad.Trans.Cofree as COFREEF

type E = Int
type Term = Maybe

annotate :: E -> Fix Term -> COFREEF.Cofree Term E
annotate = curry (ana coalg)
  where
    coalg :: (E, Fix Term) -> COFREEF.CofreeF Term E (E, Fix Term)
    coalg (environment, Fix term) = environment COFREEF.:< fmap ((,) 
environment) term

以及确切的错误消息：

Couldn't match type ‘COFREEF.CofreeF Term E (E, Fix Term)’
                 with ‘Compose
                         Data.Functor.Identity.Identity
                         (COFREEF.CofreeF Maybe Int)
                         (E, Fix Term)’
  Expected type: (E, Fix Term)
                 -> Base (COFREEF.Cofree Term E) (E, Fix Term)
    Actual type: (E, Fix Term)
                 -> COFREEF.CofreeF Term E (E, Fix Term)
• In the first argument of ‘ana’, namely ‘coalg’
  In the first argument of ‘curry’, namely ‘(ana coalg)’
  In the expression: curry (ana coalg)

Answer 1

Cofree只是类型同义词

newtype CofreeT f w a = CofreeT { runCofreeT :: w (CofreeF f a (CofreeT f w a)) }
type Cofree f = CofreeT f Identity

CofreeT有一个Base实例：

type instance Base (CofreeT f w a) = Compose w (CofreeF f a)
-- Base (Cofree f a) ~ Base (CofreeT f Identity a) ~ Compose Identity (CofreeF f a)

问题中的等式在道德上是等价的，但这还不够好。

用Compose . Identity

包裹住您的余数

annotate :: E -> Fix Term -> Cofree Term E
annotate = curry $ ana $ Compose . Identity . coalg
  where
    coalg :: (E, Fix Term) -> CofreeF Term E (E, Fix Term)
    coalg (environment, Fix term) = environment :< fmap (environment,) term

Haskell类型实例无法解析

1 个答案: