0 
module Main where

data Toy b next =
    Output b next
  | Bell next
  | Done

data FixE f e = Fix (f (FixE f e)) | Throw e

-- The working monadic function
catch :: (Functor f) => FixE f e1 -> (e1 -> FixE f e2) -> FixE f e2
catch (Fix x) f = Fix (fmap (`catch` f) x)
catch (Throw e) f = f e

-- Type error
applicate_fixe :: (Functor f) => FixE f (e1 -> e2) -> FixE f e1 -> FixE f e2
applicate_fixe a b = a `catch` (`fmap` b)

-- Type error
applicate_fixe' :: (Functor f) => FixE f (e1 -> e2) -> FixE f e1 -> FixE f e2
applicate_fixe' (Throw f) b = fmap f b
applicate_fixe' (Fix f) b = Fix (fmap (`applicate_fixe` b) f)

main :: IO()
main = print "Hello."

C:\!Various_Exercises\Haskell_Exercises\Free_Monad_Stuff\test.hs: 15, 33
Could not deduce (Functor (FixE f)) arising from a use of `fmap'
from the context (Functor f)
  bound by the type signature for
             applicate_fixe :: Functor f =>
                               FixE f (e1 -> e2) -> FixE f e1 -> FixE f e2
  at test.hs:14:19-76
In the second argument of `catch', namely `(`fmap` b)'
In the expression: a `catch` (`fmap` b)
In an equation for `applicate_fixe':
    applicate_fixe a b = a `catch` (`fmap` b)
C:\!Various_Exercises\Haskell_Exercises\Free_Monad_Stuff\test.hs: 18, 31
Could not deduce (Functor (FixE f)) arising from a use of `fmap'
from the context (Functor f)
  bound by the type signature for
             applicate_fixe' :: Functor f =>
                                FixE f (e1 -> e2) -> FixE f e1 -> FixE f e2
  at test.hs:17:20-77
In the expression: fmap f b
In an equation for applicate_fixe':
    applicate_fixe' (Throw f) b = fmap f b

我将结束本教程，试图找出 Free Monad，作为练习，我也在尝试执行 Applicative 函数。老实说，我不确定这些错误应该是什么意思。

此外，我不确定 的类型签名data FixE f e = Fix (f (FixE f e)) | Throw e应该是什么。起初我认为f (FixE f e)应该是一个元组，但它似乎是一个参数，因此该(FixE f e)部分实际上是第一个的类型参数f。但如果是这种情况，f内部不应该FixE f e也需要一个类型参数吗？

编辑：

applicate_fixe :: (Functor f) => FixE f (e1 -> e2) -> FixE f e1 -> FixE f e2
applicate_fixe (Fix f) b = Fix (fmap (`applicate_fixe` b) f) -- Works as the f argument in fmap is a functor
applicate_fixe (Throw f) (Fix b) = fmap f b -- The b is of type f (FixE f e1) so it is clearly a functor and yet the type system rejects it.

在任何事情之前，我不明白最后一部分。还有到底应该定义什么函子的实例？f在上面的定义中应该已经有了这个约束。

Edit2：也许你的意思是 FixE 应该有一个 Functor 实例。

instance Functor f => Functor (FixE f) where
  fmap f (Fix x) = fmap f x -- Type error
  fmap f (Throw e) = Throw (f e)

这是我最好的镜头，但它抱怨第一行中的 f 型太僵硬了。

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1 回答 1

1 
module Main where

data Toy b next =
    Output b next
  | Bell next
  | Done

instance Functor f => Functor (FixE f) where
  fmap f (Fix x) = Fix (fmap (fmap f) x)
  fmap f (Throw e) = Throw (f e)

data FixE f e = Fix (f (FixE f e)) | Throw e

-- Monadic function
catch :: (Functor f) => FixE f e1 -> (e1 -> FixE f e2) -> FixE f e2
catch (Fix x) f = Fix (fmap (`catch` f) x)
catch (Throw e) f = f e

-- Applicative function
applicativeFixE :: (Functor f) => FixE f (e1 -> e2) -> FixE f e1 -> FixE f e2
applicativeFixE (Fix f) b = Fix (fmap (`applicativeFixE` b) f)
applicativeFixE (Throw f) b = fmap f b

main :: IO()
main = print "Hello."

根据评论，我设法实现了应用功能。我不完全确定为什么上述方法与我尝试过的一些替代方法相比仍然有效，但它会进行类型检查。

感谢大家。

于 2016-10-01T06:56:30.057 回答