haskell – 使用具有RankNTypes的函数组合时出现类型错误

2023年4月6日 225次阅读

参见英文答案 >
runST and function composition                                    3个

我对以下程序感到困惑.

{-# LANGUAGE RankNTypes #-}
newtype A a = A ( forall f. Applicative f => f a )

good :: a -> A a
good x = A $pure $x

bad  :: a -> A a
bad  x = A . pure $x

在尝试编译时,我收到此错误消息,抱怨不好：

Couldn't match type `f0 a'
               with `forall (f :: * -> *). Applicative f => f a'
Expected type: f0 a -> A a
  Actual type: (forall (f :: * -> *). Applicative f => f a) -> A a
Relevant bindings include
  x :: a (bound at huh.hs:8:6)
  bad :: a -> A a (bound at huh.hs:8:1)
In the first argument of `(.)', namely `A'
In the expression: A . pure

为什么功能好的类型检查,而ghc拒绝接受功能不好？我能做些什么来使后一版本有效？据我所知,这两个例子应该是等价的.

最佳答案 正如在几条评论中所解释的那样,问题在于类型系统无法预测这是一种有效的类型,即使它是.在
this answer中有一个提示,您可以指定类型.明确地解决问题.

此代码段有效：

-- Type specialized composition
(.!) :: ((forall f. Applicative f => f b) -> c) ->
        (a -> (forall f. Applicative f => f b)) -> a -> c
(.!) f g x = f(g x)

-- Use the new version of composition
notBad  :: a -> A a
notBad  x = A .! pure $x
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