Question

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

尝试编译时，收到此错误消息，抱怨bad：

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

为什么函数good类型检查，而ghc拒绝接受函数bad？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。

Answer 1

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中有一条提示，您可以明确指定.的类型来解决问题。

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

使用具有RankNTypes的函数组合时键入错误

1 个答案: