Question

let f = show.sum.map read.words
f "1 2"

有效。

show.sum.map read.words "1 2"

我收到错误：

<interactive>:19:19:
    Couldn't match expected type ‘a -> [String]’
                with actual type ‘[String]’
    Relevant bindings include
      it :: a -> String (bound at <interactive>:19:1)
    Possible cause: ‘words’ is applied to too many arguments
    In the second argument of ‘(.)’, namely ‘words "1 2"’
    In the second argument of ‘(.)’, namely ‘map read . words "1 2"’
Prelude> :t show.sum.map
show.sum.map
  :: (Num [b], Show b, Foldable ((->) [a])) => (a -> b) -> String
Prelude> show.sum.map read.words "1 2"

    <interactive>:21:19:
        Couldn't match expected type ‘a -> [String]’
                    with actual type ‘[String]’
        Relevant bindings include
          it :: a -> String (bound at <interactive>:21:1)
        Possible cause: ‘words’ is applied to too many arguments
        In the second argument of ‘(.)’, namely ‘words "1 2"’
        In the second argument of ‘(.)’, namely ‘map read . words "1 2"’
    Prelude>

我想知道为什么？（show.sum.map read.words）“1 2”也可以工作。

Answer 1

问题在于应用程序比应用程序绑定得更紧密，而应用程序实际上具有任何Haskell运算符的最高优先级。这意味着您的代码实际解析

 show.sum.map read.(words "1 2")

不进行类型检查，因为words "1 2"的类型为String，而不是.组成的内容。为了解决这个问题，您可以使用$运算符，该运算符专门用作应用程序，但优先级最低而不是最高优先级。

 show.sum.map read.words $ "1 2"

将按预期工作，因为它具有正确的关联。这通常是为什么你会经常看到foo . bar . baz $ quux的Haskell代码（至少是我的Haskell代码）中的模式。

混淆了haskell中的复合函数

1 个答案: