Haskell:Lambda 函数 - 错误 - 逃脱其范围

Question

我定义了以下模块:

module ComplexCompose where

isEven x =
 if x `rem` 2 == 0 then
  (True, "Is Even")
 else
  (False, "Is Odd")


negateIt x = ( not x, "negated")



composer x =
 (c, b ++ ", " ++ d) where (a,b) = isEven x
                           (c,d) = negateIt a

对 Composer 进行以下修改可以正常工作:

composerV1 x f g =
 (c, b ++ ", " ++ d) where (a,b) = f x
                           (c,d) = g a

我想让composer函数返回一个由f和g组成的lamda。我试过这个:

composerV2 f g =
     \x -> (c, b ++ ", " ++ d) where (a,b) = f x
                                     (c,d) = g a

这不起作用。这个版本有什么问题吗？

编译器输出:

 • Couldn't match expected type ‘t0 -> (t5, t4)’
                  with actual type ‘t3’
        because type variables ‘t4’, ‘t5’ would escape their scope
      These (rigid, skolem) type variables are bound by
        the inferred type of
        a :: t5
        b :: t4
        at complex-compose.hs:27:34-44
    • In the expression: f x
      In a pattern binding: (a, b) = f x
      In an equation for ‘c4’:
          c4 f g
            = \ x -> (c, b ++ ", " ++ d)
            where
                (a, b) = f x
                (c, d) = g a
    • Relevant bindings include
        a :: t5 (bound at complex-compose.hs:27:35)
        b :: t4 (bound at complex-compose.hs:27:37)
        f :: t3 (bound at complex-compose.hs:26:4)
        c4 :: t3 -> (t2 -> (t1, [Char])) -> t -> (t1, [Char])
          (bound at complex-compose.hs:26:1)

顺便说一句，以下简单的函数可以工作:

fn f g = \x = g(f x)

Answer 1

如果我不在左侧写下 x，

composerV2 会为我进行类型检查:

composerV2 f g x =
     (c, b ++ ", " ++ d) where (a,b) = f x
                               (c,d) = g a

问题在于，在 where 子句中，标识符 x 实际上不在范围内。错误消息有点令人困惑，但如果添加类型签名，就会变得精确，无论如何，这是一个很好的做法。

或者，您也可以使用 let 绑定(bind)。

请注意，composerV2 f g = let ... in\x -> ... 和 composerV2 f g x = ... 在语义上是相同的，因为您可以部分应用第二个版本。

编辑:使用 let 绑定(bind)的版本如下所示:

composerV2' f g = \x ->
  let (a,b) = f x
      (c,d) = g a
  in (c, b ++ ", " ++ d)