haskell - 为什么这个函数不能进一步减少 eta？

Question

我有一个简单的功能:

calculate :: Int -> Int -> Int
calculate x y = sum . filter even $ enumFromTo x y

可以简化为

calculate x = sum . filter even . enumFromTo x

但不能进一步简化为

calculate = sum . filter even . enumFromTo

这是为什么？

我期望它能够编译，但我却收到以下错误:

    • Couldn't match type: Int -> [Int]
                     with: [Int -> Int]
      Expected: Int -> [Int -> Int]
        Actual: Int -> Int -> [Int]
    • Probable cause: ‘enumFromTo’ is applied to too few arguments
      In the second argument of ‘(.)’, namely ‘enumFromTo’
      In the second argument of ‘(.)’, namely ‘filter even . enumFromTo’
      In the expression: sum . filter even . enumFromTo
   |
10 | calculate'' = sum . filter even . enumFromTo
   |                                   ^^^^^^^^^^

Answer 1

核心问题是:

calculate x = (sum . filter even . enumFromTo) x -- could be eta-reduced
calculate x = sum . filter even . (enumFromTo x) -- cannot be eta-reduced

如果您想消除 x ，您必须首先将其置于一个位置，使其成为最外层函数的唯一参数。 (在本例中，最外层函数是 (.) ，而不是 enumFromTo 。)例如，让我们写 s g 。对于 sum . g和f g对于 filter even . g 。然后:

sum . filter even . enumFromTo x
=
s (filter even . enumFromTo x)
=
s (f (enumFromTo x))
=
s ((f . enumFromTo) x)
=
(s . f . enumFromTo) x
=
((sum .) . (filter even .) . enumFromTo) x

这是函数对唯一参数 x 的应用。 ，所以现在我们可以进行 eta-reduce。

calculate = (sum .) . (filter even .) . enumFromTo