haskell - 为什么这个函数的pointfree版本看起来像这样？

Question

我一直在玩 Haskell，包括练习以无点形式编写函数。这是一个示例函数:

dotProduct :: (Num a) => [a] -> [a] -> a
dotProduct xs ys = sum (zipWith (*) xs ys)

我想以无点形式编写此函数。这是我在其他地方找到的一个例子:

dotProduct = (sum .) . zipWith (*)

但是，我不明白为什么无点形式看起来像 (sum .) . zipWith (*)而不是 sum . zipWith (*) .为什么 sum 在括号中并且有 2 个组合运算符？

Answer 1

dotProduct xs ys = sum (zipWith (*) xs ys)             -- # definition

dotProduct xs    = \ys -> sum (zipWith (*) xs ys)      -- # f x = g <=> f = \x -> g
                 = \ys -> (sum . (zipWith (*) xs)) ys  -- # f (g x) == (f . g) x
                 = sum . (zipWith (*) xs)              -- # \x -> f x == f
                 = sum . zipWith (*) xs                -- # Precedence rule

dotProduct       = \xs -> sum . zipWith (*) xs         -- # f x = g <=> f = \x -> g
                 = \xs -> (sum .) (zipWith (*) xs)     -- # f * g == (f *) g
                 = \xs -> ((sum .) . zipWith (*)) xs   -- # f (g x) == (f . g) x
                 = (sum .) . zipWith (*)               -- # \x -> f x == f

(sum .)是一个部分。它被定义为

(sum .) f = sum . f

任何二元运算符都可以这样写，例如 map (7 -) [1,2,3] == [7-1, 7-2, 7-3] .