重构 Haskell lambda 函数

Question

我有一个 lambda 函数 ((:) . ((:) x)) ，我将其传递给 foldr ，如下所示： foldr ((:) . ( (:) x)) [] xs 其中 xs 是二维列表。我想重构以使其更清晰（这样我可以更好地理解它）。我想它会像这样完成：

foldr (\ element acc -> (element:acc) . (x:acc)) [] xs

但这给了我错误：

ex.hs:20:84: error:
    • Couldn't match expected type ‘a0 -> b0’ with actual type ‘[[a]]’
    • Possible cause: ‘(:)’ is applied to too many arguments
      In the second argument of ‘(.)’, namely ‘(x : acc)’
      In the expression: (element : acc) . (x : acc)
      In the first argument of ‘foldr’, namely
        ‘(\ element acc -> (element : acc) . (x : acc))’
    • Relevant bindings include
        acc :: [[a]] (bound at ex.hs:20:60)
        element :: [a] (bound at ex.hs:20:52)
        xs :: [[a]] (bound at ex.hs:20:30)
        x :: [a] (bound at ex.hs:20:28)
        prefixes :: [a] -> [[a]] (bound at ex.hs:20:1)
   |
20 | prefixes = foldr (\x xs -> [x] : (foldr (\ element acc -> (element:acc) . (x:acc)) [] xs)) []
   |  

编辑：我围绕此片段的所有相关代码是

prefixes :: Num a => [a] -> [[a]]
prefixes = foldr (\x acc -> [x] : (foldr ((:) . ((:) x)) [] acc)) []

及其调用是：

prefixes [1, 2, 3]

如何重构 lambda ((:) 。((:) x )) 包含它的两个参数？

Answer 1

您可以逐步将其转换为 lambda。

(:) . ((:) x)
\y -> ((:) . (((:) x)) y  -- conversion to lambda
\y -> (:) (((:) x) y)     -- definition of (.)
\y -> (:) (x : y)         -- rewrite second (:) using infix notation
\y z -> (:) (x : y) z     -- add another parameter
\y z -> (x : y) : z       -- rewrite first (:) using infix notation