重構Haskell lambda個函數

Question

我有一個 lambda function ((:). ((:) x))像這樣傳遞給foldr ： foldr ((:). ((:) x)) [] xs其中xs是一個二維列表。 我想重構以使其更清晰（這樣我可以更好地理解它）。 我想它會像這樣完成：

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

但這給了我錯誤：

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)) []
   |  

編輯：我圍繞這個片段的所有相關代碼是

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

它的調用是：

prefixes [1, 2, 3]

我如何重構 lambda ((:). ((:) x))以包含其 arguments？

Answer 1

您可以逐步將其轉換為 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