我如何重構這個代碼

Question

我試圖重構以下（工作）代碼（優秀NICTA Haskell course的一部分）：

instance Extend ListZipper where 
    f <<= z = 
    ListZipper (unfoldr ((<$>) (\z' -> (f z', z')) . toOptional . moveLeft) z) 
       (f z) 
       (unfoldr ((<$>) (\z' -> (f z', z')) . toOptional . moveRight) z) 

以下幾點：

instance Extend ListZipper where 
    f <<= z = ListZipper (go moveLeft) (f z) (go moveRight) 
     where go f = unfoldr ((<$>) (\z' -> (f z', z')) . toOptional . f) z 

但不幸的是，我得到以下編譯錯誤：

src/Course/ListZipper.hs:669:25: 
    Couldn't match type `b' with `MaybeListZipper a' 
     `b' is a rigid type variable bound by 
      the type signature for 
      <<= :: (ListZipper a -> b) -> ListZipper a -> ListZipper b 
      at src/Course/ListZipper.hs:669:3 
    Expected type: List b 
     Actual type: List (MaybeListZipper a) 
    In the return type of a call of `go' 
    In the first argument of `ListZipper', namely `(go moveLeft)' 
    In the expression: ListZipper (go moveLeft) (f z) (go moveRight) 

我錯過了一些小東西或者更重要的東西嗎？

在慣用的Haskell中，這樣的重構是否可能（甚至是理想的）？

Answer 1

您正在使用新的f遮蓋原始f。嘗試重命名第二個f，如

instance Extend ListZipper where 
    f <<= z = ListZipper (go moveLeft) (f z) (go moveRight) 
     where go g = unfoldr ((<$>) (\z' -> (f z', z')) . toOptional . g) z 

Answer 2

您在這裏使用f，它指的是<<=的第一個操作數。

(unfoldr ((<$>) (\z' -> (f z', z')) . toOptional . moveLeft) z) 

當你在這裏使用它時，它指的是傳遞給go的函數。

where go f = unfoldr ((<$>) (\z' -> (f z', z')) . toOptional . f) z 

選擇一個不同的名字！