我有一个像这样的数据结构
data ShoppingList a
= Empty
| Item a
| Listofitems [ShoppingList a]
deriving (Show)
我正在为此编写fmap
instance Functor ShoppingList where
fmap f Empty = Empty
fmap f (Item i) = Item (f i)
fmap f (Listofitems [Empty]) = Empty
fmap f (Listofitems ((Item a):Listofitems [as])) = Listofitems $ (fmap f (Item a)) (fmap f Listofitems [as])
这是我到目前为止所写的内容,但尚未编译,请您帮助我了解这里的问题是什么,解释很不错。 我遇到的两个错误
src\Ml.hs:19:33: error:
* Couldn't match expected type `[ShoppingList a]'
with actual type `ShoppingList a0'
* In the pattern: Listofitems [as]
In the pattern: (Item a) : Listofitems [as]
In the pattern: Listofitems ((Item a) : Listofitems [as])
* Relevant bindings include
a :: a (bound at src\Ml.hs:19:30)
f :: a -> b (bound at src\Ml.hs:19:8)
fmap :: (a -> b) -> ShoppingList a -> ShoppingList b
(bound at src\Ml.hs:16:3)
|
19 | fmap f (Listofitems ((Item a):Listofitems [as])) = Listofitems $ (fmap f (Item a)) (fmap f Listofitems [as])
| ^^^^^^^^^^^^^^^^
src\Ml.hs:19:68: error:
* Couldn't match expected type `b -> [ShoppingList b]'
with actual type `ShoppingList b'
* The function `fmap' is applied to three arguments,
but its type `(a -> b) -> ShoppingList a -> ShoppingList b'
has only two
In the second argument of `($)', namely
`(fmap f (Item a)) (fmap f Listofitems [as])'
In the expression:
Listofitems $ (fmap f (Item a)) (fmap f Listofitems [as])
* Relevant bindings include
f :: a -> b (bound at src\Ml.hs:19:8)
fmap :: (a -> b) -> ShoppingList a -> ShoppingList b
(bound at src\Ml.hs:16:3)
|
19 | fmap f (Listofitems ((Item a):Listofitems [as])) = Listofitems $ (fmap f (Item a)) (fmap f Listofitems [as])
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
基本上,如果我有一个列表= [苹果项目,空,[香蕉项目,空]] 我想要fmap(++ M)列表返回[Item AppleM .Empty。[Item BananaM,Empty]]
答案 0 :(得分:2)
首先,您的数据类型过于复杂。项目列表确实应该是一个列表,因此无论如何,您都应该在使用列表时定义data ShoppingList' a = ShoppingList' [a]
。无需嵌套购物清单。
但是,如果您需要那样的话,这是一个解决方案。注意我假设您不需要ShoppingLists列表,因为您的数据定义中已经有一个列表。所以你可以打电话
--fmap :: (a -> b) -> ShoppingList a -> ShoppingList b
fmap _ Empty = Empty
fmap f (Item i) = Item (f i)
fmap f (Listofitems ls) = Listofitems $ map (fmap f) ls
>>fmap (++ "M") $ Listofitems [Item "Apple", Empty, Listofitems [Item "Banana", Empty]]
Listofitems [Item "AppleM",Empty,Listofitems [Item "BananaM",Empty]]
注意:
还请记住,如果需要性能,字符串附加操作对于长字符串来说效率不高