在Haskell中重載通用函數

Question

我遇到了一個問題，我認爲我可能在Haskell知識上存在差距。我試圖實現一個名爲after的函數，該函數將被賦予一個項目或一個列表，並顯示它後面的內容。

after "sample" 'a'應該返回「mple」。

after "sample" "am"應該返回「ple」。

我知道如何這兩個函數定義爲after和afterList，但我想做一個泛型函數來處理這兩個

after :: (Eq a) => [a] -> a

和

after :: (Eq a) => [a] -> [a]

是這樣的功能可能嗎？我在此嘗試是：

{-# LANGUAGE MultiParamTypeClasses #-} 
sub :: (Eq a) => [a] -> Int -> [a] 
sub [] _ = [] 
sub _ 0 = [] 
sub (x:xs) c = sub xs (c - 1) 

pos :: (Eq a) => [a] -> a -> Int 
pos [] _ = 0 
pos (x:xs) c 
    | x == c = 0 
    | otherwise = 1 + pos xs c 

class (Eq a) => Continuous [a] a where 
    after x c = sub x (pos x c) 

instance (Eq a) => Continuous [a] [a] where 
    after [] _ = [] 
    after x c 
    | take (length c) x == c = sub x ((length c)+1) 
    | otherwise = after (tail x) c 

但返回的

test.hs:13:28: 
    Unexpected type `[a]' where type variable expected 
    In the declaration of `Continuous [a] a' 
Failed, modules loaded: none. 

這樣的錯誤，是我的做法根本性的缺陷？怎樣才能實現泛型函數重載，帶或不帶類型類？

Answer 1

您的方法非常正確，但您需要正確寫出它。

{-# LANGUAGE FlexibleInstances  #-} 

class Continuous list part where 
    after :: list -> part -> list 

instance (Eq a) => Continuous [a] a where 
    after x c = sub x (pos x c) 

instance (Eq a) => Continuous [a] [a] where 
    after [] _ = [] 
    after x c 
    | take (length c) x == c = sub x ((length c)+1) 
    | otherwise = after (tail x) c 

請注意，儘管您的實現看起來不起作用。但他們現在進行類型檢查。