哈斯克爾 - 更多的類型推斷問題

Question

我有以下表現：

getCount :: (Num a) => a -> [a] 
getCount int = foldl 
     processOneCount 
     [0,0,0,0,0,0,0,0,0,0] 
     (map (singleDigitCount) (map (digitToInt) (show int))) 

，我也得到了以下錯誤：

Couldn't match expected type `a' against inferred type `Int' 
    `a' is a rigid type variable bound by 
     the type signature for `getCount' 
     at C:\Users\RCIX\Desktop\Haskell Code\test.hs:23:17 
    Expected type: [a] 
    Inferred type: [Int] 
In the expression: 
    foldl 
     processOneCount 
     [0, 0, 0, 0, ....] 
     (map (singleDigitCount) (map (digitToInt) (show int))) 
In the definition of `getCount': 
    getCount int 
       = foldl 
        processOneCount 
        [0, 0, 0, ....] 
        (map (singleDigitCount) (map (digitToInt) (show int))) 

然而，當我做了:t [0,0,0,0,0,0,0,0,0,0]我回來[0,0,0,0,0,0,0,0,0,0] :: (Num t) => [t]。那麼爲什麼我不能在第一個表達式中使用它？

Answer 1

您使用的是digitToInt，它返回一個Int，而不是輸入類型。

Answer 2

查克是正確的。爲了避免弄亂你的代碼，你可以使用.運營商添加所需的功能：

(map (singleDigitCount) (map (fromIntegral . digitToInt) (show int))) 

這是假設該singleDigitCount和processOneCount也任意數字類型的工作。