哈斯克尔 - 更多的类型推断问题

Question

我有以下表现：

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))) 

，我也得到了以下错误：

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，当我做了:t [0,0,0,0,0,0,0,0,0,0]我回来[0,0,0,0,0,0,0,0,0,0] :: (Num t) => [t]。那么为什么我不能在第一个表达式中使用它？

Answer 1

您使用的是digitToInt，它返回一个Int，而不是输入类型。

Answer 2

查克是正确的。为了避免弄乱你的代码，你可以使用.运营商添加所需的功能：

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

这是假设该singleDigitCount和processOneCount也任意数字类型的工作。