Haskell - 更多类型的推理问题

Question

我有以下表达式：

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)))

我收到以下错误：

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)))

当我做:t [0,0,0,0,0,0,0,0,0,0]时，我会回来[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呢？

Answer 1

你正在使用digitToInt，它返回一个Int，而不是输入类型。

Answer 2

查克是对的。为避免使代码混乱，您可以使用.运算符添加所需的函数：

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

这假设singleDigitCount和processOneCount也适用于任意数字类型。