(Num (Int -> Int)) 没有因使用句法否定而产生的实例

No instance for (Num (Int -> Int)) arising from a use of syntactic negation

我已将以下 Haskell 代码写入 return [[Int]]

的主要和次要对角线
getDiagonal' :: [[Int]] -> Int -> (Int -> Int) -> [Int]
getDiagonal' [] _ _ = []
getDiagonal' (x:xs) i fn = i' : getDiagonal' xs (fn i) fn
  where i' = head $ drop i x

getPrimaryDiagonal :: [[Int]] -> [Int]
getPrimaryDiagonal x = getDiagonal' x 0 (+1)

getSecondaryDiagonal :: [[Int]] -> [Int]
getSecondaryDiagonal x = getDiagonal' x ((length x) - 1) (+(-1))

然而,我本以为最后一行可能是下面的,使用 (-)(+)

相同的方式
getSecondaryDiagonal x = getDiagonal' x ((length x) - 1) (-1)

但是,这不起作用,当我这样做时,我得到了

Main.hs:27:59: error:
    • No instance for (Num (Int -> Int))
        arising from a use of syntactic negation
        (maybe you haven't applied a function to enough arguments?)
    • In the third argument of ‘getDiagonal'’, namely ‘(- 1)’
      In the expression: getDiagonal' x ((length x) - 1) (- 1)
      In an equation for ‘getSecondaryDiagonal’:
          getSecondaryDiagonal x = getDiagonal' x ((length x) - 1) (- 1)

为什么 (-) 会产生该错误?

(-1) 被解释为 负数 ,据我所知,它是 non-binary 运算符的唯一例外. (-1) 因此不是减一的函数。

您可以为此使用 subtract :: Num a => a -> a -> a

getSecondaryDiagonal x = getDiagonal' x ((length x) - 1) <b>(subtract 1)</b>

或者您可以使用 flip :: (a -> b -> c) -> b -> a -> c ,这就是 subtract 的实现方式:

getSecondaryDiagonal x = getDiagonal' x ((length x) - 1) <b>(flip (-) 1)</b>