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我正在关注Learn you a Haskell for great good，我已经实施了take'：

take' :: (Ord i, Num i) => i -> [a] -> [a]
take' n _
  | n <= 0 = []
take' _ [] = []
take' n (x:xs) = x: take' (n-1) xs

测试功能时：

take' -2 [2]

而不是得到一个空列表，我有这样的消息：

Non type-variable argument in the constraint: Num (i -> [a] -> [a])
    (Use FlexibleContexts to permit this)
    When checking that ‘it’ has the inferred type
      it :: forall i a …

我有以下操作:

Prelude> mod (3 - 12) 7

结果我有5个.

为什么结果是5？

当我尝试这样的事情时:

Prelude> mod -9 7

然后我有错误:

<interactive>:6:1: error:
    • Non type-variable argument
        in the constraint: Num (t -> a -> a -> a)
      (Use FlexibleContexts to permit this)
    • When checking the inferred type
        it :: forall a t.
              (Num (t -> a -> a -> a), Num (a -> a -> a), Num t, Integral a) =>
              a -> a -> a

为什么？

我忘了提,我刚开始学习哈斯克尔.