Question

我无法理解为什么功能：

repli :: [a] -> Int -> [a]
repli xs n = concatMap (replicate n) xs

不能改写为：

repli :: [a] -> Int -> [a]
repli [] _ = []
repli (x:xs) n = (take n $ repeat x) : repli xs n

或

repli :: [a] -> Int -> [a]
repli [] _ = []
repli (x:xs) n = (replicate n x) : repli xs n

Ghci抱怨道：

Couldn't match expected type ‘a’ with actual type ‘[a]’
  ‘a’ is a rigid type variable bound by
      the type signature for repli :: [a] -> Int -> [a]
      at 99questions.hs:41:10
Relevant bindings include
  xs :: [a] (bound at 99questions.hs:43:10)
  x :: a (bound at 99questions.hs:43:8)
  repli :: [a] -> Int -> [a] (bound at 99questions.hs:42:1)
In the first argument of ‘(:)’, namely ‘(replicate n x)’
In the expression: (replicate n x) : repli xs n

我不明白为什么，因为做了所有类型的计算，结果证明没问题。 repeat x为[a]，因此take n为[a]。所以它不应该抱怨。

Answer 1

(:)的签名是a -> [a] -> [a]。因此，您不能在运营商的两侧都有列表。这是导致错误的原因。

您可以改为使用(++)，其中包含签名[a] -> [a] -> [a]。

Haskell，类型匹配问题

1 个答案: