Question

我有下面的代码很好用

replacePointsWithZeros :: Eq a => a -> a -> [a] -> [a]
replacePointsWithZeros _ _ [] = []
replacePointsWithZeros replacee replacer (x:xs)
   | replacee == x = replacer:replacePointsWithZeros replacee replacer xs
   | otherwise = x:replacePointsWithZeros replacee replacer xs

但是我想用默认值替换replacee和replacer参数，所以本质上我想要的是

replacePointsWithZeros :: Eq a => [a] -> [a]
replacePointsWithZeros [] = []
replacePointsWithZeros (x:xs)
   | '.' == x = '0':replacePointsWithZeros xs
   | otherwise = x:replacePointsWithZeros xs

但是，当我尝试使用某种消息来抱怨类型时，我并不十分了解，因为我对Haskell非常陌生。我在做什么错，我该如何解决？

我假设问题出在行

replacePointsWithZeros :: Eq a => [a] -> [a]

以及我使用a代替Char但我将a切换为Char的事实，Eq语句出现问题

Answer 1

自从您编写'.' == x以来，这意味着x是Char，因此(x:xs)是Char的列表，因此{{ 1}}或[Char]。

此外，您将String和'0':写为输出，因此这意味着输出也包含x:的列表。因此，这意味着Char的签名是：

replacePointsWithZeros

通过将条件放在映射函数中，可以使以上内容变得更懒：

replacePointsWithZeros :: String -> String
replacePointsWithZeros [] = []
replacePointsWithZeros (x:xs)
   | '.' == x = '0':replacePointsWithZeros xs
   | otherwise = x:replacePointsWithZeros xs

，我们可以使用replacePointsWithZeros :: String -> String replacePointsWithZeros [] = [] replacePointsWithZeros (x:xs) = f x : replacePointsWithZeros xs where f '.' = '0' f x = x函数代替显式递归：

map

用另一个默认值Haskell替换默认值

1 个答案: