haskell - 用另一个默认值 Haskell 替换默认值

Question

我有以下代码可以正常工作

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 

但我想用默认值替换 replacee 和 replacer 参数，所以基本上我想要的是

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

但是，当我尝试它时，它会用一条消息提示类型，我不太明白，因为我对 Haskell 很陌生。我做错了什么，我该如何解决？

我假设问题出在这条线上

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

事实上，我使用 a 而不是 Char 但如果我将 a 切换到 Char 我会遇到 Eq 语句的问题

Answer 1

既然你写 '.' == x ，这意味着 x是 Char ，因此 (x:xs)是 Char 的列表s，所以[Char] ，或 String .

此外你写'0':和 x:作为输出，因此这意味着输出是 Char 的列表s 也是。因此，这意味着 replacePointsWithZeros 的签名是:

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

您可以通过将条件放在映射函数中来使上述更懒惰:

replacePointsWithZeros :: String -> String
replacePointsWithZeros [] = []
replacePointsWithZeros (x:xs) = f x : replacePointsWithZeros xs
    where f '.' = '0'
          f x = x

我们可以使用 map函数而不是显式递归:

replacePointsWithZeros :: String -> String
replacePointsWithZeros = map f
    where f '.' = '0'
          f x = x