无法理解为什么可能在 Haskell 中键入签名

Question

您好，我无法理解我的函数所需的类型签名。

-- findPassword :: Map.Map Hash Passwd -> Int -> Hash -> Maybe Passwd
findPassword rTable width hashVal = do
    let usefulHashes = take (width+1) (iterate (pwHash.pwReduce) hashVal)
    let hashesInMap = [i | i <- usefulHashes, Map.member i rTable]
    let goodPass = [ rTable Map.! j | j <- hashesInMap]
    let findPass = listToMaybe [ helper k hashVal width | k <- goodPass, (helper k hashVal width) /= "" ] 
    return findPass
    where
        helper :: Passwd -> Hash -> Int -> Passwd
        helper passW hashVal width
            | (pwHash passW) == hashVal   = passW
            | width == 0                  = ""
            | otherwise                   = helper (pwReduce (pwHash passW)) hashVal (width-1)

在这个函数中，我使用一个 table，它是哈希值 (Int32) 的映射作为密码 (Strings) 的键，并尝试在 table 中找到给定的哈希。找到我要查找的密码后，我使用 listToMaybe 和 return 值作为可能的密码。但是，当我运行收到此错误时：

* Couldn't match type `Maybe Passwd' with `[Char]'
      Expected type: Maybe Passwd
        Actual type: Maybe (Maybe Passwd)
    * In a stmt of a 'do' block: return result
      In the expression:
        do let usefulHashes
                 = take (width + 1) (iterate (pwHash . pwReduce) hashVal)
           let hashesInMap = ...
           let goodPass = ...
           let findPass = ...
           ....
      In an equation for `findPassword':
          findPassword rTable width hashVal
            = do let usefulHashes = ...
                 let hashesInMap = ...
                 let goodPass = ...
                 ....
            where
                turntoMaybe :: [Passwd] -> Maybe Passwd
                turntoMaybe list = listToMaybe list
                helper :: Passwd -> Hash -> Int -> Passwd
                helper passW hashVal width
                  | (pwHash passW) == hashVal = passW
                  | width == 0 = ""
                  | otherwise = helper (pwReduce (pwHash passW)) hashVal (width - 1)
   |
78 |     return result

所以我的问题是为什么它需要一个 Maybe (Maybe Passwd) 值？如果我将类型签名切换为 Maybe (Maybe Passwd) 它可以工作，但输出预期是双 Just Just "Passwd"。如果我删除类型签名，它就可以正常工作。这只是项目其余部分的一小部分，所以如果需要任何说明，请告诉我。

Answer 1

Haskell 不是 Java。您不应该以 return 结束所有内容。要使其正常工作，只需将 return findPass 更改为 findPass。您还应该考虑删除多余的 do 块，只使用 where 而不是所有 let。

无法理解为什么可能在 Haskell 中键入签名

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