Question

无法弄清楚如何使下面的代码正常工作。我尝试了所有可能的typeigs。

代码

type SerialNumber = (String, Int)
serialList :: Map.Map String SerialNumber
serialList = Map.fromList [("belt drive",("BD",0001))
                          ,("chain drive",("CD",0002))
                          ,("drive pulley",("DP",0003))
                          ,("drive sprocket",("DS",0004))
                          ]
findSerial :: Ord k => k -> Map.Map k a -> Maybe a
findSerial input = Map.lookup input
outOfContext (Just (a, b)) = (a, b)
getSerialFromUser :: IO ()
getSerialFromUser = do
                    putStr "Lookup part: "
                    input <- getLine
                    let output = findSerial input serialList
                    putStrLn "\nFound entry: " ++ output

错误

    • Couldn't match expected type ‘[()]’
                  with actual type ‘Maybe SerialNumber’
    • In the second argument of ‘(++)’, namely ‘output’
      In a stmt of a 'do' block: putStrLn "\nFound entry: " ++ output
      In the expression:
        do putStr "Lookup part: "
           input <- getLine
           let output = findSerial input serialList
           putStrLn "\nFound entry: " ++ output
   |
62 |                     putStrLn "\nFound entry: " ++ output

在没有书籍的情况下尝试我在Haskell的第一步，请保持温柔。

Answer 1

在Haskell函数中，应用程序保持关联，表示表达式：

putStrLn "\nFound entry: " ++ output

解析为

(putStrLn "\nFound entry: ") ++ output

但是您可能希望将其解析为：

putStrLn ("\nFound entry: " ++ output)

这就是为什么您需要显式指定括号或使用$运算符的原因：

putStrLn $ "\nFound entry: " ++ output

但是请注意，output是SerialNumber类型的，但是(++)需要两个列表作为参数。因此，您需要修改要为程序指定的行为。

结合Maybe，IO和上下文外部的功能会导致错误

1 个答案: