Question

我定义了以下类型：

   data Taula = Taula String [[String]]

我创建了以下实例：

instance Show Taula   
   where 
     show  Taula [( n , i , c )]  = show  Taula [( n , i , c )]++ "\n" 
     show ( l : ls )  = show l ++  "\n"  ++ show ls

但它会产生以下错误：

  Equations for `show' have different numbers of arguments
  lastHaskell.hs:184:18-82
  lastHaskell.hs:185:18-63
  In the instance declaration for `Show Taula'

然后尝试将问题分为以下几种方法：

  instance Show mostrarTaula where
      show (Taula sep []) = ""
      show (Taula sep (k:lls) ) =  show ([tractar sep k]) ++ "\n" 
                                 ++ show (mostrarTaula (Taula sep lls))

--["Hola","Que","fas"] --> "HolakkQuekkfas"
 tractar :: String -> [String] -> String
 tractar sep [] = ""
 tractar sep par = (head par) ++ sep ++ tractar sep (drop 1 par) 

  mostrarTaula :: Taula -> [String] --[[String]] = lls i k=[String]
  mostrarTaula (Taula sep []) = [""]
 mostrarTaula (Taula sep (k:lls) ) =  [tractar sep k] ++ (mostrarTaula(Taula sep lls))

我能做什么？我尝试打印由[（String，String，String）]组成的数据类型？我想以表格形式显示数据。每个元组后都有\ n。

输出示例：

first element tuple1 Second element tuple1 third element tuple1 first element tuple2 Second element tuple2 third element tuple2 first element tuple3 Second element tuple3 third element tuple3

Answer 1

我不完全明白什么是想要的输出，但也许你可以适应这个：

instance Show Taula where
  show (Taula sep []) = ""
  show (Taula sep (k:lls)) = tractar sep k ++ "\n" ++ show (Taula sep lls)

实例显示特定数据

1 个答案: