如何在 Haskell 中转换 [[Char]] -> Char？

Question

我正在尝试以一种漂亮的文本格式输出一堆城市，每个城市都在一个新行开始。每条数据用逗号隔开。但是我得到了错误，cannot match expected type Char to actual type [[Char]]。例如

("New York", (1,1), [5, 4, 3, 2])

将输出为：

New York, (1,1), [5, 4, 3, 2]

到目前为止，这是我的代码：

type Name = String
type Coordinates = (Int, Int)
type Pop = Int
type TotalPop = [Pop]
type Place = (Name, Coordinates, TotalPop)
places :: [Place]
places = [("New York", (1,1), [5, 4, 3, 2]),
        ("Washington DC", (1, 1), [4, 3, 2, 1]),
        ("Los Angeles", (3,3), [7, 6, 5, 4])]

placesToString :: [Place] -> IO ()
placesToString ps =  putStrLn (showPlaces ps)

showPlaces :: [Place] -> String
showPlaces ps = [showPlace p ++ "\n" | p <- ps] where
        showPlace p = [showData d | d <- p] where
            showData (w, (x,y), z) = w ++ ", (" ++ show x ++ ", " ++ show y ++ "), [" ++ unwords (map show z) ++ "]"

完整错误如图：

template.hs:14:27: error:
    • Couldn't match type ‘(Name, Coordinates, TotalPop)’
                     with ‘[([Char], (a0, a1), [a2])]’
      Expected type: [([Char], (a0, a1), [a2])]
        Actual type: Place
    • In the first argument of ‘showPlace’, namely ‘p’
      In the first argument of ‘(++)’, namely ‘showPlace p’
      In the expression: showPlace p ++ "\n"
   |
14 | showPlaces ps = [showPlace p ++ "\n" | p <- ps] where
   |                           ^

template.hs:14:32: error:
    • Couldn't match type ‘Char’ with ‘[Char]’
      Expected type: [[Char]]
        Actual type: [Char]
    • In the second argument of ‘(++)’, namely ‘"\n"’
      In the expression: showPlace p ++ "\n"
      In the expression: [showPlace p ++ "\n" | p <- ps]
   |
14 | showPlaces ps = [showPlace p ++ "\n" | p <- ps] where
   |

提前感谢您提供的任何帮助:)

Answer 1

我使用 intercalate from Data.List 使其工作（插入类似于其他语言中的 Array.join）

import Data.List (intercalate)

showPlaces :: [Place] -> String
showPlaces ps = intercalate "\n" [showData d | d <- ps]
  where
    showData (w, (x, y), z) = w ++ ", (" ++ show x ++ ", " ++ show y ++ "), [" ++ unwords (map show z) ++ "]"

如何在 Haskell 中转换 [[Char]] -> Char？

How can I convert [[Char]] -> Char in Haskell?

haskell

types

compiler-errors

type-conversion