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

Question

I'm trying to output a bunch of cities in a nice text format, each one starting on a new line. With each piece of data comma separated. But I'm getting the error, cannot match expected type Char to actual type [[Char]]. For example

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

would be outputted as:

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

Here's my code for it so far:

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) ++ "]"

The full error is as shown:

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
   |

Thanks in advance for any help given :)

The full error text would also help. We shouldn't guess at which line the error occurs. — arrowd
– arrowd, Commented Apr 22, 2021 at 10:45
Maybe you need to concat just after both your list comprehensions, e.g. ... = concat [showPlace p ++ "\n" | p <- ps]. You need to convert your list-of-strings into a single string. — chi
– chi, Commented Apr 22, 2021 at 10:45
a bit OT (sorry) but just out of curiosity: is this some kind of course-work (there are quite a few questions around this very problem/data floating around here lately ;)) - are you fine with saying which university this course is taking place at? — Random Dev
– Random Dev, Commented Apr 22, 2021 at 11:16

cdimitroulas · Accepted Answer · 2021-04-22 11:38:32Z

1

I got it working using intercalate from Data.List (intercalate is similar to Array.join in other languages)

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) ++ "]"

answered Apr 22, 2021 at 11:38

cdimitroulas

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