Haskell: Couldn't match type `[Char]' with `Char'

Question

I want a program that takes a text and changes some words, in particular it should write "Leonardo" every time it reads "Chiara" and vice-versa. This is my code:

changeText [] = []
changeText (x:xs)
  | x == "C" = isChiara x xs
  | x == "L" = isLeo x xs
  | otherwise = x ++ changeText x

isChiara x xs
  | nome == "hiara" = "Leonardo" ++ changeText (drop 5 xs)
  | otherwise = x ++ changeText xs
  where nome = take 5 xs

isLeo x xs
  | nome == "eonardo" = "Chiara" ++ changeText (drop 7 xs)
  | otherwise = x ++ changeText xs
  where nome = take 7 xs

However, when I try to run it I get this error:

main.hs:10:1: error:
    Couldn't match type `[Char]' with `Char'
    Expected type: [Char] -> [Char]
      Actual type: [[Char]] -> [Char]

main.hs:17:13: error:
    * Couldn't match type `Char' with `[Char]'
      Expected type: [[Char]]
        Actual type: [Char]
    * In the second argument of `(==)', namely `"hiara"'
      In the expression: nome == "hiara"
      In a stmt of a pattern guard for
                     an equation for `isChiara':
        nome == "hiara"

main.hs:17:49: error:
    * Couldn't match type `[Char]' with `Char'
      Expected type: [Char]
        Actual type: [[Char]]
    * In the first argument of `changeText', namely `(drop 5 xs)'
      In the second argument of `(++)', namely `changeText (drop 5 xs)'
      In the expression: "Leonardo" ++ changeText (drop 5 xs)

main.hs:18:33: error:
    * Couldn't match type `[Char]' with `Char'
      Expected type: [Char]
        Actual type: [[Char]]
    * In the first argument of `changeText', namely `xs'
      In the second argument of `(++)', namely `changeText xs'
      In the expression: x ++ changeText xs

main.hs:22:13: error:
    * Couldn't match type `Char' with `[Char]'
      Expected type: [[Char]]
        Actual type: [Char]
    * In the second argument of `(==)', namely `"eonardo"'
      In the expression: nome == "eonardo"
      In a stmt of a pattern guard for
                     an equation for `isLeo':
        nome == "eonardo"

main.hs:22:49: error:
    * Couldn't match type `[Char]' with `Char'
      Expected type: [Char]
        Actual type: [[Char]]
    * In the first argument of `changeText', namely `(drop 7 xs)'
      In the second argument of `(++)', namely `changeText (drop 7 xs)'
      In the expression: "Chiara" ++ changeText (drop 7 xs)

main.hs:23:33: error:
    * Couldn't match type `[Char]' with `Char'
      Expected type: [Char]
        Actual type: [[Char]]
    * In the first argument of `changeText', namely `xs'
      In the second argument of `(++)', namely `changeText xs'
      In the expression: x ++ changeText xs
Failed, modules loaded: none.

What is going wrong? I am completly new to Haskell and I have tried to look for answers to similar problems but it is really, really hard for me to understand what's going on.

Answer 1

A string in Haskell is a list of Char; Haskell denotes its type as [Char] .

If you write 'C' , then Haskell will interpret this as the Char C .
If you write "C" , then Haskell will interpret this as a single-character string "C".
And Haskell is pedantic about using the right type.

---

So to give you the more specific solution:

The problem originates in the first part.

changeText (x:xs)
  | x == "C" = isChiara x xs
  | x == "L" = isLeo x xs
  | otherwise = x ++ changeText xs

The x is a Char, but you compare it to "C" or "L" - which Haskell interprets as a string of 1 character.

Make it:

changeText (x:xs)
  | x == 'C' = isChiara x xs
  | x == 'L' = isLeo x xs
  | otherwise = x ++ changeText xs

And the error message should be gone.

Answer 2

Since you already solved this, let me show a reasonably short alternative

changeText :: String -> String
changeText [] = []
changeText ('C':'h':'i':'a':'r':'a':xs) = "Leonardo" ++ changeText xs
changeText ('L':'e':'o':'n':'a':'r':'d':'o':xs) = "Chiara" ++ changeText xs
changeText (x:xs) = x : changeText xs