Data type conversion in list comprehension (Haskell)

Question

I'm making a simple Tic-Tac-Toe game to learn Haskell. Each cell in the board has either a player value (X or O) or a positional value if the square is empty (letters A-I, such that:

A|B|C
D|E|F
G|H|I

)

To start, I've created a simple function that takes an empty board (which looks like the one above) and a String (a letter "A" through "I"), and returns a board where the letter entered is replaced with an "X". So if I were to call newBoard blankBoard "C", I would expect this to return:

A|B|X
D|E|F
G|H|I

If I want to keep the Square data type as it is, how could I make this code work:

data Square = A | B | C | D | E | F | G | H | I | X | O deriving (Read)
instance Show Square where
  show A = "A"
  show B = "B"
  show C = "C"
  show D = "D"
  show E = "E"
  show F = "F"
  show G = "G"
  show H = "H"
  show I = "I"
type Board = [[Square]]

blankBoard :: Board
blankBoard = [[A,B,C],[D,E,F],[G,H,I]]

newBoard :: Board -> String -> Board
newBoard board letter = [[ if show sq == letter then X else sq | sq <- row ]| row <- board]

I'm getting this error, and not sure why:

[[*** Exception: myfunctions.hs:(386,3)-(394,14): Non-exhaustive patterns in function show

You are trying to use show on an X or an O, which is something that you haven't defined. Is there a reason that you don't just use deriving Show for your Square? — JHobern
That did the trick, thanks! I didn't realize that X was being shown. — Mark Karavan
Enabling warnings would have found this bug at compile time. — chi
@JHobern, can you make your comment an answer, so OP can accept? Also, OP, why are you using == on String, not on Square, simply deriving Eq? — Tobi Nary

JHobern JHobern · Accepted Answer · 2015-11-17T19:31:49

The error Non-exhaustive patterns in function means that you have tried to use a function with an argument for which it wasn't defined. In this case, you have tried to use show on either an X or an O, whereas you have only defined it for A through I.

You could fix this by adding:

show X = "X"
show O = "O"

However, since your show function doesn't do anything special at all, you would be better off just using deriving Show, and letting Haskell create the function for you.

Data type conversion in list comprehension (Haskell)

1 Answers