How does one override show for a newtype?

Question

I want to override the default integer constructors in Haskell so they produce strings (mostly for curiosity, but temporarily to make a nice input alternative for LaTeX's \frac{}{} inconvenience).

I wanted to be able to use the language itself, instead of a special parser, but I guess that's probably not going to work out...

module Main where

import Prelude hiding ((+))

newtype A = A Int deriving (Eq, Show, Num)
default (A)

(+) :: A -> (A -> String)
(A a) + (A b) = (show a) ++ " + " ++ (show b)

main2 = 3+4

main :: IO ()
main = putStrLn main2

The problem with the above is that the + function only works for (A, A) instead of (A, String), etc. If one simply leaves out the pattern match "(A a)" and writes "a" instead, then the show() function prepends "A " so "3" becomes "A 3" instead of just "3".

I want to override Show for A, but it seems to be quite a headache...

Answer 1

If you want your own Show instance for A, then just don't derive it and make your own instance:

newtype A = A Int deriving (Eq, Num)

instance Show A where
  show (A a) = show a

Then you can write something like:

(+) :: (Show a, Show b) => a -> b -> String
a + b = show a ++ " + " ++ show b

Of course, if you are defining your own + operator like that, then I don't think your problem requires the newtype A declaration:

module Main where

import Prelude hiding ((+))

(+) :: (Show a, Show b) => a -> b -> String
a + b = show a ++ " + " ++ show b

aSum = 3 + 4

main :: IO ()
main = putStrLn aSum

Answer 2

override the default integer constructors in Haskell so they produce strings

So this is done by defining a Num instance for String. Then (+) can be used as String -> String -> String.

A super quick example:

{-# LANGUAGE TypeSynonymInstances #-}

module A where

instance Num String where (+) = (++)

{-

*A> "hello" + "world"
"helloworld"

-}

Write a fromIntegral method to get functions from integer literals to strings (e.g. 1 --> "1").

For a more general, more disciplined approach to lifting lists of Num values to Num, see the Hinze approach to streams as Num, http://hackage.haskell.org/package/hinze-streams

Answer 3

Is this what you are trying to do? Create a numeric type so that you can write expressions in Haskell, and then just print them and have them come out as LaTeX math strings?

module Main where

import Data.Ratio

data LaTeXmath = E Precedence String
    deriving (Eq)

data Precedence = Pterm | Pmul | Padd | Pexp
    deriving (Show, Eq, Ord, Bounded)

expr :: Precedence -> LaTeXmath -> String
expr p (E q s) | p >= q    = s
               | otherwise = "\\left(" ++ s ++ "\\right)"

instance Num LaTeXmath where
    a + b = E Padd (expr Padd a ++ " + " ++ expr Padd b)
    a - b = E Padd (expr Padd a ++ " - " ++ expr Padd b)
    a * b = E Pmul (expr Pmul a ++ " "   ++ expr Pmul b)

    negate a = E Pterm (" -" ++ expr Pterm a)
    abs    a = E Pterm (" |" ++ expr Pexp a ++ "| ")
    signum a = E Pterm (" \\signum (" ++ expr Pexp a ++ ") ")

    fromInteger i = E Pterm (show i)

instance Fractional LaTeXmath where
    a / b = E Pterm ("\\frac{" ++ expr Pexp a ++ "}{" ++ expr Pexp b ++ "}")

    fromRational r = fromInteger num / fromInteger denom
        where num = numerator r
              denom = denominator r

instance Show LaTeXmath where
    show a = "\\[" ++ expr Pexp a ++ "\\]"

sym :: String -> LaTeXmath
sym x = E Pterm x

anExample :: LaTeXmath
anExample = sym "y" / (recip 2 * ( 3 + sym "x" + 2 * sym "y" ) )

main :: IO ()
main = print anExample

This is complicated by the logic required to handle precedence so that parentheses are inserted correctly. The example prints out:

\[\frac{y}{\frac{1}{2} \left(3 + x + 2 y\right)}\]