
Use
value :: a -> Integer
On 8/18/07, DavidA
Hi,
I am trying to implement quadratic fields Q(sqrt d). These are numbers of the form a + b sqrt d, where a and b are rationals, and d is an integer.
In an earlier attempt, I tried data QF = QF Integer Rational Rational (see http://www.polyomino.f2s.com/david/haskell/hs/QuadraticField.hs.txt) The problem with this approach is that it's not really type-safe: I can attempt to add a + b sqrt 2 to c + d sqrt 3, whereas this should be a type error because 2 /= 3.
So I thought I'd have a go at doing it with phantom types. In effect I'd be using phantom types to simulate dependent types. Here's the code:
{-# OPTIONS_GHC -fglasgow-exts #-}
import Data.Ratio
class IntegerType a where value :: Integer
data Two instance IntegerType Two where value = 2
data Three instance IntegerType Three where value = 3
data QF d = QF Rational Rational deriving (Eq)
instance IntegerType d => Show (QF d) where show (QF a b) = show a ++ " + " ++ show b ++ " sqrt " ++ show value
instance IntegerType d => Num (QF d) where QF a b + QF a' b' = QF (a+a') (b+b') negate (QF a b) = QF (-a) (-b) QF a b * QF c d = QF (a*c + b*d*value) (a*d + b*c) fromInteger n = QF (fromInteger n) 0
The problem is, this doesn't work. GHC complains: The class method `value' mentions none of the type variables of the class IntegerType a When checking the class method: value :: Integer In the class declaration for `IntegerType'
Is what I'm trying to do reasonable? If no, what should I be doing instead? If yes, why doesn't GHC like it?
Thanks, David
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