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Asked: 2026-06-03T03:10:38+00:00

I would like to define several versions of a thing but with different types

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I would like to define several versions of a thing but with different types to enhance type safety in my program. For example I have several types of bivariate values which I want to be instances of Num but all should be different types. So what I have done is creating a newtype with one type variable and declared new types based on that. However, I find it slightly annoying that I now have to use both constructors all the time. Is there any way around this? 

{-# LANGUAGE GeneralizedNewtypeDeriving #-}

newtype Bivar t = Bivar (t,t) deriving (Show, Eq)

instance (Num t) => Num (Bivar t) where 
  (+) (Bivar (x1,y1)) (Bivar (x2,y2)) = Bivar (x1+x2, y1+y2)
  (-) (Bivar (x1,y1)) (Bivar (x2,y2)) = Bivar (x1-x2, y1-y2)
  (*) (Bivar (x1,y1)) (Bivar (x2,y2)) = Bivar (x1*x2, y1*y2)
  abs (Bivar (x1,y1)) = Bivar (abs x1, abs y1)
  fromInteger i = Bivar (fromInteger i, fromInteger i)
  signum (Bivar (x1,y1)) = Bivar (signum x1, signum y1)

newtype BivarNode = BivarNode (Bivar Int) deriving (Show, Eq, Num)
newtype BivarVal = BivarVal (Bivar Double) deriving (Show, Eq, Num)
newtype HBivarVal = HBivarVal (Bivar Double) deriving (Show, Eq, Num)

-- This is annoying:
a1 = BivarVal (Bivar (1.0, 2.0))
a2 = HBivarVal (Bivar (1.0, 2.0))
b = BivarNode (Bivar (1,2))

-- is there a way so that I can write it this way?
aa1 = BivarVal (1.0, 2.0)
aa2 = HBivarVal (1.0, 2.0)
bb = BivarNode (1,2)

thank you!

EDIT:

To expand on the original question, I would also like to use the type names on pattern matches, something along the lines of

myFunction :: HBivarVal -> Double
myFunction (HBivarVal (Bivar (x,y))) = x

is that possible too?

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1 Answer

  1. Editorial Team

    Why not make all of them a newtype, even Bivar?

    instance (Num a, Num b) => Num (a, b) where
    (a, b) + (a', b') = (a+a', b+b')
    (a, b) * (a', b') = (a*a', b*b')
    (a, b) - (a', b') = (a-a', b-b')
    fromInteger i = (fromInteger i, fromInteger i)
    abs    (a, b) = (abs    a, abs    b)
    signum (a, b) = (signum a, signum b)

newtype Bivar t   = Bivar     (t     , t     ) deriving (Show, Eq, Num)
newtype BivarNode = BivarNode (Int   , Int   ) deriving (Show, Eq, Num)
newtype BivarVal  = BivarVal  (Double, Double) deriving (Show, Eq, Num)
newtype HBivarVal = HBivarVal (Double, Double) deriving (Show, Eq, Num)
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