Can you overload + in haskell?

Question

While I\'ve seen all kinds of weird things in Haskell sample code - I\'ve never seen an operator plus being overloaded. Is there something special about it?

Let\'s

Answer 1

Yes

(+) is part of the Num typeclass, and everyone seems to feel you can't define (*) etc for your type, but I strongly disagree.

newtype Pair a b = Pair (a,b)  deriving (Eq,Show) 

I think Pair a b would be nicer, or we could even just use the type (a,b) directly, but...

This is very much like the cartesian product of two Monoids, groups, rings or whatever in maths, and there's a standard way of defining a numeric structure on it, which would be sensible to use.

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

Now we've overloaded (+) in an obvious way, but also gone the whole hog and overloaded (*) and all the other Num functions in the same, obvious, familiar way mathematics does it for a pair. I just don't see the problem with this. In fact I think it's good practice.

*Main> Pair (3,4.0) + Pair (7, 10.5)
Pair (10,14.5)
*Main> Pair (3,4.0) + 1    -- *
Pair (4,5.0)

* - Notice that fromInteger is applied to numeric literals like 1, so this was interpreted in that context as Pair (1,1.0) :: Pair Integer Double. This is also quite nice and handy.