[Haskell] type inference & instance extensions

[Malcolm Wallace]

Doug McIlroy <doug at cs.dartmouth.edu> wrote:

>         instance Num a => Num (a,a) where
>                 (x,y) * (u,v) = (x*u-y*v, x*v+y*u)
> 
> Unfortunately, type inference isn't strong enough to cope with
> 
>         (1,1)*(1,1)


I'm guessing it is because
    (fromInteger 1, fromInteger 1) :: (Num a, Num b) => (a,b)

So you want to force the two components to be the same type:
    let x = 1 in (x,x)*(x,x)

Or for differing component values:
    let x = 1 in (x,2`asTypeOf`x)*(3`asTypeOf`x,4`asTypeOf`x)

Regards,
    Malcolm