It is impossible.You can make a new functor class using contraint kinds that allows what you want. There is probably a package out there already that does!Sections 2.2 and 5.1 have the relevant stuff. I realise this is a bit err, dense for the beginners list. There are probably better references out there.BenOn Mon, 4 Jan 2016 at 14:30 martin <martin.drautzburg@web.de> wrote:Am 01/04/2016 um 11:45 AM schrieb Imants Cekusins:
>> a newtype for ordered lists
>
> why not:
> newtype Ordlist a = Ordlist [a]
All nice and dandy, but at first you already need an Ord constraint for your smart constructor
-- and a ctor:
ordList::(Ord a) => [a]-> OrdList a
ordList = OrdList . sort
but this is still not the main problem. When you try to define a Functor instance, you'd be tempted to do this (at least
I was):
instance Functor OrdList where
fmap f (OrdList xs) = OrdList $ sort $ map f xs
but you can't do this, because of: "No instance for (Ord b) arising from a use of ‘sort’", where b is the return type of
f :: (a->b). This does make sense, the function has to return something which can be sorted.
So my question is: is it impossible to write a functor instance for ordered lists? It appears so, because a Functor does
not impose any constraints of f. But my knowledge is quite limited and maybe a well-set class constraint can fix things.
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