[Haskell-cafe] Swapping type arguments

[Alexander V Vershilov]

Hello, Alexey.

As far as I know this is not directly possible, at least in 7.8. However
you could workaround it with a newtypes. (I've changed a bit definition of T,
because otherwise there is no way to write MonadTrans instance (it seems)):

import Data.Bifunctor
import Control.Monad.Trans

data T b m a = T (m (Either b a))

instance MonadTrans (T b) where
  lift g = T (return . Right =<< g)

newtype TF m b a = TF { unTF :: T b m a}

instance Functor m => Bifunctor (TF m) where
  bimap f g (TF (T z)) = TF (T (fmap (bimap f g) z))

In order to easily use bimap and friends you can introduce some helpers
functions like:

withTF :: (TF m b a -> TF m b c) -> T b m a -> T b m c
withTF f = unTF . f . TF

or look at more heavyweight solutions.

--
Best regards,
  Vershilov Alexander


On 1 May 2015 at 22:32, Alexey Uimanov <s9gf4ult at gmail.com> wrote:
> This question probably asked already.
>
> I have a type
>
> data T m b a = T (m (b, a))
>
> and it should be instance of `Bifunctor` and `MonadTrans` same time. But
> there is a problem:
>
> instance Bifunctor (T m)
>
> is ok, but
>
> instance MonadTrans (T ...
>
> is not, because first argument is `m` but we need `b`. Type can be rewritten
> like
>
> data T b m a = T (m (b, a))
>
> and instance of MonadTrans is ok:
>
> instance MonadTrans (T b)
>
> but how to define Bifunctor?
>
> instance Bifunctor (T ...
>
> I did not found how to workaround this. Is there any type magic for such
> cases?
>
> _______________________________________________
> Haskell-Cafe mailing list
> Haskell-Cafe at haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
>



-- 
Alexander