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 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@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe

-- Alexander