You will feel like this was obvious in hindsight.
fmap :: (a -> b) -> f a -> f b
fmap (a -> b) -> EitherT m c a -> EitherT m c b
Now follow the types in your code.
m :: EitherT m c a
mv :: Either c a
return mv :: EitherT m c a -- <- you are back to the wrong type.
How can you instead return EitherT m c b?
On Thu, Apr 12, 2018 at 6:25 AM, mike h
Hi, I’m trying to write EitherT from first principles and I’m stuck at the first hurdle - Functor. I’m looking for a hint rather than a complete answer :)
This is what I have
newtype EitherT m a b = EitherT {runEitherT :: m (Either a b)} instance Monad m => Functor (EitherT m a) where ---- fmap :: (a -> b) -> f a -> f b fmap f m = EitherT $ do mv <- runEitherT m case mv of Left _ -> return mv Right rv -> return $ Right (f rv)
and here is the compilers view Phrase.hs:31:25: error: • Couldn't match type ‘b’ with ‘a1’ ‘b’ is a rigid type variable bound by the type signature for: fmap :: forall a1 b. (a1 -> b) -> EitherT m a a1 -> EitherT m a b at Phrase.hs:27:5-8 ‘a1’ is a rigid type variable bound by the type signature for: fmap :: forall a1 b. (a1 -> b) -> EitherT m a a1 -> EitherT m a b at Phrase.hs:27:5-8 Expected type: m (Either a a1) Actual type: m (Either a b) • In the expression: return $ Right (f rv) In a case alternative: Right rv -> return $ Right (f rv) In a stmt of a 'do' block: case mv of Left _ -> return mv Right rv -> return $ Right (f rv) • Relevant bindings include rv :: a1 (bound at Phrase.hs:31:19) mv :: Either a a1 (bound at Phrase.hs:28:9) m :: EitherT m a a1 (bound at Phrase.hs:27:12) f :: a1 -> b (bound at Phrase.hs:27:10) fmap :: (a1 -> b) -> EitherT m a a1 -> EitherT m a b (bound at Phrase.hs:27:5)
what I think I need to do is fmap over the right value after pulling if out of the monad m by doing mv <- runEitherT m
these lines from the compiler are particularly confusing Expected type: m (Either a a1) Actual type: m (Either a b)
as I believe f is f:: a1->b
So just hints please and I expect I’ll have another duh moment.
Thanks
Mike
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