[Haskell-cafe] Applicative and Monad transformers

[Ralf Hinze]

Hi Jeremy,

> I have seen it said that all Monads are Applicative Functors because
> you can just do:
> 
> instance (Monad f, Applicative f) => Applicative (ReaderT r f) where
>     pure = return
>     (<*>) = ap
> 
> However, that instance for ReaderT does not exhibit the properties I
> was hoping for.

OK, let's calculate. Here are the necessary definitions for the reader
monad (not the monad transformer).

    m >>= k  =  \ x -> k (m x) x

    u <*> v  =  \ x -> (u x) (v x)

    return a  =  pure a  =  \ x -> a

So, <*> is the S combinator and pure is the K combinator.

    u >>= \ f -> v >>= \ a -> return (f a)
  = { definition of >>= }
    \ x -> (\ f -> v >>= \ a -> return (f a)) (u x) x
  = { beta }
    \ x -> (v >>= \ a -> return ((u x) a)) x
  = { definition of >>= }
    \ x -> (\ a -> return ((u x) a)) (v x) x
  = { beta }
    \ x -> return ((u x) (v x)) x
  = { definition of return }
    \ x -> (u x) (v x)
  = { definition of <*> }
    u <*> v

Yes, both definitions are, in fact, equal.

So, what went wrong in your program? Observe that the first instance
declaration can be simplified to

    instance (Monad f) => Applicative (ReaderT r f) where
      pure = return
      (<*>) = ap

which is suspicious.

Hope this helps, Ralf