StateT mf <*> StateT mx

mf :: s -> m (a -> b, s)

We have (s :: s) in scope. We can apply mf to s no problem. Now what do we do with mx? We also need to give it (_ :: s). We could give it the same s we gave mf, but that would be wrong. It would ignore any state changes in mf.

So the correct option is to use the s value returned by mf. But there's no way to get it out of the monad. You could use the functor properties of m to get

fmap (\(s, f) -> fmap (\(s, a) -> (s, f a)) (mx s)) (mf s)

So you still need a monadic join to get the desired result.

The applicative instance doesn't really help as far as I can tell because it doesn't let you do anything to things of the form

Which is what StateT is made of.

On Feb 15, 2017, at 11:01 AM, Laurent Christophe <lachrist@vub.ac.be> wrote:

Hi guys, the way `StateT` are implemented as `Applicative` have been buggling my mind for some time.

https://hackage.haskell.org/package/transformers-0.5.2.0/docs/src/Control.Monad.Trans.State.Lazy.html#line-201

instance (Functor m, Monad m) => Applicative (StateT s m) where

pure a = StateT $ \ s -> return (a, s)

StateT mf <*> StateT mx = StateT $ \ s -> do

(f, s') <- mf s

(x, s'') <- mx s'

return (f x, s'')

Using dependant monadic computations, this implementation cannot be expressed in term of applicative.

This explains why we cannot have `instance (Applicative m) => Applicative (State s m)`.

However using real monadic style computations for implementing `<*>` buggles my mind.

Moreover `liftA2 (<*>)` can be used to generically compose applicative functors so why monads are needed?

https://www.haskell.org/haskellwiki/Applicative_functor#Applicative_transfomers

Any inputs would be greatly appreciated!

Cheers,

Laurent

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