Re: [Haskell-cafe] Are all monads functions?

31 Dec 2011

      Thanks for the explanation on free monads, it's interesting.

But still, I maintain my previous view. I could clarify that by saying that
(e.g. for Maybe) we could separate it in two types, Maybe itself and its
monad:

-- The plain Maybe type
data Maybe a = Just a | Nothing

-- The MaybeMonad
newtype MaybeMonad a = MM ( () -> Maybe a )

That's what using Maybe as a monad semantically means, doesn't it?

It's just that as I said before, the function () -> Maybe a is useless,
thus making the whole type MaybeMonad entirely equivalent to Maybe.

So Roman, as I understand it, a free monad can only be a stateless monad.
So *again*, FreeM is equivalent to :

data FreeM f a = Return a
               | Bind (* () ->* f (FreeM f a) )

You cannot express StateT for instance with a functor and FreeM, can you?

But if you change FreeM to be:

data FreeM f s a = Return a
               | Bind (* s ->* f (FreeM f a) )

Then maybe it becomes possible...

2011/12/31 Roman Cheplyaka 
...
* Yves Parès  [2011-12-31 13:09:37+0100]
...
One thought occur to me recently when explaining the concept of Monad to
non-haskellers: internally, all standard Monads are newtypes wrapping
functions:
StateT is, WriterT is, ContT is. Even IO and ST are, both conceptually
and
in their implementation by GHC.
ParsecT (not part of mtl, but still) is. And so on...
Writer(T) is not a function, it's just a tuple.
Jerzy already mentioned [] and Maybe.
Another example is a free monad generated by any polynomial functor.
This subsumes Maybe and (almost) [], as described here:
http://blog.omega-prime.coRoman.uk/?p=34http://blog.omega-prime.co.uk/?p=34
To summarise, since there are only so many ways to form types in Haskell
(sum, product and exponentiation (functions)), it's no surprise that
functions do occur often, but that's not something fundamental to monads.
--
Roman I. Cheplyaka :: http://ro-che.info/
-- 
The ⊥ is a lie.