[Haskell-cafe] Re: what is inverse of mzero and return?

[Daniel Fischer]

Am Samstag, 22. Januar 2005 10:09 schrieb Ashley Yakeley:
> In article <41F211FB.3030706 at imperial.ac.uk>,
>
>  Keean Schupke <k.schupke at imperial.ac.uk> wrote:
> > This fits the above description, but I don't see how the following can
> > be true:
> >
> >     (mplus a b) >>= c = mplus (a >>= c) (b >>= c)
>
> Try it (and my test code) with [], which is an instance of MonadPlus.
> mplus is defined as (++) for [].

Well, mathematically, a MonadPlus m is a functor into the category of monoids.
With the appropriate options, we can write

instance Monad m => Functor m where
   fmap = liftM

-- so we see that any Monad is a functor (if >>= is properly defined) --

instance MonadPlus m => Monoid (m a) where
   mempty = mzero
   mappend = mplus

and feed that into ghc or hugs. Only the monoid Maybe a is not very nice (nor 
is the monoid IO a),since the second argument of the composition is in 
general ignored. Actually we can make every nonempty set a monoid in this 
way, choose an element x1 and define the composition (#) by
   x1 # x = x,
   y   # z = y, if y /= x1.
Now the second argument to (>>=) is an arbitrary function into some monoid and 
there is no reason why (>>= c) should be a morphism in the category of 
monoids -- for the trivial monoids just defined, it would naturally be simply 
c itself. It is a nice feature of lists that (>>= c) is a homomorphism there, 
but that is only so because (>>=) is appropriately defined.
However, (>> k) is, if I see it correctly, a monoid-homomorphism in all these 
cases  -- though somewhat boring.

So I think, rather than separating mplus, one should think about whether it is 
sensible to make Maybe and IO instances of MonadPlus in the first place.
I don't know nearly enough of the innards of Haskell to form a valuable 
opinion of that, but perhaps somebody could enlighten me?

Regards,
Daniel