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

[Jules Bean]

On 25 Jan 2005, at 11:49, Keean Schupke wrote:

> Jules Bean wrote:
>
>> A monad T is a (endo)functor T : * -> * where * is the category of 
>> types, together with a multiplication mu and a unit eta.
>
> So, * is the category of Types, and functions on type (which map 
> values to values), and T is
> an endofunctor (mapping functions on type to functions on type).

T is an endofunctor (mapping functions on types to functions on types 
*and* types to types) : functors map not only morphisms but also 
objects.

>
> How does this affect the IO monad though?
>
>    m >>= (\a -> mzero) === mzero
>
> If we consider the state monad, surely the above makes no comment on 
> what
> the final state should be, only the final value returned...
>
> Or is MonadPlus not definable on State monads?

I don't know. The reason I don't know, is I can't actually find written 
down the laws that MonadPlus is 'supposed' to obey. I agree with the OP 
(ashley, IIRC) that mzero and the IO monad behave in a surprising way.

I would think that the closest I could get to a 'sane' definition of 
MonadPlus for State Monads is something which for mzero goes into a 
'special state' representing exception (with undefined for a value, I 
suppose), and for mplus tries the left branch, if that goes into the 
exception state then tries the right branch.

> If it is then considering IO === ST RealWorld, would imply that the 
> actions
> of the IO monad are not important as long as the final value returned 
> is
> mzero?

Well, mzero isn't a return value in the IO monad, it's an exception.  
But yes, I agree with you that the (plausible) laws I have seen for 
MonadPlus seem to say that mzero should ignore the actions. But this in 
practice is not how IO behaves.

Jules