[Haskell-cafe] Re: Re: A free monad theorem?

[Andrea Rossato]

Il Sun, Sep 03, 2006 at 12:26:25AM +0200, Tomasz Zielonka ebbe a scrivere:
> On Sat, Sep 02, 2006 at 09:51:26PM +0200, Benjamin Franksen wrote:
> > However, in order to 'run' (i.e. finally actually use) a monadic value
> > that involves an application of bind, the latter would have to supply
> > some argument to its second argument (which is a function),
> 
> If I didn't see above, only the following text...
> 
> > so there must be some way to 'get an a out of' the first argument m.
> 
> I would answer: Yes, there is a method in Monad which has exactly this
> purpose: >>=
> 

I tried to describe this stuff in the link below, the best I could.

The text is still in part messing but you should be able to understand
where I'm headed just from the code...

http://www.haskell.org/haskellwiki/Meet_Bob_The_Monadic_Lover

andrea