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

Andrea Rossato mailing_list at istitutocolli.org
Sat Sep 2 18:55:23 EDT 2006

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...



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