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

[Daniel Fischer]

Am Sonntag, 3. September 2006 15:39 schrieb Lennart Augustsson:
> Well, bind is extracting an 'a'.  I clearly see a '\ a -> ...'; it
> getting an 'a' so it can give that to g.  Granted, the extraction is
> very convoluted, but it's there.
>
> 	-- Lennart

But

instance Monad (Cont r) where
	return = flip id
	(>>=) = (. flip) . (.)  
	-- or would you prefer (>>=) = (.) (flip (.) flip) (.) ?

if we write it points-free. No '\a -> ...' around.
And, being more serious, I don't think, bind is extracting an 'a' from m.
How could it? m does not produce a value of type a, like a (State f) does 
(if provided with an initial state), nor does it contain values of type a, 
like [] or Maybe maybe do. And to my eyes it looks rather as though the 
'\a -> ...' tells us that we do _not_ get an 'a' out of m, it specifies to 
which function we will eventually apply m, namely 'flip g k'.
But I've never really understood the Continuation Monad, so if I'm dead wrong,
would you kindly correct me?

And if anybody knows a nontrivial but not too advanced example which could 
help understanding CPS, I'd be glad to hear of it.

Cheers,
Daniel

>
> On Sep 2, 2006, at 19:44 , Udo Stenzel wrote:
> > Benjamin Franksen wrote:
> >> Sure. Your definition of bind (>>=):
> >> ...
> >> applies f to something that it has extracted from m, via
> >> deconstructor
> >> unpack, namely a. Thus, your bind implementation must know how to
> >> produce
> >> an a from its first argument m.
> >
> > I still have no idea what you're driving at, but could you explain how
> > the CPS monad 'extracts' a value from something that's missing
> > something
> > that's missing a value (if that makes sense at all)?
> >
> > For reference (newtype constructor elided for clarity):
> >> type Cont r a = (a -> r) -> r
> >>
> >> instance Monad (Cont r) where
> >> 	return a = \k -> k a
> >> 	m >>= g = \k -> m (\a -> g a k)
> >
> > Udo.
> > --
> > Streitigkeiten dauerten nie lange, wenn nur eine Seite Unrecht hätte.
> > 	-- de la Rochefoucauld
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