[Haskell-cafe] Re: Re: A free monad theorem?
Lennart Augustsson
lennart at augustsson.net
Sun Sep 3 09:39:22 EDT 2006
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
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.
> --
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> -- de la Rochefoucauld
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