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 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
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