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. -- 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