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

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)

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