[Haskell-cafe] Re: Re: A free monad theorem?
u.stenzel at web.de
Sat Sep 2 19:44:25 EDT 2006
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)
Streitigkeiten dauerten nie lange, wenn nur eine Seite Unrecht hätte.
-- de la Rochefoucauld
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