That makes sense. But why does
instance Monad m => ArrowApply (Kleisli m)
show that a Monad can do anything an ArrowApply can (and the two are
thus equivalent)?
On Tue, May 28, 2013 at 5:17 PM, Tom Ellis
On Tue, May 28, 2013 at 04:42:35PM +0200, Johannes Gerer wrote:
By the same argument, could'nt I say, that any type class (call it AnyClass) can do everything a Monad can:
instance AnyClass m => Monad (Cokleilsi m ())
That doesn't say that AnyClass can do anything a Monad can. "AnyClass m => Monad m" would say that, but that's not what you've got.
What you've got is that "Cokleisli m ()" i.e. "(->) m ()" is a Monad for any "m". This is not surprising. The implementation is the same as the Reader monad.
Check out the instance implementations for "Monad (Reader r)" and "Monad (CoKleisli w a)". You will find they are the same.
http://hackage.haskell.org/packages/archive/mtl/1.1.0.2/doc/html/src/Control... http://hackage.haskell.org/packages/archive/comonad/3.0.0.2/doc/html/src/Con...
Tom
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