Re: [Haskell-cafe] Why Kleisli composition is not in the Monad signature?

24 Oct 2012

      As a side note, I think a direct superclass of Functor for Monad is not
a good idea, just sayin'

class Functor f where
  fmap :: (a -> b) -> f a -> f b

class Functor f => Apply f where
  (<*>) :: f (a -> b) -> f a -> f b

class Apply f => Bind f where
  (=<<) :: (a -> f b) -> f a -> f b

class Apply f => Applicative f where
  unit :: a -> f a

class (Applicative f, Bind f) => Monad f where

Same goes for Comonad (e.g. [] has (=<<) but not counit)
... and again for Monoid, Category, I could go on...

On 17/10/12 04:14, David Thomas wrote:
...
I think the version below (with a Functor or Applicative superclass)
is clearly the right answer if we were putting the prelude together
from a clean slate.  You can implement whichever is easiest for the
particular monad, use whichever is most appropriate to the context
(and add optimized versions if you prove to need them).  I see no
advantage in any other specific proposal, except the enormous
advantage held by the status quo that it is already written, already
documented, already distributed, and already coded to.
Regarding mathematical "purity" of the solutions, "this is in every
way isomorphic to a monad, but we aren't calling it a monad because we
are describing it a little differently than the most common way to
describe a monad in category theory" strikes me as *less*
mathematically grounded than "we are calling this a monad because it
is in every way isomorphic to a monad."
On Tue, Oct 16, 2012 at 7:03 AM, AUGER Cédric  wrote:
...
So I think that an implicit question was why wouldn't we have:
class Monad m where
  return :: a -> m a
  kleisli :: (a -> m b) -> (b -> m c) -> (a -> m c)
  bind = \ x f -> ((const x) >=> f) ()
  join = id>=>id :: (m (m a) -> m a)
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Tony Morris
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