有名な論文Idioms are oblivious, arrows are meticulous, monads are promiscuousによると、arrows の表現力 (追加の型クラスなし) は厳密に applicative functor と monads の間のどこかになければなりませArrowApply
んApplicative
:紙は「静的矢印」と呼んでいます。ただし、この「静的」性が意味する制限が何であるかは明確ではありません。
問題の 3 つの型クラスをいじってみると、Applicative Functor と Arrow の間の同等性を構築することができましMonad
たArrowApply
。この構築は正しいですか?(矢の法則のほとんどは、飽きる前に証明済みです)。Arrow
とApplicative
はまったく同じということではないでしょうか。
{-# LANGUAGE TupleSections, NoImplicitPrelude #-}
import Prelude (($), const, uncurry)
-- In the red corner, we have arrows, from the land of * -> * -> *
import Control.Category
import Control.Arrow hiding (Kleisli)
-- In the blue corner, we have applicative functors and monads,
-- the pride of * -> *
import Control.Applicative
import Control.Monad
-- Recall the well-known result that every monad yields an ArrowApply:
newtype Kleisli m a b = Kleisli{ runKleisli :: a -> m b}
instance (Monad m) => Category (Kleisli m) where
id = Kleisli return
Kleisli g . Kleisli f = Kleisli $ g <=< f
instance (Monad m) => Arrow (Kleisli m) where
arr = Kleisli . (return .)
first (Kleisli f) = Kleisli $ \(x, y) -> liftM (,y) (f x)
instance (Monad m) => ArrowApply (Kleisli m) where
app = Kleisli $ \(Kleisli f, x) -> f x
-- Every arrow arr can be turned into an applicative functor
-- for any choice of origin o
newtype Arrplicative arr o a = Arrplicative{ runArrplicative :: arr o a }
instance (Arrow arr) => Functor (Arrplicative arr o) where
fmap f = Arrplicative . (arr f .) . runArrplicative
instance (Arrow arr) => Applicative (Arrplicative arr o) where
pure = Arrplicative . arr . const
Arrplicative af <*> Arrplicative ax = Arrplicative $
arr (uncurry ($)) . (af &&& ax)
-- Arrplicatives over ArrowApply are monads, even
instance (ArrowApply arr) => Monad (Arrplicative arr o) where
return = pure
Arrplicative ax >>= f =
Arrplicative $ (ax >>> arr (runArrplicative . f)) &&& id >>> app
-- Every applicative functor f can be turned into an arrow??
newtype Applicarrow f a b = Applicarrow{ runApplicarrow :: f (a -> b) }
instance (Applicative f) => Category (Applicarrow f) where
id = Applicarrow $ pure id
Applicarrow g . Applicarrow f = Applicarrow $ (.) <$> g <*> f
instance (Applicative f) => Arrow (Applicarrow f) where
arr = Applicarrow . pure
first (Applicarrow f) = Applicarrow $ first <$> f