Everyone knows State
is a monad:
import Control.Monad
newtype State s a = State {runState :: s -> (a, s)}
instance Functor (State s) where
fmap = liftM
instance Applicative (State s) where
pure x = State (\s -> (x, s))
(<*>) = ap
instance Monad (State s) where
State f >>= k = State $ \s -> let
(x, s2) = f s
State g = k x
in g s2
But is it an arrow as well? Here's my attempt of implementing instance Arrow State
:
import Control.Arrow
import Control.Category
instance Category State where
id = State (\s -> (s, s))
State f . State g = State $ \x -> let
(y, x2) = g x
(z, _) = f y
in (z, x2)
instance Arrow State where
arr f = State (\s -> (f s, s))
State f *** State g = State $ \(s, t) -> let
(x1, s1) = f s
(y1, t1) = g t
in ((x1, y1), (s1, t1))
instance ArrowChoice State where
State f +++ State g = State $ \s -> case s of
Left s2 -> let
(x, s3) = f s2
in (Left x, Left s3)
Right s2 -> let
(y, s3) = g s2
in (Right y, Right s3)
instance ArrowApply State where
app = State (\k@(State f, s) -> (fst (f s), k))
instance ArrowLoop State where
loop (State f) = State $ \s -> let
((x, d), (s2, _)) = f (s, d)
in (x, s2)
I've confirmed that the Category
, Arrow
, ArrowChoice
and ArrowApply
instances are correct, but I'm unsure about the ArrowLoop
instance. And there is no QuickCheck for arrow-related classes. Are these instances really correct?
The category instance is not correct. It fails the right identity law:
f . id = f
But, in your implementation of .
and id
,
(f . id) s ==
(fst (f s) , s) /=
f s