... и педантично выровнял:
($) :: (a -> b) -> a -> b
(<$>) :: Functor f => (a -> b) -> f a -> f b
(<*>) :: Applicative f => f (a -> b) -> f a -> f b
(=<<) :: Monad m => (a -> m b) -> m a -> m b
(&) :: a -> (a -> b) -> b -- Data.Function
(<&>) :: Functor f => f a
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