Implement State Monad transformer in Haskell from scratch

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When I was studying Monad Transformer, I decided to create StateT s m a from scratch with instances for Functor, Applicative and Monad.



This is what I have:



newtype StateT s m a = StateT { runStateT :: (s -> m (a, s)) }

instance Functor m => Functor (StateT s m) where
-- fmap :: (a -> b) -> StateT s m a -> StateT s m b
-- which is (a -> b) -> (s -> m (a, s)) -> (s -> m (b, s))
f `fmap` (StateT x) = StateT $ s -> fmap run (x s)
where run (a, s) = (f a, s)

instance Monad m => Applicative (StateT s m) where
-- pure :: a -> StateT s m a
pure a = StateT $ s -> pure (a, s)
-- <*> :: f (a -> b) -> f a -> f b
-- which is StateT s m (a -> b) -> StateT s m a -> State s m b
k <*> x = StateT $ s -> do
(f, s1) <- runStateT k s -- :: m ((a -> b), s)
(a, s2) <- runStateT x s1
return (f a, s2)

instance (Monad m) => Monad (StateT s m) where
return a = StateT $ s -> return (a, s)
-- >>= :: StateT s m a -> (a -> StateT s m b) -> StateT s m b
(StateT x) >>= f = StateT $ s -> do
(v, s') <- x s
runStateT (f v) s'


My original intention is to implement Functor (StateT s m) with Functor m restriction, Applicative (StateT s m) with Applicative m restriction, and Monad (StateT s m) withMonad m) restriction. However I couldn't do the Applicative case and had to use Monad m restriction instead. Is there a way to do it with Applicative m?



Thank you in advance.









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    up vote
    0
    down vote

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    When I was studying Monad Transformer, I decided to create StateT s m a from scratch with instances for Functor, Applicative and Monad.



    This is what I have:



    newtype StateT s m a = StateT { runStateT :: (s -> m (a, s)) }

    instance Functor m => Functor (StateT s m) where
    -- fmap :: (a -> b) -> StateT s m a -> StateT s m b
    -- which is (a -> b) -> (s -> m (a, s)) -> (s -> m (b, s))
    f `fmap` (StateT x) = StateT $ s -> fmap run (x s)
    where run (a, s) = (f a, s)

    instance Monad m => Applicative (StateT s m) where
    -- pure :: a -> StateT s m a
    pure a = StateT $ s -> pure (a, s)
    -- <*> :: f (a -> b) -> f a -> f b
    -- which is StateT s m (a -> b) -> StateT s m a -> State s m b
    k <*> x = StateT $ s -> do
    (f, s1) <- runStateT k s -- :: m ((a -> b), s)
    (a, s2) <- runStateT x s1
    return (f a, s2)

    instance (Monad m) => Monad (StateT s m) where
    return a = StateT $ s -> return (a, s)
    -- >>= :: StateT s m a -> (a -> StateT s m b) -> StateT s m b
    (StateT x) >>= f = StateT $ s -> do
    (v, s') <- x s
    runStateT (f v) s'


    My original intention is to implement Functor (StateT s m) with Functor m restriction, Applicative (StateT s m) with Applicative m restriction, and Monad (StateT s m) withMonad m) restriction. However I couldn't do the Applicative case and had to use Monad m restriction instead. Is there a way to do it with Applicative m?



    Thank you in advance.









    share
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      When I was studying Monad Transformer, I decided to create StateT s m a from scratch with instances for Functor, Applicative and Monad.



      This is what I have:



      newtype StateT s m a = StateT { runStateT :: (s -> m (a, s)) }

      instance Functor m => Functor (StateT s m) where
      -- fmap :: (a -> b) -> StateT s m a -> StateT s m b
      -- which is (a -> b) -> (s -> m (a, s)) -> (s -> m (b, s))
      f `fmap` (StateT x) = StateT $ s -> fmap run (x s)
      where run (a, s) = (f a, s)

      instance Monad m => Applicative (StateT s m) where
      -- pure :: a -> StateT s m a
      pure a = StateT $ s -> pure (a, s)
      -- <*> :: f (a -> b) -> f a -> f b
      -- which is StateT s m (a -> b) -> StateT s m a -> State s m b
      k <*> x = StateT $ s -> do
      (f, s1) <- runStateT k s -- :: m ((a -> b), s)
      (a, s2) <- runStateT x s1
      return (f a, s2)

      instance (Monad m) => Monad (StateT s m) where
      return a = StateT $ s -> return (a, s)
      -- >>= :: StateT s m a -> (a -> StateT s m b) -> StateT s m b
      (StateT x) >>= f = StateT $ s -> do
      (v, s') <- x s
      runStateT (f v) s'


      My original intention is to implement Functor (StateT s m) with Functor m restriction, Applicative (StateT s m) with Applicative m restriction, and Monad (StateT s m) withMonad m) restriction. However I couldn't do the Applicative case and had to use Monad m restriction instead. Is there a way to do it with Applicative m?



      Thank you in advance.









      share













      When I was studying Monad Transformer, I decided to create StateT s m a from scratch with instances for Functor, Applicative and Monad.



      This is what I have:



      newtype StateT s m a = StateT { runStateT :: (s -> m (a, s)) }

      instance Functor m => Functor (StateT s m) where
      -- fmap :: (a -> b) -> StateT s m a -> StateT s m b
      -- which is (a -> b) -> (s -> m (a, s)) -> (s -> m (b, s))
      f `fmap` (StateT x) = StateT $ s -> fmap run (x s)
      where run (a, s) = (f a, s)

      instance Monad m => Applicative (StateT s m) where
      -- pure :: a -> StateT s m a
      pure a = StateT $ s -> pure (a, s)
      -- <*> :: f (a -> b) -> f a -> f b
      -- which is StateT s m (a -> b) -> StateT s m a -> State s m b
      k <*> x = StateT $ s -> do
      (f, s1) <- runStateT k s -- :: m ((a -> b), s)
      (a, s2) <- runStateT x s1
      return (f a, s2)

      instance (Monad m) => Monad (StateT s m) where
      return a = StateT $ s -> return (a, s)
      -- >>= :: StateT s m a -> (a -> StateT s m b) -> StateT s m b
      (StateT x) >>= f = StateT $ s -> do
      (v, s') <- x s
      runStateT (f v) s'


      My original intention is to implement Functor (StateT s m) with Functor m restriction, Applicative (StateT s m) with Applicative m restriction, and Monad (StateT s m) withMonad m) restriction. However I couldn't do the Applicative case and had to use Monad m restriction instead. Is there a way to do it with Applicative m?



      Thank you in advance.







      haskell monads





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