Re: [Haskell-beginners] State Monad vs. fold

...

...

...

...

martin writes:

Can someone give me some insights when the State Monad is beneficial and where a fold is the better choice.

Looking at the type of a fold: foldr :: (a -> b -> b) -> b -> [a] -> b If we juggle the arguments we get: foldr :: (a -> b -> b) -> [a] -> b -> b And if we imagine State b () actions, we can directly rewrite this as: foldrS :: (a -> State b ()) -> [a] -> State b () Which generalizes to: foldrS :: MonadState b m => (a -> m ()) -> [a] -> m () Which is roughly the same thing as using mapM_ over our State monad: mapM_ :: Monad m => (a -> m b) -> [a] -> m () In other words, these two forms in our example say the same thing: foldr f b xs execState (mapM_ f' xs) b With the only difference being the types of f and f': f : a -> b -> b f' : a -> State b () The other question you asked is when to choose one over the other. Since they are equivalent, it's really up to you. I tend to prefer using a fold over State, to keep things on the level of functions and values, rather than pulling in monads and possibly monad transformers unnecessarily. But it's a very good thing to hold these isomorphisms in your mind, since you can then freely switch from one representation to another as need be. This is true of a lot of type equivalences throughout Haskell, where the more things you can see as being essentially the same, the more freedom you have to find the best abstraction for a particular context. John