[Haskell-cafe] Retrieving inner state from outside the
transformer
Ryan Ingram
ryani.spam at gmail.com
Sat Aug 1 15:17:13 EDT 2009
I am not sure I entirely understand your question; it sounds like you
are confused and thus your question is a bit confused. So instead,
I'll explain in a bit more detail.
A common pattern in Haskell is that you have a type that you want to
perform some operations on, and then afterwards you "observe" the type
to convert it to some simpler type that no longer has those
operations.
To see why the "observation" is important, consider this type:
> newtype StupidT s m a = StupidT ()
Would you believe that StupidT is a state monad transformer?
> instance Monad StupidT s m where
> return _ = StupidT ()
> StupidT () >>= f = StupidT ()
> instance MonadState s (StupidT s m) where
> get = StupidT ()
> put _ = StupidT ()
> instance MonadTrans (StupidT s) where
> lift _ = StupidT ()
StupidT follows all the laws for these typeclasses, because they all
have to do with observational equality of different sets of
operations. Since all operations are equal, the laws all trivially
hold. For example:
Proof: get >>= put = return () (this is one of the laws every
MonadState needs to fulfill)
get >>= put
apply (>>=)
= StupidT ()
unapply return
= return ()
All of the other laws are proved similarily.
Obviously, you can't do much with StupidT; the thing that makes StateT
useful is its observation function "runStateT"
runStateT :: StateT s m a -> (s -> m (a,s))
which converts from (StateT s m a), into (s -> m (a,s)).
You'll often find that the most elegant implementation of a type in
Haskell is to use the observation function's return type as the
representation type of the object! So since we want runStateT with
this particular type, we just make StateT hold that type:
> newtype StateT s m a = StateT { runStateT :: s -> m (a,s) }
Now, of course, we have to implement all the operations we care about
(return, (>>=), get, put, lift), such that they obey the laws those
operations are supposed to fulfill, but implementing the observation
function is trivial!
So, how does the plumbing convert the "s" you give it into a "m (a,s)"
result? Simple, there is no plumbing. You just call the function!
Of course, there is some plumbing in the implementation of the
operations on the transformer:
> instance Monad m => Monad (StateT s m) where
> return a = StateT $ \s -> return (a,s)
> m >>= f = StateT $ \s0 -> do
> (a, s1) <- runStateT m s0
> (b, s2) <- runStateT (f a) s1
> return (b,s2)
>
> instance Monad m => MonadState s (StateT s m) where
> get = StateT $ \s -> return (s,s)
> put s = StateT $ \_ -> return ((), s)
>
> instance MonadTrans (StateT s) where
> lift m = StateT $ \s -> do
> a <- m
> return (a,s)
You should notice that all this makes runStateT "just work"; there's
no need to call additional code to get the (s -> m (a,s)) back out of
the StateT. You should also notice that these are the *only*
type-correct, non-_|_-using implementations for most of the functions;
the only places where we could make a semantic error that wasn't also
a type error are putting the wrong states through (put) and (>>=). By
choosing a representation that matches the observable value we want,
implementing the operations becomes much simpler!
-- ryan
On Sat, Aug 1, 2009 at 11:06 AM, <phil at beadling.co.uk> wrote:
> Thanks very much for both replies.
> I think I get this now.
> Simply, my choice of evaluation functions (evalStateT, execStateT and
> execState) ensured that the states are not returned. It was obvious.
> I can get this working, but I have one more more question to make sure I
> actually understand this.
> Below is a very simple and pointless example I wrote to grasp the
> concept. This returns ((1,23),21) which is clear to me.
> import Control.Monad.State
> myOuter :: StateT Int (State Int) Int
> myOuter = StateT $ \s -> do p <- get
> return (s,p+s+1)
> main :: IO()
> main = do let innerMonad = runStateT myOuter 1
> y = runState innerMonad 21
> print y
> Thus we are saying that a=(1,23) and s=21 for the state monad, and that a=1
> and s=23 for the state
> transformer. That is the return value of the state monad is the (a,s) tuple of the transformer and it's own state is of course 21.
> This got me thinking - the return value's type of the state monad is dictated by the evaluation function used
> on the state transformer - it could be a, s, or (a,s) depending which function is used. Thus if I edit the code to to:
> do let innerMonad = evalStateT myOuter 1
> I get back (1,21) - which is the problem I had - we've lost the
> transformer's state.
> Look at the Haskell docs I get:
> evalStateT :: Monad m => StateT s m a -> s -> m a
> runStateT :: s -> m (a, s)
> So the transformer valuation functions are returning a State monad
> initialized with either a or (a,s).
> Now I know from messing around with this that the initialization is the
> return value, from the constructor:
> newtype State s a = State {
> runState :: s -> (a, s)
> }
> Am I right in assuming that I can read this as:
> m (a,s_outer) returned from runStateT is equivalent to calling the
> constructor as (State s_inner) (a,s_outer)
> This makes sense because in the definition of myOuter we don't specify the
> return value type of the inner monad:
> myOuter :: StateT Int (State Int) Int
>
> The problem is whilst I can see that we've defined the inner monad's return
> value to equal the *type* of the transformer's evaluation function, I'm
> loosing the plot trying to see how the *values* returned by the transformer
> are ending up there. We haven't specified what the state monad actually
> does?
> If I look at a very simple example:
> simple :: State Int Int
> simple = State $ \s -> (s,s+1)
> This is blindly obvious, is I call 'runState simple 8', I will get back
> (8,9). Because I've specified that the return value is just the state.
> In the more original example, I can see that the 'return (s,p+s+1)' must
> produce a state monad where a=(1,23), and the state of this monad is just
> hardcoded in the code = 21.
> I guess what I'm trying to say is - where is the plumbing that ensures that
> this returned value in the state/transformer stack is just the (a,s) of the
> transformer?
>
> I have a terrible feeling this is a blindly obvious question - apologies if
> it is!
>
> Thanks again!
>
> Phil.
>
>
> On 31 Jul 2009, at 04:39, Ryan Ingram wrote:
>
> StateT is really simple, so you should be able to figure it out:
>
> runStateT :: StateT s m a -> s -> m (a,s)
> runState :: State s a -> s -> (a,s)
>
> So if you have
> m :: StateT s1 (StateT s2 (State s3)) a
>
> runStateT m :: s1 -> StateT s2 (State s3) (a,s)
>
> \s1 s2 s3 -> runState (runStateT (runStateT m s1) s2) s3)
> :: s1 -> s2 -> s3 -> (((a,s1), s2), s3)
>
>
>
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