This is the State Monad code I am trying to figure out
data State a = State (Int -> (a, Int))
instance Monad State where
return x = State (\c -> (x, c))
State m >>= f = State (\c ->
case m c of { (a, acount) ->
case f a of State b -> b acount})
getState = State (\c -> (c, c))
putState count = State (\_ -> ((), count))
instance Show State where -- doesn't work
show (State a) = show a -- doesn't work
I am trying to make State as instance of Show so that I can see the action of getState and putState count on the ghci prompt.
Any tutorial or link to State Monad material would be nice too.
Here's is a Show instance that can help see what's going on:
instance Show a => Show (State a) where
show (State f) = show [show i ++ " => " ++ show (f i) | i <- [0..3]]
Then you can do:
*Main> getState
["0 => (0,0)","1 => (1,1)","2 => (2,2)","3 => (3,3)"]
*Main> putState 1
["0 => ((),1)","1 => ((),1)","2 => ((),1)","3 => ((),1)"]
In Haskell, typeclasses classify only types of the same kind. Monad classifies types of kind * -> *, while Show classifies types of kind *. Your State type has kind * -> * which is why there isn't a problem with your Monad instance, but there is a problem with your Show instance. State is called a "type constructor" because it consumes a type in order to produce another type. Think of it sort of like function application at the type level. You can therefore apply a specific type and make instances of that:
instance Show (State Char) where
show _ = "<< Some State >>"
Now, this isn't a very useful instance, try something like Sjoerd's suggestion to get a more meaningful Show instance. Notice that his version uses a generic type with a constraint.
instance (Show a) => Show (State a) where
-- blah blah
The generic type is a, and the constraint is (Show a) =>, in other words, a itself must be an instance of Show.
Here's a great (and personally, my favorite) explanation of the State Monad: Learn You A Haskell. (Also a great resource for learning Haskell in general).
You may have noticed that functions aren't part of the Show typeclass in Haskell. And since State is basically just a newtype wrapper for functions of certain types, you can't make a (meaningful) State instance of Show.
Here's code using the State Monad from LYAH:
import Control.Monad.State -- first, import the state monad
pop :: State Stack Int
pop = State $ \(x:xs) -> (x,xs)
push :: Int -> State Stack ()
push a = State $ \xs -> ((),a:xs)
stackManip :: State Stack Int
stackManip = do
push 3
a <- pop
pop
And here's that code in action from ghci:
*Main> runState stackManip [1,2,3]
(1,[2,3])
The fst of the tuple is the result, and the snd of the tuple is the (modified) state.
Related
I have a situation where I want to interleave a free monadic DSL with state, but interpret the state mid-program. Below is a simplified example with only logging and state. More generally the problem is that we need to perform some effects to obtain the initial state, so it can't be done during the final interpretation (runProgram below). Using tagless final and StateT this is simple. The same is probably true using an extensible effect system like polysemy. I couldn't figure out how to do this using coproducts from Data types à la carte while only interpreting part of the union. Perhaps there's an elegant way using coproducts.
I decided to try using FreeT after seeing similar examples here in Haskell and here in Scala. However, the situation is slightly different: the base monad is their custom DSL, rather than some state monad. As far as I can tell, I need state to be the base in order to be able to “eliminate” it in the middle of the program.
Here’s my broken attempt (full example in this gist):
runProgram :: IO ()
runProgram = iterM interpret program
data Log next = Log String next | Noop next deriving Functor
-- Might be the problem, applicative instance using dummy constructor needed for iterT
instance Applicative Log where pure = Noop
type Program a = FreeT (State Int) (Free Log) a
myLog :: String -> Free Log ()
myLog s = liftF $ Log s ()
interpret :: Log (IO a) -> IO a
interpret = \case
Log s next -> putStrLn s >> next
Noop next -> next
-- Program with only logging, state interpreted in the middle
program :: Free Log ()
program = do
myLog "Before stateful program"
finalState <- eliminateState 3 statefulProgram
myLog $ "Final state: " ++ show finalState
eliminateState :: Int -> Program Int -> Free Log Int
eliminateState initialState = iterT (`evalState` initialState)
-- Combines logging and state, but the state doesn't stick. Why?
statefulProgram :: Program Int
statefulProgram = do
lift $ myLog "In stateful program, modifying state"
liftF $ modify (+1)
updatedState <- liftF get
lift $ myLog ("In stateful program, updated state is: " ++ show updatedState)
return updatedState
The state doesn't stick here, output is:
Before stateful program
In stateful program, modifying state
In stateful program, updated state is: 3
Final state: 3
I suspect there's something fundamentally not sound with this approach and that the bodged together DSL Applicative instance is a symptom of that. But I'm not sure what it is. So two questions:
What is the problem with this particular implementation?
How can this situation be modeled using free monads?
As for
What is the problem with this particular implementation?
The FreeT monad transformer is like an onion, with multiple nested layers of the functor parameter (here, State Int).
The function that is passed to iterT is used to "peel" the functorial layers. The problem is that there's no communication between the successive State Int layers. (`evalState` initialState) is used to run each layer separately, always with the same initial state, always discarding the final state for the layer, as evalState is wont to do.
Answering the narrow question (from your comment) of how to write the handler for StateF. (I’m on my phone, so I’ve compiled this code only in my head, but it should give you the general idea.)
data StateF s a = Get (s -> a) | Put s a deriving Functor
data (f :+: g) a = L (f a) | R (g a) deriving Functor
runState :: Functor f => s -> Free (StateF :+: f) a -> Free f (a, s)
runState s (Pure x) = Pure (x, s)
runState s (Free (L (Get k))) = runState s (k s)
runState _ (Free (L (Put s m))) = runState s m
runState s (Free (R f)) = Free (runState s <$> f)
In the R case, you want to leave this layer of the free monad intact — it’s some unknown non-StateF effect. So we just recursively run the rest of the computation inside the f functor using <$>.
You could probably write this code using iter(M), if you don’t like the explicit recursion, but I think you’d need to use StateT to do that.
So, I have this datatype (it's from here: https://wiki.haskell.org/IO_Semantics):
data IO a = Done a
| PutChar Char (IO a)
| GetChar (Char -> IO a)
and I thought of writing a StateMonad instance for it. I have already written Monad and Applicative instances for it.
instance MonadState (IO s) where
get = GetChar (\c -> Done c)
put = PutChar c (Done ())
state f = Done (f s)
I don't think I fully understand what state (it was named 'modify' before) is supposed to do here.
state :: (s -> (a, s)) -> m a
I also have messed up with declarations. I don't really understand what is wrong, let alone how to fix it. Would appreciate your help.
Expecting one more argument to ‘MonadState (IO s)’
Expected a constraint,
but ‘MonadState (IO s)’ has kind ‘(* -> *) -> Constraint’
In the instance declaration for ‘MonadState (IO s)’
As I mentioned in the comments, your type doesn't really hold any state, so a StateMonad instance would be nonsensical for it.
However, since this is just an exercise (also based on the comments), I guess it's ok to implement the instance technically, even if it doesn't do what you'd expect it to do.
First, the compiler error you're getting tells you that the MonadState class actually takes two arguments - the type of the state and the type of the monad, where monad has to have kind * -> *, that is, have a type parameter, like Maybe, or list, or Identity.
In your case, the monad in question (not really a monad, but ok) is IO, and the type of your "state" is Char, since that's what you're getting and putting. So the declaration has to look like this:
instance MonadState Char IO where
Second, the state method doesn't have the signature (s -> s) -> m s as you claim, but rather (s -> (a, s)) -> m a. See its definition. What it's supposed to do is create a computation in monad m out of a function that takes a state and returns "result" plus new (updated) state.
Note also that this is the most general operation on the State monad, and both get and put can be expressed in terms of state:
get = state $ \s -> (s, s)
put s = state $ \_ -> ((), s)
This means that you do not have to implement get and put yourself. You only need to implement the state function, and get/put will come from default implementations.
Incidentally, this works the other way around as well: if you define get and put, the definition of state will come from the default:
state f = do
s <- get
let (a, s') = f s
put s'
return a
And now, let's see how this can actually be implemented.
The semantics of the state's parameter is this: it's a function that takes some state as input, then performs some computation based on that state, and this computation has some result a, and it also may modify the state in some way; so the function returns both the result and the new, modified state.
In your case, the way to "get" the state from your "monad" is via GetChar, and the way in which is "returns" the Char is by calling a function that you pass to it (such function is usually referred to as "continuation").
The way to "put" the state back into your "monad" is via PutChar, which takes the Char you want to "put" as a parameter, plus some IO a that represents the computation "result".
So, the way to implement state would be to (1) first "get" the Char, then (2) apply the function to it, then (3) "put" the resulting new Char, and then (3) return the "result" of the function. Putting it all together:
state f = GetChar $ \c -> let (a, c') = f c in PutChar c' (Done a)
As further exercise, I encourage you to see how get and put would unfold, starting from the definitions I gave above, and performing step-by-step substitution. Here, I'll give you a couple first steps:
get = state $ \s -> (s, s)
-- Substituting definition of `state`
= GetChar $ \c -> let (a, c') = (\s -> (s, s)) c in PutChar c' (Done a)
-- Substituting (\s -> (s, s)) c == (c, c)
= GetChar $ \c -> let (a, c') = (c, c) in PutChar c' (Done a)
= <and so on...>
MonadState takes two arguments, in this order:
the type of the state
the monad
In this case the monad is IO:
instance MonadState _ IO where
...
And you need to figure out what goes in place of the underscore
I'm quite new to Haskell and want to use makeLenses from Control.Lens and class constraints together with type synonyms to make my functions types more compact (readable?).
I've tried to come up with a minimal dummy example to demonstrate what I want to achieve and the example serves no other purpose than this.
At the end of this post I've added an example closer to my original problem if you are interested in the context.
Minimal example
As an example, say I define the following data type:
data State a = State { _a :: a
} deriving Show
, for which I also make lenses:
makeLenses ''State
In order to enforce a class constraint on the type parameter a used by the type constructor State I use a smart constructor:
mkState :: (Num a) => a -> State a
mkState n = State {_a = n}
Next, say I have a number of functions with type signatures similar to this:
doStuff :: Num a => State a -> State a
doStuff s = s & a %~ (*2)
This all works as intended, for example:
test = doStuff . mkState $ 5.5 -- results in State {_a = 11.0}
Problem
I've tried to use the following type synonym:
type S = (Num n) => State n -- Requires the RankNTypes extensions
, together with:
{-# LANGUAGE RankNTypes #-}
, in an attempt to simplify the type signature of doStuff:
doStuff :: S -> S
, but this gives the following error:
Couldn't match type `State a0' with `forall n. Num n => State n'
Expected type: a0 -> S
Actual type: a0 -> State a0
In the second argument of `(.)', namely `mkState'
In the expression: doStuff . mkState
In the expression: doStuff . mkState $ 5.5
Failed, modules loaded: none.
Question
My current knowledge of Haskell is not sufficient to understand what causes the above error. I hope someone can explain what causes the error and/or suggest other ways to construct the type synonym or why such a type synonym is not possible.
Background
My original problem looks closer to this:
data State r = State { _time :: Int
, _ready :: r
} deriving Show
makeLenses ''State
data Task = Task { ... }
Here I want to enforce the type of _ready being an instance of the Queue class using the following smart constructor:
mkState :: (Queue q) => Int -> q Task -> State (q Task)
mkState t q = State { _time = t
, _ready = q
}
I also have a number of functions with type signatures similar to this:
updateState :: Queue q => State (q Task) -> Task -> State (q Task)
updateState s x = s & ready %~ (enqueue x) & time %~ (+1)
I would like to use a type synonym S to be able to rewrite the type of such functions as:
updateState :: S -> Task -> S
, but as with the first minimal example I don't know how to define the type synonym S or whether it is possible at all.
Maybe there is no real benefit in trying to simplify the type signatures?
Related reading
I've read the following related questions on SO:
Class constraints for data records
Are type synonyms with typeclass constraints possible?
This might also be related but given my current understanding of Haskell I cannot really understand all of it:
Unifying associated type synonyms with class constraints
Follow-up
It's been a while since I've had the opportunity to do some Haskell. Thanks to #bheklilr I've now managed to introduce a type synonym only to hit the next type error I'm still not able to understand. I've posted the following follow-up question Type synonym causes type error regarding the new type error.
You see that error in particular because of the combination of the . operator and your use of RankNTypes. If you change it from
test = doStuff . mkState $ 5.5
to
test = doStuff $ mkState 5.5
or even
test = doStuff (mkState 5.5)
it will compile. Why is this? Look at the types:
doStuff :: forall n. Num n => State n -> State n
mkState :: Num n => n -> State n
(doStuff) . (mkState) <===> (forall n. Num n => State n -> State n) . (Num n => n -> State n)
Hopefully the parentheses help make it clear here, the n from forall n. Num n ... for doStuff is a different type variable from the Num n => ... for mkState because the scope of the forall only extends to the end of the parentheses. So these functions can't actually compose because the compiler sees them as separate types! There are actually special rules for the $ operator specifically for using the ST monad precisely for this reason, just so you can do runST $ do ....
You may be able to accomplish what you want easier using GADTs, but I don't believe lens' TemplateHaskell will work with GADT types. However, you can write your own pretty easily in this case, so it isn't that big of a deal.
A further explanation:
doStuff . mkState $ 5.5
is very different than
doStuff $ mkState 5.5
In the first one, doStuff says that for all Num types n, its type is State n -> State n, whereas mkState says for some Num type m, its type is m -> State m. These two types are not the same because of the "for all" and "for some" quantifications (hence ExistentialQuantification), since composing them would mean that for some Num m you can produce all Num n.
In the doStuff $ mkState 5.5, you have the equivalent of
(forall n. Num n => State n -> State n) $ (Num m => State m)
Notice that the type after the $ is not a function because mkState 5.5 is fully applied. So this works because for all Num n you can do State n -> State n, and you're providing it some Num m => State m. This works intuitively. Again, the difference here is the composition versus application. You can't compose a function that works on some types with a function that works on all types, but you can pass a value to a function that works on all types ("all types" here meaning forall n. Num n => n).
I often end up in a situation where it's very convenient to be using the State monad, due to having a lot of related functions that need to operate on the same piece of data in a semi-imperative way.
Some of the functions need to read the data in the State monad, but will never need to change it. Using the State monad as usual in these functions works just fine, but I can't help but feel that I've given up Haskell's inherent safety and replicated a language where any function can mutate anything.
Is there some type-level thing that I can do to ensure that these functions can only read from the State, and never write to it?
Current situation:
iWriteData :: Int -> State MyState ()
iWriteData n = do
state <- get
put (doSomething n state)
-- Ideally this type would show that the state can't change.
iReadData :: State MyState Int
iReadData = do
state <- get
return (getPieceOf state)
bigFunction :: State MyState ()
bigFunction = do
iWriteData 5
iWriteData 10
num <- iReadData -- How do we know that the state wasn't modified?
iWRiteData num
Ideally iReadData would probably have the type Reader MyState Int, but then it doesn't play nicely with the State. Having iReadData be a regular function seems to be the best bet, but then I have to go through the gymnastics of explicitly extracting and passing it the state every time it's used. What are my options?
It's not hard to inject the Reader monad into State:
read :: Reader s a -> State s a
read a = gets (runReader a)
then you could say
iReadData :: Reader MyState Int
iReadData = do
state <- ask
return (getPieceOf state)
and call it as
x <- read $ iReadData
this would allow you to build up Readers into larger read-only sub-programs and inject them into State only where you need to combine them with mutators.
It's not hard to extend this to a ReaderT and StateT at the top of your monad transformer stack (in fact, the definition above works exactly for this case, just change the type). Extending it to a ReaderT and StateT in the middle of the stack is harder. You basically need a function
lift1 :: (forall a. m0 a -> m1 a) -> t m0 a -> t m1 a
for every monad transformer t in the stack above the ReaderT/StateT, which isn't part of the standard library.
I would recommend wrapping up the State monad in a newtype and defining a MonadReader instance for it:
{-# LANGUAGE GeneralizedNewtypeDeriving #-}
{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE FlexibleContexts #-}
import Control.Applicative
import Control.Monad.State
import Control.Monad.Reader
data MyState = MyState Int deriving Show
newtype App a = App
{ runApp' :: State MyState a
} deriving
( Functor
, Applicative
, Monad
, MonadState MyState
)
runApp :: App a -> MyState -> (a, MyState)
runApp app = runState $ runApp' app
instance MonadReader MyState App where
ask = get
local f m = App $ fmap (fst . runApp m . f) $ get
iWriteData :: MonadState MyState m => Int -> m ()
iWriteData n = do
MyState s <- get
put $ MyState $ s + n
iReadData :: MonadReader MyState m => m Int
iReadData = do
MyState s <- ask
return $ s * 2
bigFunction :: App ()
bigFunction = do
iWriteData 5
iWriteData 10
num <- iReadData
iWriteData num
This is certainly more code that #jcast's solution, but it follows the the tradition of implementing your transformer stack as a newtype wrapper, and by sticking with constraints instead of solidified types you can make strong guarantees about the use of your code while providing maximum flexibility for re-use. Anyone using your code would be able to extend your App with transformers of their own while still using iReadData and iWriteData as intended. You also don't have to wrap every call to a Reader monad with a read function, the MonadReader MyState functions are seamlessly integrated with functions in the App monad.
Excellent answers by jcast and bhelkir, with exactly the first idea I thought of—embedding Reader inside State.
I think it's worthwhile to address this semi-side point of your question:
Using the State monad as usual in these functions works just fine, but I can't help but feel that I've given up Haskell's inherent safety and replicated a language where any function can mutate anything.
That's a potential red flag, indeed. I've always found that State works best for code with "small" states that can be contained within the lifetime of a single, brief application of runState. My go-to example is numbering the elements of a Traversable data structure:
import Control.Monad.State
import Data.Traversable (Traversable, traverse)
tag :: (Traversable t, Enum s) => s -> t a -> t (s, a)
tag i ta = evalState (traverse step ta) init
where step a = do s <- postIncrement
return (s, a)
postIncrement :: Enum s => State s s
postIncrement = do result <- get
put (succ result)
return result
You don't directly say so, but you make it sound you may have a big state value, with many different fields being used in many different ways within a long-lived runState call. And perhaps it does need to be that way for your program at this point. But one technique for coping with this might be to write your smaller State actions so that they only use narrower state types than the "big" one and then embed these into the larger State type with a function like this:
-- | Extract a piece of the current state and run an action that reads
-- and modifies only that piece.
substate :: (s -> s') -> (s' -> s -> s) -> State s' a -> State s a
substate extract replace action =
do s <- get
let (s', a) = runState action (extract s)
put (replace s' s)
return a
Schematic example
example :: State (A, B) Whatever
example = do foo <- substate fst (,b) action1
bar <- substate snd (a,) action2
return $ makeWhatever foo bar
-- Can only touch the `A` component of the state
action1 :: State A Foo
action1 = ...
-- Can only touch the `B` component of the state
action2 :: State B Bar
action2 = ...
Note that the extract and replace functions constitute a lens, and there are libraries for that, which may even already include a function like this.
I am building some product monads from Control.Monad.Product package. For reference, the product monad type is:
newtype Product g h a = Product { runProduct :: (g a, h a) }
Its monad instance is:
instance (Monad g, Monad h) => Monad (Product g h) where
return a = Product (return a, return a)
Product (g, h) >>= k = Product (g >>= fst . runProduct . k, h >>= snd . runProduct . k)
Product (ga, ha) >> Product (gb, hb) = Product (ga >> gb, ha >> hb)
source: http://hackage.haskell.org/packages/archive/monad-products/3.0.1/doc/html/src/Control-Monad-Product.html
Problem I
I build a simple monad that is a product of two State Int Monads, However, when I try to access the underlying state next:
ss :: Product (State Int) (State Int) Int
ss = do
let (a,b) = unp $ P.Product (get,get) :: (State Int Int,State Int Int)
return 404
You see get just creates another State Int Int, and I am not sure how to actually get the value of the underlying state, how might I do that? Note I could potentially runState a and b to get the underlying value, but this solution doesn't seem very useful since the two states' initial values must be fixed a priori.
Question II.
I would really like to be able to create a product monad over states of different types, ie:
ss2 :: Product (State Int) (State String) ()
ss2 = do
let (a,b) = unp $ P.Product (get,get) :: (State Int Int,State Int String)
return ()
But I get this type error:
Couldn't match expected type `String' with actual type `Int'
Expected type: (State Int Int, State String String)
Actual type: (StateT Int Identity Int,
StateT String Identity Int)
Because I presume the two get must return the same type, which is an unfortunate restriction. Any thoughts on how to get around this?
The solution is to use a state monad with a product of your states:
m :: State (Int, String) ()
Then you can run an operation that interacts with one of the two fields of the product using zoom and _1/_2 from the lens library, like this:
m = do
n <- zoom _1 get
zoom _2 $ put (show n)
To learn more about this technique, you can read my blog post on lenses which goes into much more detail.
It can't be done the way you want. Suppose there would be a way how to get the current state out of the left monad. Then you'd have a function of type
getLeft :: Product (State a) (State b) a
which is isomorphic to (State a a, State b a).
Now we can choose to throw away the left part and run only the right part:
evalState (snd (runProduct getLeft)) () :: a
So we get an inhabitant of an arbitrary type a.
In other words, the two monads inside Product are completely independent. They don't influence each other and can be run separately. Therefore we can't take a value out of one and use it in another (or both of them).