State的`put`和`get`函数

Question

I'n looking at the State Monad 's put and get :

ghci> :t get
get :: MonadState s m => m s

ghci> :t runState
runState :: State s a -> s -> (a, s)

ghci> runState get [1,2,3]
([1,2,3],[1,2,3])

In get 's type signature: MonadState sm => ms , how does [1,2,3] have a type of MonadState sm ? It's not clear to me what the types of s and m are.

Also, can you please say more as to how to use put ?

ghci> :t put put :: MonadState sm => s -> m ()

Overall, it seems that I don't understand what MonadState sm is. Could you please explain with put and get examples?

Answer 1

MonadState sm is a typeclass constraint, not a type. The signature:

get :: MonadState s m => m s

Says that for some monad m storing some state of type s , get is an action in m that returns a value of type s . This is pretty abstract, so let's make it more concrete with the less-overloaded version of State from transformers :

get :: State s s
put :: s -> State s ()

Now say we want to use State to keep a simple counter. Let's use execState instead of runState so we can just pay attention to the final value of the state. We can get the value of the counter:

> execState get 0
0

We can set the value of the counter using put :

> execState (put 1) 0
1

We can set the state multiple times:

> execState (do put 1; put 2) 0
2

And we can modify the state based on its current value:

> execState (do x <- get; put (x + 1)) 0
1

This combination of get and put is common enough to have its own name, modify :

> execState (do modify (+ 1)) 0
1

> execState (do modify (+ 2); modify (* 5)) 0
10

MonadState is the class of types that are monads with state. State is an instance of that class:

instance MonadState s (State s) where
  get = Control.Monad.Trans.State.get
  put = Control.Monad.Trans.State.put

So are StateT (the state monad transformer, which adds state to another monad) and various others. This overloading was introduced so that if you're using a stack of monad transformers, you don't need to explicitly lift operations between different transformers. If you're not doing that, you can use the simpler operations from transformers .

---

Here's another example of how to use State to encapsulate a map of variables with Data.Map :

import Control.Monad.Trans.State
import qualified Data.Map as M

action = do
  modify (M.insert "x" 2)        -- x = 2
  modify (M.insert "y" 3)        -- y = 3
  x <- gets (M.! "x")
  y <- gets (M.! "y")
  modify (M.insert "z" (x + y))  -- z = x + y
  modify (M.adjust (+ 2) "z")    -- z += 2
  gets (M.! "z")                 -- return z

main = do
  let (result, vars) = execState action M.empty

  putStr "Result: "
  print result

  putStr "Vars: "
  print vars

Answer 2

In get's type signature: MonadState sm => ms, how does [1,2,3] have a type of MonadState sm? It's not clear to me what the types of s and m are.

ghci> runState get [1,2,3]

The function runState takes two arguments: the first is the State action to run, and the second is the initial state. So, since the initial state is [1,2,3] which is a list of integers (*), the state type s is just [Integer] .

(*) Actually, [1,2,3] :: Num a => [a] before GHCi defaults it, but for simplicity's sake let's use [Integer] as GHCi does.

Hence, we see that runState is specialized to

runState :: State [Integer] a -> [Integer] -> (a, [Integer])

Now, about the first argument:

get :: MonadState s m => m s

We must have ms = State sa because we are passing it to runState which requires such type. Hence:

runState :: State [Integer] a -> [Integer] -> (a, [Integer])
get :: MonadState s m => m s
with m s = State [Integer] a

The latter equation can be simplified as follows:

runState :: State [Integer] a -> [Integer] -> (a, [Integer])
get :: MonadState s m => m s
with m = State [Integer]
and  s = a

Substituting s :

runState :: State [Integer] a -> [Integer] -> (a, [Integer])
get :: MonadState a m => m a
with m = State [Integer]

Substituting m :

runState :: State [Integer] a -> [Integer] -> (a, [Integer])
get :: MonadState a (State [Integer]) => State [Integer] a

Now, the constraint MonadState a (State [Integer]) is satisfied only when a = [Integer] . This is tricky to see, since the MonasState type class exploits a functional dependency to enforce that every monad in that class has only one related state type. This is also made more complex from State being a wrapper around StateT . Anyway, we get:

runState :: State [Integer] a -> [Integer] -> (a, [Integer])
get :: MonadState a (State [Integer]) => State [Integer] a
with a = [Integer]

So,

runState :: State [Integer] [Integer] -> [Integer] -> ([Integer], [Integer])
get :: MonadState [Integer] (State [Integer]) => State [Integer] [Integer]

And since the constraint is satisfied:

runState :: State [Integer] [Integer] -> [Integer] -> ([Integer], [Integer])
get :: State [Integer] [Integer]

And now we can see the involved ground types.