State Monad
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The State Monad by Example
This is a short tutorial on the state monad. Emphasis is placed on intuition. The types have been simplified to protect the innocent.
Another longer walkthrough of the state monad can be found in the wiki book section Understanding monads/State.
Contents 
1 Foundations
1.1 Primitives
runState (return 'X') 1
('X',1)
Comments:
return 'X' :: State Int Char runState (return 'X') :: Int > (Char, Int) initial state = 1 :: Int final value = 'X' :: Char final state = 1 :: Int result = ('X', 1) :: (Char, Int)
runState get 1
(1,1)
Comments:
get :: State Int Int runState get :: Int > (Int, Int) initial state = 1 :: Int final value = 1 :: Int final state = 1 :: Int
runState (put 5) 1
((),5)
Comments:
put 5 :: State Int () runState (put 5) :: Int > ((),Int) initial state = 1 :: Int final value = () :: () final state = 5 :: Int
1.2 Combinations
Because (State s) forms a monad, values can be combined together with (>>=) or do{}.
runState (do { put 5; return 'X' }) 1
('X',5)
Comments:
do { put 5; return 'X' } :: State Int Char runState (do { put 5; return 'X' }) :: Int > (Char,Int) initial state = 1 :: Int final value = 'X' :: Char final state = 5 :: Int
postincrement = do { x < get; put (x+1); return x } runState postincrement 1
(1,2)
predecrement = do { x < get; put (x1); get } runState predecrement 1
(0,0)
1.3 Other Functions
runState (modify (+1)) 1
((),2)
runState (gets (+1)) 1
(2,1)
evalState (gets (+1)) 1
2
execState (gets (+1)) 1
1
2 Implementation
At its heart, a value of type (State s a) is a function from initial state s to final value a and final state s: (a,s). These are usually wrapped, but shown here unwrapped for simplicity.
Return leaves the state unchanged and sets the result:
 ie: (return 5) 1 > (5,1) return :: a > State s a return x s = (x,s)
Get leaves state unchanged and sets the result to the state:
 ie: get 1 > (1,1) get :: State s s get s = (s,s)
Put sets the result to () and sets the state:
 ie: (put 5) 1 > ((),5) put :: s > State s () put x s = ((),x)
The helpers are simple variations of these primitives:
modify :: (s > s) > State s () modify f = do { x < get; put (f x) } gets :: (s > a) > State s a gets f = do { x < get; return (f x) }
EvalState and execState just select one of the two values returned by runState. EvalState returns the final result while execState returns the final state:
evalState :: State s a > s > a evalState act = fst . runState act execState :: State s a > s > s execState act = snd . runState act
Combining two states is the trickiest bit in the whole scheme. To combine do { x < act1; act2 x } we need a function which takes an initial state, runs act1 to get an intermediate result and state, feeds the intermediate result to act2 and then runs that action with the intermediate state to get a final result and a final state:
(>>=) :: State s a > (a > State s b) > State s b (act1 >>= fact2) s = runState act2 is where (iv,is) = runState act1 s act2 = fact2 iv
3 Complete and Concrete Example 1
Simple example that demonstrates the use of the standard Control.Monad.State monad. Its a simple string parsing algorithm.
module StateGame where import Control.Monad.State  Example use of State monad  Passes a string of dictionary {a,b,c}  Game is to produce a number from the string.  By default the game is off, a C toggles the  game on and off. A 'a' gives +1 and a b gives 1.  E.g  'ab' = 0  'ca' = 1  'cabca' = 0  State = game is on or off & current score  = (Bool, Int) type GameValue = Int type GameState = (Bool, Int) playGame :: String > State GameState GameValue playGame [] = do (_, score) < get return score playGame (x:xs) = do (on, score) < get case x of 'a'  on > put (on, score + 1) 'b'  on > put (on, score  1) 'c' > put (not on, score) _ > put (on, score) playGame xs startState = (False, 0) main = print $ evalState (playGame "abcaaacbbcabbab") startState
4 Complete and Concrete Example 2
 a concrete and simple example of using the State monad import Control.Monad.State  non monadic version of a very simple state example  the State is an integer.  the value will always be the negative of of the state type MyState = Int valFromState :: MyState > Int valFromState s = s nextState :: MyState>MyState nextState x = 1+x type MyStateMonad = State MyState  this is it, the State transformation. Add 1 to the state, return 1*the state as the computed value. getNext :: MyStateMonad Int getNext = state (\st > let st' = nextState(st) in (valFromState(st'),st') )  advance the state three times. inc3::MyStateMonad Int inc3 = getNext >>= \x > getNext >>= \y > getNext >>= \z > return z  advance the state three times with do sugar inc3Sugared::MyStateMonad Int inc3Sugared = do x < getNext y < getNext z < getNext return z  advance the state three times without inspecting computed values inc3DiscardedValues::MyStateMonad Int inc3DiscardedValues = getNext >> getNext >> getNext  advance the state three times without inspecting computed values with do sugar inc3DiscardedValuesSugared::MyStateMonad Int inc3DiscardedValuesSugared = do getNext getNext getNext  advance state 3 times, compute the square of the state inc3AlternateResult::MyStateMonad Int inc3AlternateResult = do getNext getNext getNext s<get return (s*s)  advance state 3 times, ignoring computed value, and then once more inc4::MyStateMonad Int inc4 = do inc3AlternateResult getNext main = do print (evalState inc3 0)  3 print (evalState inc3Sugared 0)  3 print (evalState inc3DiscardedValues 0)  3 print (evalState inc3DiscardedValuesSugared 0)  3 print (evalState inc3AlternateResult 0)  9 print (evalState inc4 0)  4