Does the state get passed as a variable, with each >> and if so could someone show me how?
It does indeed. It's helpful to look at a toy implementation of the State monad.
newtype State s a = State { runState :: s -> (a,s) }
instance Monad (State s) where
return a = State $ \s -> (a, s)
State act >>= k = State $ \s ->
let (a, s') = act s
in runState (k a) s'
get :: State s s
get = State $ \s -> (s, s)
put :: s -> State s ()
put s = State $ \_ -> ((), s)
modify :: (s -> s) -> State s ()
modify f = get >>= \x -> put (f x)
When you bind using >>=
or >>
the accumulated state is threaded through as an argument to the function on the right hand side.
When you run execState
or evalState
it then just extracts either the resulting value or the state from the resulting tuple.
execState :: State s a -> s -> s
execState act = snd . runState act
evalState :: State s a -> s -> a
evalState act = fst . runState act