|Subject:||Re: [Haskell-cafe] Newbie: State monad example questions|
|Date:||Wed, 21 May 2008 14:31:09 -0400|
On Wed, May 21, 2008 at 11:10 AM, Dmitri O.Kondratiev <dokondr@xxxxxxxxx> wrote:
But how will 'g1' actually get delivered from 'makeRandomValueST g1' to invocation of 'getAny' I don't yet understand!
It may be easier to understand the state passing if you remove the do notation and replace get, put and return with their definition in the instance declarations (Monad and MonadState).
getAny :: (Random a) => State StdGen a
getAny = do g <- get
(x,g') <- return $ random g
get = State $ \s -> (s, s) -- copy the state as a return value and pass state
put s = State $ \_ -> ((), s) -- return unit, ignore the passed state and replace it with the state given as parameter.
return a = State $ \s -> (a, s) -- return given value and pass state.
getAnyNoSugar :: (Random a) => State StdGen a
getAnyNoSugar = (State $ \s -> (s, s)) >>= \g ->
(State $ \s -> (random g, s)) >>= \(x,g') ->
(State $ \_ -> ((), g')) >>
(State $ \s -> (x, s))
The function is still useable this way and the state transformations should be a bit more visible. The first element of the tuple is the value that will be used to call the next function (of type Monad m => a -> m b). The second element of the tuple is the state and the (>>=) operator will handle passing it between actions.
Desugaring the (>>=) and (>>) operators would give you something like this (I replaced `s` with `y` in the `put` and `return` desugaring and simplified it):
State $ \s = let
(g, s') = (\y -> (y,y)) s
((x,g'), s'') = (\y -> (random g, y)) s'
(_, s''') = (\_ -> ((), g')) s''
in (x, s''')
Which is explict state passing between function calls. Extract the State using `runState`, run it with an initial state and it should give you the expected result.
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