Re: [Haskell-cafe] Newbie: State monad example questions

["Dmitri O.Kondratiev"]

Jules,

Stupid question, please bear with me:

x :: Int -- x declared, but not constructed
x = 1 -- x constructed

s1 :: 
State StdGen a -- s1 declared, yes, but why s1 is *also already constructed* ?

On Wed, May 21, 2008 at 6:54 PM, Jules Bean <jules@xxxxxxxxxxxxxxx> wrote:

Dmitri O.Kondratiev wrote:

Thanks everybody for your help!
Oliver,  you provided an excellent write-up  on  State  monad without  going  into 'scary' :) details, great work indeed!
Alas,  in this case I need the details, and in particular the most scary ones!

So let's start with fundamental and most intriguing  (to me) things:

getAny :: (Random a) => State StdGen a
getAny = do g <- get -- magically get the current StdGen

First line above declares a data type:

State StdGen a

which is constructed with the function:

State {runState :: (StdGen -> (a, StdGen))}

Q1: Where in the example (http://www.haskell.org/all_about_monads/examples/example15.hs) data of this type *actually gets constructed* ?

Actually get constructed?

It gets constructed by >>= and return, both of which construct state objects:

instance Monad (State s) where

   return a = State $ \s -> (a, s)

   m >>= k  = State $ \s -> let
       (a, s') = runState m s
       in runState (k a) s'


How do >>= and return get called? Well you can see explicit calls to return. The >>= is implicit in the way do-notation is desugared.

getAny = do g      <- get

           let (x,g') = random g
           put g'

           return x

rewrites to

getAny = get >>= \g -> ( let (x,g') = random g in (put g' >> return x) )

where I have added some not strictly necessary ()s and taken the liberty of changing the confusing "a <- return x" idiom to "let a = x".

So the *actually gets constructed* part is that use of >>= .

HTH,

Jules

-- 
Dmitri O. Kondratiev
dokondr@xxxxxxxxx
http://www.geocities.com/dkondr
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