Dmitri,
Excellent questions. There's one step you're missing. Most of your
questions revolve around 'foo <- bar' constructs within a monad. I
would suggest that you review the de-sugaring rules at
http://en.wikibooks.org/wiki/Haskell/Syntactic_sugar#Do_and_proc_notation
and see if that helps you out some. The best process would be for you
to 1.) De-sugar this function completely and 2.) look at bind (denoted
as >>=), and substitute it in.
Hope this helps!
2008/5/19 Dmitri O.Kondratiev
I am trying to understand State monad example15 at: http://www.haskell.org/all_about_monads/html/statemonad.html
Example 15 uses getAny that I don't understand at all how it works:
getAny :: (Random a) => State StdGen a getAny = do g <- get (x,g') <- return $ random g put g' return x
Questions: 1) random has type: random :: (Random a, RandomGen g) => g -> (a, g)
and for State monad:
return a = State (\s -> (a, s))
then: return (random g) = State (\s -> ((a,g), s))
Is it correct?
2) What x and g' will match to in: do ... (x,g') <- return $ random g
which, as I understand equals to: do ... (x,g') <- State (\s -> ((a,g), s))
What x and g' will match to in the last expression?
3) In general, in do expression (pseudo): do { x <- State (\s -> (a, s)); ...}
What x will refer to? Will x stand for a whole lambda function: \s -> (a, s) ?
4) How 'g <- get' works in this function (getAny) ? 5) Why we need 'put g'?
Thanks!
_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe