Marc Weber wrote: > I've tried as an exercise to learn how to use the state monad to create > a tree this way: > > createTree :: Int -> Int -> (Tree Int, Int) > createTree 4 = runState $ State $ \s -> (Node s [] , s+1) -- stop at level 4 > createTree level = runState (do item <- State $ (\s -> (s,s+1)) > forest <- State $ (\s -> foldr (\_ (for, n) -> > let (l, n') = (createTree (level + 1) n) in (l:for,n')) > ([], s) > (replicate level > ()) ) > return $ Node item (reverse forest) )
Isn't the whole point of the State Monad *not* to thread the state
through every function explicitly? It should probably look like this
(untested code):
createTree :: Int -> Int -> (Tree Int, Int)
createTree = runState . createTree'
bump :: State Int Int
bump = do s <- get ; put $! s+1 ; return s
createTree' :: Int -> State (Tree Int)
createTree' 4 = do s <- bump ; return $ Node s []
createTree' level = do item <- bump
forest <- replicateM (createTree' $ level+1) level
return $ Node item forest
or even
createTree' level = liftM2 Node bump
(replicateM (createTree' $ level+1) level)
Udo.
--
Two rules get you through life: If it's stuck and it's not supposed to
be, WD-40 it. If it's not stuck and it's supposed to be, duct tape it.
-- The Duct Tape Guys' book "WD-40"
signature.asc
Description: Digital signature
_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
