[Haskell-beginners] Re: pattern for tree traversel with a state
apfelmus at quantentunnel.de
Thu Oct 23 12:39:02 EDT 2008
Andreas-Christoph Bernstein wrote:
> Is there a pattern for tree traversal with a state ?
> I am developing a small scenegraph represented by a tree. To draw a
> scenegraph one traverses over the graph starting with a global state.
> Inner Nodes can overwrite the inherited state for their subtree (e.g.
> Transformations are accumulated). The accumulated state is then either
> used immediately to draw the geometry in the leaf nodes, or a secondary
> data structure is build. This secondary data structure (a list or a
> tree) can then be sorted for optimal drawing performance.
> So i want to do the second and create a list of all leaves with the
> accumulated global state. To illustrate my problem i appended some code.
> The idea similar applies to a scenegraph.
> So my Question is: Is there already a pattern for traversal with a state ?
Yes. I'm not sure whether state is really necessary for your problem,
i.e. whether there is a more elegant formulation, but your algorithm
fits a well-known pattern, namely the one in Data.Traversable
import qualified Control.Monad.State
data BTree a = Fork a (BTree a) (BTree a) | Leaf a deriving Show
-- main functionality
instance Traversable BTree where
traverse f (Leaf x) = Leaf <$> f x
traverse f (Fork x l r) = Fork <$>
f x <*> traverse f l <*> traverse f r
-- derived examples
instance Foldable BTree where
foldMap = foldMapDefault
instance Functor BTree where
fmap = fmapDefault
flattenTree = toList
-- state example
data StateMod = ModInt | ModString | ModNop deriving Show
type State = (Int, String)
modState :: StateMod -> State -> State
modState ModInt (x,w) = (x+1,w)
modState ModNop s = s
modState ModString (x,w) = (x,'b':w)
startState = (0,"a")
newTree :: BTree StateMod -> BTree State
newTree = flip evalState startState
. Data.Traversable.mapM (modify' . modState)
modify' f = Control.Monad.State.modify f >> Control.Monad.State.get
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