Re: [Haskell-cafe] Trouble understanding records and existential types

This is how I would write getLeaves, based on your GADT:

data IsLeaf data IsBranch

newtype Node = Node { getNode :: (forall c. ANode c) }

data ANode :: * -> * where Branch :: String -> String -> (ANode a,ANode b) -> [Node] -> ANode IsBranch Leaf :: String -> String -> ANode IsLeaf

getLeaves :: ANode a -> [ANode IsLeaf] getLeaves (Branch _ _ (l1,l2) rest) = getLeaves l1 ++ getLeaves l2 ++ concatMap (getLeaves.getNode) rest getLeaves x@(Leaf {}) = [x]

Brian Hulley wrote:

On Thursday, January 25, 2007 7:08 AM, John Ky wrote:

...

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On 1/25/07, Brandon S. Allbery KF8NH wrote: I'm probably missing something, but:

(a) Why not:

data ANode = Branch { name :: String, description :: String, children :: [AnyNode] } | Leaf { name :: String, value :: String } -- this reuse

Would I be able to this?

getLeaves :: ANode -> [Leaf]

If not, is it the case that people generally don't bother and do this instead?

getLeaves :: ANode -> [ANode]

As has been pointed out, Leaf is a data constructor not a type so you'd have to use [ANode]. Inspired by the problem I tried a GADT:

data IsLeaf data IsBranch

data ANode a where Branch :: String -> String -> [forall b. ANode b] -> ANode IsBranch Leaf :: String -> String -> ANode IsLeaf

getLeaves :: ANode IsBranch -> [ANode IsLeaf] getLeaves (Branch _ _ ls) = leaves ls

leaves :: [forall b. ANode b] -> [ANode IsLeaf] leaves (l@(Leaf _ _) : ls) = l : leaves ls leaves (Branch _ _ ls : lls) = leaves ls ++ leaves lls

but unfortunately the above code generates the following error by GHC6.6:

Couldn't match expected type `forall b. ANode b' against inferred type `ANode a' In the pattern: Leaf _ _ In the pattern: (l@(Leaf _ _)) : ls In the definition of `leaves': leaves ((l@(Leaf _ _)) : ls) = l : (leaves ls)

Just out of curiosity, does anyone know why the above code doesn't compile ie why is the inferred type for the pattern:

Leaf _ _

(ANode a) and not (ANode IsLeaf)?

Thanks, Brian.