```Am Dienstag, 6. Mai 2008 22:40 schrieb patrik osgnach: > Brent Yorgey ha scritto: > > On Tue, May 6, 2008 at 8:20 AM, patrik osgnach > > > > wrote: > >> Hi. I'm learning haskell but i'm stuck on a generic tree folding > >> exercise. i must write a function of this type > >> treefoldr::(Eq a,Show a)=>(a->b->c)->c->(c->b->b)->b->Tree a->c > >> Tree has type > >> data (Eq a,Show a)=>Tree a=Void | Node a [Tree a] deriving (Eq,Show) > >> as an example treefoldr (:) [] (++) [] (Node '+' [Node '*' [Node 'x' [], > >> Node 'y' []], Node 'z' []]) > >> must return "+âxyz" > >> any help? > >> (sorry for my bad english) > > > > Having a (Tree a) parameter, where Tree is defined as an algebraic data > > type, also immediately suggests that you should do some pattern-matching > > to break treefoldr down into cases: > > > > treefoldr f y g z Void = ? > > treefoldr f y g z (Node x t) = ? > > > > -Brent > > so far i have tried > treefoldr f x g y Void = x Yes, nothing else could be done. > treefoldr f x g y (Node a []) = f a y Not bad. But actually there's no need to treat nodes with and without children differently. Let's see: treefoldr f x g y (Node v ts) should have type c, and it should use v. We have f :: a -> b -> c x :: c g :: c -> b -> b y :: b v :: a. The only thing which produces a value of type c using a value of type a is f, so we must have treefoldr f x g y (Node v ts) = f v someExpressionUsing'ts' where someExpressionUsing'ts' :: b. The only thing we have which produces a value of type b is g, so someExpressionUsing'ts' must ultimately be g something somethingElse. Now take a look at the code and type of foldr, that might give you the idea. Cheers, Daniel > treefoldr f x g y (Node a (t:ts)) = treefoldr f x g (g (treefoldr f x g > y t) y) (Node a ts) > but it is incorrect. i can't figure out how to build the recursive call > thanks for the answer > Patrik _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@xxxxxxxxxxx http://www.haskell.org/mailman/listinfo/haskell-cafe ```