Brent Yorgey ha scritto:
On Tue, May 6, 2008 at 8:20 AM, patrik osgnach <[EMAIL PROTECTED]>
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 "+∗xyz"
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
treefoldr f x g y (Node a []) = f a y
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
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