On Sat, Feb 12, 2011 at 12:39 PM, htc2011 <jakobusbe...@gmail.com> wrote: > > Hi All, > > I am learning Haskell and can't understand the following problem. Maybe > somebody could advise me on a solution? > > Using GHCI, I have the following definition of a BST: > data Ord a => BST a = EmptyBST | Node ( BST a ) a ( BST a ) deriving (Show) > > I want to determine the number of leaves that a tree using the above > definition has: > numLeaves :: Ord a => BST a -> Int > numLeaves EmptyBST = 0 > numLeaves (Node b a c) > | b == EmptyBST = 1 + numLeaves c > | c == EmptyBST = 1 + numLeaves b > | otherwise = numLeaves b + numLeaves c > > However whenever I load my haskell file containing the above code into GHCI, > I get the following error: > > Could not deduce (Eq (BST a)) from the context (Ord a) > arising from a use of `==' at a8.hs:17:3-15 > Possible fix: > add (Eq (BST a)) to the context of > the type signature for `numLeaves' > or add an instance declaration for (Eq (BST a)) > In the expression: b == EmptyBST > In a stmt of a pattern guard for > the definition of `numLeaves': > b == EmptyBST > In the definition of `numLeaves': > numLeaves (Node b a c) > | b == EmptyBST = 1 + numLeaves c > | c == EmptyBST = 1 + numLeaves b > | otherwise = numLeaves b + numLeaves c > > Could anybody explain to me what this means? / How to get around this? > > Thank you for your time!
Hi, You are comparing two BST instances in the second expression for numLeaves without declaring the Eq instance. numLeaves (Node b a c) will bind b and c to two BST instances. When you are comparing b and c with EmptyBST an error is raised. To solve this, you'll have to declare an Eq instance, just like you've declared a Show one: data Ord a => BST a = EmptyBST | Node ( BST a ) a ( BST a ) deriving (Show, Eq) This will solve it :D -- Mihai _______________________________________________ Haskell mailing list Haskell@haskell.org http://www.haskell.org/mailman/listinfo/haskell
