On 2014-04-02 12:53, Nishant wrote:
Can some explain me the error and a possibly a fix :
data Tree a = NULL | Node (Tree a ) a (Tree a)
instance Functor Tree where fmap f NULL = NULL fmap f (Node left x right) | (left == NULL) && (right == NULL) = Node left (f x) right | (left /= NULL) = Node (fmap f left) x right | (right /= NULL) = Node left x (fmap f right)
[..]
programstree.hs:32:78: Couldn't match type `a' with `b' `a' is a rigid type variable bound by the type signature for fmap :: (a -> b) -> Tree a -> Tree b at programstree.hs:31:7 `b' is a rigid type variable bound by the type signature for fmap :: (a -> b) -> Tree a -> Tree b at programstree.hs:31:7 Expected type: Tree b Actual type: Tree a In the first argument of `Node', namely `left'
The issue is that in 'fmap f (Node left x right)', 'left' has type 'Tree a'. Your 'fmap' function should yield a 'Tree b'though, i.e. the 'Node' constructor should take a 'Tree b' as well - and 'left' doesn't fit. You can fix this by using 'NULL' directly, e.g. instance Functor Tree where fmap f NULL = NULL fmap f (Node left x right) | (left == NULL) && (right == NULL) = Node NULL (f x) NULL | (left /= NULL) = Node (fmap f left) (f x) NULL | (right /= NULL) = Node NULL (f x) (fmap f right) ...note that instead of 'x' you also have to use 'f x' (which is also the whole point of a Functor instance). -- Frerich Raabe - raabe@froglogic.com www.froglogic.com - Multi-Platform GUI Testing