| OK, fmap2 works, but not fmap3. What am I not understanding? Michael import Control.Applicative -- f :: (a -> b -> c) -- fmap :: Functor f => (d -> e) -> f d -> f e sumsqr :: Int -> Int -> Int sumsqr i j = i*i+j*j -- fmap :: Functor f => f a -> f (b -> c) -- Identify d with a, and e with (b -> c) fmap2 f a b = f `fmap` a <*> b fmap3 f a b c = f `fmap` a <*> b <*> c fmap4 f a b c d = f `fmap` a <*> b <*> c <*> d -- fmap2 f a b = f <$> a <*> b -- fmap3 f a b c = f <$> a <*> b <*> c -- fmap4 f a b c d = f <$> a <*> b <*> c <*> d *Main> fmap2 sumsqr (Just 3) (Just 4) Just 25 *Main> fmap3 sumsqr (Just 3) (Just 4) (Just 5) <interactive>:1:6: Couldn't match expected type `a2 -> b' against inferred type `Int' In the first argument of `fmap3', namely `sumsqr' In the expression: fmap3 sumsqr (Just 3) (Just 4) (Just 5) In the definition of `it': it = fmap3 sumsqr (Just 3) (Just 4) (Just 5) *Main> --- On Thu, 8/26/10, Ivan Lazar Miljenovic <ivan.miljenovic@gmail.com> wrote:
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