Re: [Haskell-cafe] On to applicative

I think it works well :) But sumsqr has type Int -> Int -> Int, not Int -> Int -> Int -> Int. I.e. it does take only two arguments while fmap3 takes a function of three arguments. 2010/8/26 michael rice

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 wrote:

From: Ivan Lazar Miljenovic Subject: Re: [Haskell-cafe] On to applicative To: "michael rice" Cc: haskell-cafe@haskell.org Date: Thursday, August 26, 2010, 2:33 AM

On 26 August 2010 16:29, michael rice wrote:

...

Can you recommend an example that works?

An example of what?

The definitions of fmap2, etc. on that page look like they're correct.

-- Ivan Lazar Miljenovic Ivan.Miljenovic@gmail.com IvanMiljenovic.wordpress.com

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe