Re: [Haskell-cafe] Applicative but not Monad

Thank you for your correction. I tried your (>>=) and replaced return's definition with return = ZipList . repeat then as you said this works fine for infinite lists. Cheers, -~nwn On Sat, Oct 31, 2009 at 2:39 AM, David Menendez wrote:

On Fri, Oct 30, 2009 at 1:33 PM, Yusaku Hashimoto wrote:

...

Thanks for fast replies! Examples you gave explain why all Applicatives are not Monads to me.

And I tried to rewrite Bob's Monad instance for ZipList with (>>=).

import Control.Applicative

instance Monad ZipList where  return = ZipList . return  (ZipList []) >>= _ = ZipList []  (ZipList (a:as)) >>= f = zlHead (f a) `zlCons` (ZipList as >>= f)

This is taking the first element of each list, but you need to take the nth element. Try

 (ZipList (a:as)) >>= f = zlHead (f a) `zlCons` (ZipList as >>= zlTail . f)

-- Dave Menendez http://www.eyrie.org/~zednenem/