[Haskell-cafe] Applicative but not Monad

[Yusaku Hashimoto]

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 <dave at zednenem.com> wrote:
> On Fri, Oct 30, 2009 at 1:33 PM, Yusaku Hashimoto <nonowarn at gmail.com> 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 <dave at zednenem.com>
> <http://www.eyrie.org/~zednenem/>
>