[Haskell-cafe] Applicative but not Monad
nonowarn at gmail.com
Fri Oct 30 13:57:37 EDT 2009
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.
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>
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