[Haskell-cafe] Applicative but not Monad

[Yusaku Hashimoto]

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)

zlHead :: ZipList a -> a
zlHead (ZipList (a:_)) = a
zlCons :: a -> ZipList a -> ZipList a
zlCons a (ZipList as) = ZipList $ a:as
zlTail :: ZipList a -> ZipList a
zlTail (ZipList (_:as)) = ZipList as

I understand if this instance satisfies the laws, we can replace <$>
with `liftM` and <*> and `ap`. And I found a counterexample (correct
me if I'm wrong).

*Main Control.Monad> getZipList $ (*) <$> ZipList [1,2] <*> ZipList [3,4,5]
[3,8]
*Main Control.Monad> getZipList $ (*) `liftM` ZipList [1,2] `ap` ZipList [3,4,5]
[3,6]

Cheers,
-~nwn

On Sat, Oct 31, 2009 at 2:06 AM, Tom Davie <tom.davie at gmail.com> wrote:
> On Fri, Oct 30, 2009 at 5:59 PM, Luke Palmer <lrpalmer at gmail.com> wrote:
>>
>> On Fri, Oct 30, 2009 at 10:39 AM, Tom Davie <tom.davie at gmail.com> wrote:
>> > Of note, there is a sensible monad instance for zip lists which I
>> > *think*
>> > agrees with the Applicative one, I don't know why they're not monads:
>> > instance Monad (ZipList a) where
>> >   return = Ziplist . return
>> >   join (ZipList []) = ZipList []
>> >   join (ZipList (a:as)) = zlHead a `zlCons` join (map zlTail as)
>>
>> IIRC, that doesn't satisfy the associativity law, particularly when
>> you are joining a list of lists of different lengths.  2 minutes of
>> experimenting failed to find me the counterexample though.
>
> Cool, thanks Luke, that explains why this is available in Stream, but not in
> ZipList too.
> Bob
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