You can still use the monadic combinators, with the price of wrapping and
unwrapping in case of newtype.
newtype P a = P {unP :: a -> Bool}
liftM2'P :: (Bool -> Bool -> Bool) -> P a -> P a -> P a
liftM2'P op = (P .) . on (liftM2 op) unP
paolino
2012/7/8 Sebastián Krynski <skryn...@gmail.com>
> Ok , thanks for the answers, I understand now what liftM2 does.
> In this case would it be silly to use combinerPred (and maybe a newType
> Predicate a = a -> Bool) for the sake of readability or shoud I stick with
> a -> Bool and liftM2?
>
> thanks, Sebastián
>
>
>
> 2012/7/6 Brent Yorgey <byor...@seas.upenn.edu>
>
>> On Fri, Jul 06, 2012 at 03:17:54PM -0300, Felipe Almeida Lessa wrote:
>> > On Fri, Jul 6, 2012 at 2:11 PM, Sebastián Krynski <skryn...@gmail.com>
>> wrote:
>> > > As I was using predicates (a -> bool) , it appeared the need for
>> combining
>> > > them with a boolean operator (bool -> bool -> bool) in order to get
>> a new
>> > > predicate
>> > > combining the previous two. So I wrote my function combinerPred (see
>> code
>> > > below). While I think this is JUST ok, i'm feeling a monad in the air.
>> > > So.. where is the monad?
>> > >
>> > > combinerPred :: (a -> Bool) -> (a -> Bool) -> (Bool -> Bool ->
>> Bool) ->
>> > > (a -> Bool)
>> > > combinerPred pred1 pred2 op = \x -> op (pred1 x) (pred2 x)
>> >
>> > That's the `(->) a` monad:
>> >
>> > import Control.Applicative
>> >
>> > combinerPred :: (a -> Bool) -> (a -> Bool) -> (Bool -> Bool ->
>> > Bool) -> (a -> Bool)
>> > combinerPred pred1 pred2 op = op <$> pred1 <*> pred2
>>
>> By the way, I find it more natural to make 'op' the first argument,
>> because it is more useful to partially apply combinerPred to an
>> operation that it is to some predicates. Also, in that case
>> combinerPred is simply liftA2:
>>
>> import Control.Applicative
>>
>> combinerPred :: (Bool -> Bool -> Bool) -> (a -> Bool) -> (a -> Bool) ->
>> (a -> Bool)
>> combinerPred = liftA2
>>
>> -Brent
>>
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