[Haskell-cafe] combining predicates, noob question

[Sebastián Krynski]

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 <byorgey at 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 <skrynski at 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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