[Haskell-cafe] combining predicates, noob question
skrynski at gmail.com
Sun Jul 8 02:42:01 CEST 2012
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?
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>
> > > As I was using predicates (a -> bool) , it appeared the need for
> > > them with a boolean operator (bool -> bool -> bool) in order to get a
> > > predicate
> > > combining the previous two. So I wrote my function combinerPred (see
> > > 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
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