[Haskell-beginners] flatten comma operator
Arlen Cuss
a at unnali.com
Wed Jun 6 08:42:52 CEST 2012
By the way, is the excerpt from RWH involving liftA2 the chapter on using Parsec? If so, this may be the code snippet you refer to:
-- file: ch16/FormApp.hs a_pair :: CharParser () (String, Maybe String) a_pair = liftA2 (,) (many1 a_char) (optionMaybe (char '=' *> many a_char))
In this case, liftA2 is promoting the (,) operation to work with the two operations in the CharParser applicative functor.
(,) is of type "a -> b -> (a,b)", so without lifting, we'd end up with something like "(CharParser () String, CharParser () Maybe String)" (just a guess here).
liftA2 produces a new applicative functor action which computes each of (many1 a_char) and (optionMaybe (char '=' *> many a_char)), then gives the pure results to (,).
On Wednesday, 6 June 2012 at 4:36 PM, Arlen Cuss wrote:
> If (,) is a function that takes two elements and returns the 2-tuple, have you considered something like (,,)? :)
>
>
>
> On Wednesday, 6 June 2012 at 4:33 PM, Kees Bleijenberg wrote:
>
> > In 'Real World Haskell' I found code like LiftA2 (,) ....
> > Looks odd. But after some experimenting in winghci I found that (,) 1 2 is valid code and is equal to (1,2).
> > Now I wonder how to create (1,2,3). I think you need a join or a flatten function or ...? Join doesn't work?
> >
> > Kees
> >
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