[Haskell-beginners] Parsing an integer

[Patrick LeBoutillier]

Tim,

On Sun, Nov 21, 2010 at 7:27 AM, Tim Baumgartner
<baumgartner.tim at googlemail.com> wrote:
>
> data Token = Number Int  -- ...
> number :: Parser Token
> number = many1 digit ↠ (read >>> Number >>> return)
> -- = many1 digit ↠ return∘Number∘read
>
> But I wonder what an advanced haskeller would code instead.
> Particularly, I'd like to get rid of the 'return'.

My understanding is that to remove the return, you can lift your pure
function (Number . read) into the Parser monad using fmap or <$> from
Control.Applicative:

  number' = Number . read <$> many1 digit


A bit of Sunday morning exploration yielded that you may also like the
<$$> operator, but it is not defined in the standard libraries (see
discussion here:
http://www.haskell.org/pipermail/libraries/2010-April/013403.html) :

  (<$$>) :: Functor f => f a -> (a -> b) -> f b
  (<$$>) = flip (<$>)
  infixl 4 <$$>


  number' = many1 digit <$$> Number . read
    or
  number' = many1 digit <$$> (read >>> Number)


Patrick

>
> Thanks in advance
> Tim
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-- 
=====================
Patrick LeBoutillier
Rosemère, Québec, Canada