You can write (f . g) x as f . g $ x so for me, it's avoiding extra
parenthesis.
Mukesh
On Wed, Feb 13, 2013 at 2:53 AM, Emanuel Koczwara wrote: Hi, Dnia 2013-02-12, wto o godzinie 22:09 +0100, Martin Drautzburg pisze: On Friday, 1. February 2013 23:02:39 Ertugrul Söylemez wrote: (f . g) x = f (g x) so (f . g) x = f $ g x right? That looks like the two are pretty interchangeable. When would I prefer
one
over the other? ($) has lower precedence (it was introduced for that reason I belive). Prelude> :info ($)
($) :: (a -> b) -> a -> b -- Defined in GHC.Base
infixr 0 $ Please take a look at: http://www.haskell.org/ghc/docs/latest/html/libraries/base-4.6.0.1/Prelude.h... From the docs: "Application operator. This operator is redundant, since ordinary
application (f x) means the same as (f $ x). However, $ has low,
right-associative binding precedence, so it sometimes allows parentheses
to be omitted..." Emanuel _______________________________________________
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