----- Forwarded message from Frank Schwidom ----- Date: Sat, 24 Jan 2009 21:40:54 +0000 From: Frank Schwidom To: John Hartnup Subject: Re: [Haskell-beginners] ($) operator Message-ID: <20090124214053.GA4496@BigBox> References: <6c1ba2fc0901241137x242cf4b4w2398896af45ae694@mail.gmail.com> MIME-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Disposition: inline In-Reply-To: <6c1ba2fc0901241137x242cf4b4w2398896af45ae694@mail.gmail.com> User-Agent: Mutt/1.5.16 (2007-06-09) Status: RO On Sat, Jan 24, 2009 at 07:37:29PM +0000, John Hartnup wrote:

Hi.

I'm working through Real World Haskell, and although it's going well (I just finished the exercise to write a glob matcher without using a regex library, and I'm pleased as punch), I keep seeing the ($) operator, and I'm not sure I understand its use. If the book explains it, I've been unable to find it.

Empirically, it seems like: a $ b c d e f .. is equivalent to .. a (b c d e f)

But is that it's only purpose? To placate the LISP haters by removing parentheses?

(1 +) 2 does the same thing as (1 +) $ 2, and has the same type.

Am I missing something?

Thanks, John

-- "There is no way to peace; peace is the way" _______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

maybe can this answer your question: A: ((==) True) (1 == 1) -- won't work: ((==) True) 1 == 1 => ((==) True) ((==) 1 1) -- won't work: ((==) True) (==) 1 1 => ((==) True) $ (==) 1 1 => ((==) True) $ 1 == 1 Regards ----- End forwarded message -----