Am Samstag, 24. Januar 2009 20:37 schrieb John Hartnup:
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)
Indeed. ($) is the function application operator, f $ x = f x , so a $ b c d e f is equivalent to a (b c d e f) and a $ b c $ d e f to a (b c (d e f)).
But is that its only purpose? To placate the LISP haters by removing parentheses?
More or less. Although the idea was more to enhance readability than placating LISP haters :) It's also handy to pass to some higher order functions, although offhand I can't think of any situation where you couldn't pass id instead of ($).
(1 +) 2 does the same thing as (1 +) $ 2, and has the same type.
Am I missing something?
Thanks, John