Thanks, Brandon,
That's a very clear explanation. I remembered that it's low, but forgot that
it's this low.
Maybe it's because of my mis-understanding that in C, infix operators have
lower precedence.
Just curious, the following is not allowed in Haskell either for the same
reason.
applySkip i f ls = (take i) ls ++ $ f $ drop i ls
I read somewhere that people a couple of hundreds of years ago can manage to
express things using ($)-like notation without any parenthesis at all.
Is it possible to define an operator to achieve this which works with infix
operators? If so, then ($) and parentheses are then really equivalent.
Best,
Ting
Date: Tue, 19 Jul 2011 02:22:47 -0400
Subject: Re: [Haskell-cafe] question regarding the $ apply operator
From: allber...@gmail.com
To: tin...@hotmail.com
CC: haskell-cafe@haskell.org
On Tue, Jul 19, 2011 at 02:16, Ting Lei <tin...@hotmail.com> wrote:
I have a naive question regarding the basic use of the $ operator, and I am
confused why certain times it doesn't seem to work.
e.g.
The following works:
applySkip i f ls = (take i) ls ++ f (drop i ls)
But the following doesn't:
applySkip i f ls = (take i) ls ++ f $ drop i ls
($) has lower precedence than almost every other operator, so your "doesn't
work" translates to
> applySkip i f ls = ((take i) ls ++ f) $ (drop i ls)
--
brandon s allbery allber...@gmail.com
wandering unix systems administrator (available) (412) 475-9364 vm/sms
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