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: allbery.b@gmail.com To: tinlyx@hotmail.com CC: haskell-cafe@haskell.org On Tue, Jul 19, 2011 at 02:16, Ting Lei 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 allbery.b@gmail.com

wandering unix systems administrator (available) (412) 475-9364 vm/sms

On 19 July 2011 06:52, Ting Lei wrote:

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.

The only thing that I can think of that matches this is Reverse Polish Notation, and according to Wikipedia was about 90 years ago, not hundreds ;-) -- Ivan Lazar Miljenovic Ivan.Miljenovic@gmail.com IvanMiljenovic.wordpress.com

On 19 July 2011 09:51, Maciej Marcin Piechotka wrote:

On Tue, 2011-07-19 at 07:11 +0000, Ivan Lazar Miljenovic wrote:

...

The only thing that I can think of that matches this is Reverse Polish Notation, and according to Wikipedia was about 90 years ago, not hundreds ;-)

I may be wrong but weren't the RPN calculator slightly later then 90 years ago? In any case according to wikipedia RPN was proposed in 1954 so less then 60 years ago. (PN on the other hand was found around 90 years ago).

You're right; I mis-read the wikipedia article whilst skimming it. -- Ivan Lazar Miljenovic Ivan.Miljenovic@gmail.com IvanMiljenovic.wordpress.com

I know the Reverse Polish is not a couple of hundred years old. I have an impression of reading something about people writing natural deduction systems using only dots in place of parenthesis. And it is said that it was "natural" in those pre-historic times. That's also why I had this (mis-)conception that ($) can be used to achieve something similar. It would be marginally interesting, since it saves us one character in the mental stack. I still remember looking at the parentheses in a lisp program. Anyways, thanks for the clarifications. On 19 July 2011 09:51, Maciej Marcin Piechotka wrote:

On Tue, 2011-07-19 at 07:11 +0000, Ivan Lazar Miljenovic wrote:

...

The only thing that I can think of that matches this is Reverse Polish Notation, and according to Wikipedia was about 90 years ago, not hundreds ;-)

I may be wrong but weren't the RPN calculator slightly later then 90 years ago? In any case according to wikipedia RPN was proposed in 1954 so less then 60 years ago. (PN on the other hand was found around 90 years ago).

You're right; I mis-read the wikipedia article whilst skimming it. -- Ivan Lazar Miljenovic Ivan.Miljenovic@gmail.com IvanMiljenovic.wordpress.com

On Mon, Jul 18, 2011 at 11:52:15PM -0700, Ting Lei wrote:

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.

($) and parentheses are quite different. Anyone who tells you they are equivalent, although probably well-meaning, should be ignored. Here is everything you need to know about ($): infixr 0 $ f $ x = f x Although working out all the implications of this definition may take a bit of practice. -Brent