($) operator

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($) operator

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)

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"
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($) operator

Daniel Fischer-4
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

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($) operator

Brent Yorgey-2
In reply to this post by John Hartnup
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?

More generally, in a language with first-class and higher-order
functions like Haskell, it's simply useful to have a function which
performs function application.  For example, one place that ($) comes
in handy is when applying each function in a list of functions to a
single input value.  Using ($) you can just write something like

  map ($5) [(+1), (^2), abs]

which evaluates to [6,25,5].

-Brent
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($) operator

David Morse
They definitely dropped the ball on introducing '$'. I think you just
hit the seam between two of the authors, and it fell through the
cracks.
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($) operator

Jan Jakubuv-2
In reply to this post by John Hartnup
2009/1/24 John Hartnup <[hidden email]>:
>
> But is that it's only purpose? To placate the LISP haters by removing
> parentheses?
>

Also one important difference is that the standard application
operator (space) associates to the left while $ associates to the
right. It says how missing parenthesis are filled in.

For example

(a b c d e) stands for ((((a b) c) d) e)

while

(a $ b $ c $ d $ e) stands for (a (b (c (d e))))

Sometimes you need the former behavior sometimes the later. The $ is
useful for example when an argument to an unary function is itself a
function application. For example you can write

    filter (<4) $ map (+1) $ [100,99,1,2,3,4]

In this case you can think of $ to be like a "unix pipe".

Sincerely,
   jan.
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($) operator

Frank Schwidom
In reply to this post by John Hartnup
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

I made some Examples for better understanding:

 -- creating an $ like function:

 f g x = (g x)

 -- some tests comparing $ with f:

 5 == ($ 2) (+ 3)

 5 == (`f` 2) (+ 3)

 5 == (\x -> (($) x 2)) (+ 3)

 5 == (\x -> (f x 2)) (+ 3)

Regards
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($) operator

Brent Yorgey-2
In reply to this post by Brent Yorgey-2
On Sun, Jan 25, 2009 at 05:34:35AM +0000, John Hartnup wrote:

> 2009/1/24 Brent Yorgey <[hidden email]>:
>
> >
> >  map ($5) [(+1), (^2), abs]
> >
> > which evaluates to [6,25,5].
> >
>
> Just to check I understand correctly - you're using ($) to 'promote'
> an expression into a function, in order that it may be used in higher
> order functions?

In this case, I'm using ($) to *make* a higher-order function---($5)
is a function which takes its argument (which must be a function) and
applies it to 5.  (I'm assuming you're familiar with 'operator
sections', if you aren't I can explain that too.)

-Brent
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($) operator

Dave Bayer
In reply to this post by David Morse
2009/1/24 John Hartnup <[hidden email]>:

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

On Jan 25, 2009, at 12:25 AM, David Morse wrote:

> They definitely dropped the ball on introducing '$'. I think you just
> hit the seam between two of the authors, and it fell through the
> cracks.

Haskell isn't Cobol, but in comparison with the lambda calculus, it  
flows a bit better like a natural language. '$' helps with this.

For example, in the line

> mempty = en $ en []

...one clearly doesn't need to use $, parenthesizing (en []) has the  
same meaning. However, reading left-to-right, the $ coaxes one into  
dropping top-down into a new view of the line, partially abstracting  
out or forgetting what came before, without having to maintain a  
mental stack and brace oneself for the closing ).

I've read books on numeracy and vision, that present the same  
conclusion: As our brain evolved new systems for seeing or counting,  
the old systems didn't get thrown out, they got buried by the new  
layers, but they're still wired in. Part of our reaction to a line of  
lisp that ends in seven )))))) is no different than a pigeon or a  
horse seeing the same number of objects; we're not sure. Then for us a  
new linguistic layer kicks in, "I can see two groups of three, that's  
even" and we rise above the animals, but our single conscious core got  
hijacked to sort this out.

People take the ergonomics of cockpit design very seriously, to reduce  
the cognitive load on pilots in emergencies. On Apollo 13, they nearly  
blew it by flipping a switch the wrong way. I fly a lot, so I'm very  
grateful that the "purists" that gravitate to math and computer  
science aren't instead in ergonomic design.

I actually love lisp, or more precisely Scheme. I came back to Haskell  
for strong typing and class abstractions; Haskell more directly  
supports making a theory about what one is coding. In order to use  
Scheme, I had to defang it of its parentheses. I didn't believe that  
any of the web proposals for this saw actual use, so I rolled my own  
preprocessor, using indentation to indicate outline structure, and  
using '|' roughly as Haskell uses '$'. There are other issues one  
inevitably faces, e.g. missing starts to outline levels, but I came to  
love '$' as Haskell uses it.

Some people develop a point-free style in Haskell, e.g.

> blah foo . blah . blah foo foo . blah

and different people write code that piles up uses of '$' e.g.

> blah foo $ blah $ blah foo foo $ blah

With decent syntax coloring the second line can be given as light a  
touch as the first.

Ultimately, '$' helps mitigate a left / right tension in Haskell that  
exists in all languages, and in algebra itself.

In algebra, "f g" can mean either "f then g" or "g then f". It means  
"g then f" in any situation crippled by the ancestry of the function  
composition "f(g(x))", and it means "f then g" any time one has the  
free choice of easier ergonomics in a culture that reads left to right.

Quick, ten times, work out out the permutation cycle product

> (1 2 3) (2 3 4 5) (3 4 5 6) (4 5 6 2) (5 6 1 2) (1 2 3) (2 3 4 5)  
> (3 4 5 6) (4 5 6 2) (5 6 1 2)

under each assumption for the order of operations. One wraps around  
through parentheses either way, but using the "f then g" convention,  
all other scanning is consistently left to right, not flitting back  
and forth like a deranged water bug.

Now try composing a complex set of type signatures for Haskell  
functions. Do you feel like a deranged water bug? The arrows go the  
wrong way, don't they! You'd be playing a nice game of dominoes,  
canceling off neighbors and so forth, if the arrows went the other way.

So Haskell doesn't have its left / right tension completely worked  
out, but I'd say '$' isn't part of the problem, it's part of the  
solution.