Comments on: Point-Free style: What is it good for?

By: Scala: Post-Functional, Post-Modern, or Just Perl++?

Scala: Post-Functional, Post-Modern, or Just Perl++? — Sun, 07 Mar 2010 03:31:12 +0000

[...] paragon of this way of programming is point-free programming, where you are quite literally building up a mega-function that describes how your program works, [...]

By: OJ

OJ — Tue, 17 Nov 2009 01:26:23 +0000

Hi DK,

The short answer is: Yes, you should use the same method that I used above There might be some other tricks that you have to follow along the way, but refactoring to point-free is essentially the same process (which is what lambdabot does).

Have you tried to change those functions to point-free ?

Cheers
OJ

By: DK

DK — Sun, 15 Nov 2009 10:38:18 +0000

Hi, Thank you very much for such a great tutorial. I am also a noob to Haskell.
I have a question: how will you point free the following ones?
f x y z = (x y) y z
g x y z = x z (y z)
filter p = foldr (x xs -> if p x then (x:xs) else xs) []

Of course, I can do it using some tool like lambdadot, but is there any method that I can follow and do it manually?

Thanks.
DK

By: OJ

OJ — Sat, 11 Jul 2009 22:53:05 +0000

@Petro: don’t worry mate, we are all here to learn! I’m certainly no expert.

If you see “(Num a)=> a” that means that “a” has to be an instance of the class “Num”. This is how you enforce certain type constraints on your function parameters.

The reason this solves the problem is because Haskell’s type-inference system chooses one of the Num types for us (Integer in this case). Putting the type signature in prevents it from doing that.

Hope that helps!

By: Petronilho

Petronilho — Sat, 11 Jul 2009 17:05:26 +0000

Thanks for the assistance, OJ. Don’t kill me, lol.
I’m just learning the language, and I didn’t put the funtion type signature.

I don’t even know what the => means yet. :P

By: OJ

OJ — Fri, 10 Jul 2009 02:13:49 +0000

There is most definitely a way of doing it (there’s probably many ways). Off the top of my head I came up with this:

double :: (Num a) => a -> a
double = sum . take 2 . repeat
 
dub :: (Num a) => a -> a
dub = ((*)2)

Note that the type signatures are important. I would argue this is a case where point-free doesn’t add any value. It’s not like it’s difficult to understand what your function is doing. Use of point-free isn’t a catch-all solution. You should use it where it makes sense.

Wait for the more able Haskellers to see your question. They’ll no doubt come up with a better solution

Cheers!

By: Petronilho

Petronilho — Fri, 10 Jul 2009 01:42:59 +0000

Okay, so I want to make a function that doubles its number argument.
The non-point-free way accepts Integer or Double.
The point-free way works with Integer but doesn’t work on Double.

By: OJ

OJ — Thu, 02 Jul 2009 21:51:25 +0000

@Philipp: Cheers mate. It took me a while to get what it was I like about Point-Free, now I know why too

Thanks for the comment.

By: Philipp Meier

Philipp Meier — Thu, 02 Jul 2009 12:41:43 +0000

Great stuff. I always liked point-free. Now I know why.

By: OJ

OJ — Wed, 01 Jul 2009 11:33:25 +0000

@Frederick: Wow, that was a really interesting stuff! Cheers. This stuff is very new to me, so information like this is really valuable.

It’s good to see that what I’m talking has actually got some history. It validates the learning in some way and makes me feel good about what I’ve written.

Thanks for taking the time to comment