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Today's Topics:
1. Re: How to Update Cabal? (Daniel Trstenjak)
2. Re: A little explanation! (Michael Orlitzky)
3. Re: A little explanation! follow up (Dimitri DeFigueiredo)
4. Re: How to Update Cabal? (Norbert Melzer)
5. Re: A little explanation! follow up (Brent Yorgey)
6. Re: A little explanation! (Kyle Murphy)
----------------------------------------------------------------------
Message: 1
Date: Wed, 30 Apr 2014 18:22:18 +0200
From: Daniel Trstenjak <[email protected]>
To: [email protected]
Subject: Re: [Haskell-beginners] How to Update Cabal?
Message-ID: <20140430162218.GA28575@machine>
Content-Type: text/plain; charset=us-ascii
Hi Ari,
On Wed, Apr 30, 2014 at 12:18:08PM -0400, Ari King wrote:
> Is it best to clone the cabal git repo and install as described here? In which
> case, should I remove the existing cabal installation from "/usr/bin"?
It's just:
cabal install cabal-install
Greetings,
Daniel
------------------------------
Message: 2
Date: Wed, 30 Apr 2014 13:19:02 -0400
From: Michael Orlitzky <[email protected]>
To: [email protected]
Subject: Re: [Haskell-beginners] A little explanation!
Message-ID: <[email protected]>
Content-Type: text/plain; charset=ISO-8859-1
On 04/30/2014 11:29 AM, Gilberto Melfe wrote:
> Hello everybody !
>
> Could someone explain me exactly how this function works?
>
> elementAt_w'pf = (last .) . take . (+ 1)
>
> It's a posted solution to: 99 Haskell problems, problem 3.
>
> I'm having trouble with the "(last .) ." thing!
>
It's not your fault, that solution is retarded. Once you remove all of
the intentional obfuscation, it looks like,
elementAt_w'pf n xs = last (take (n + 1) xs)
which is easy to understand. You can un-retard it step-by-step:
elementAt_w'pf = (last .) . take . (+ 1)
<=> elementAt_w'pf n = ((last .) . take . (+ 1)) n
= (last .) ((take . (+ 1)) n)
= (last .) (take (n + 1))
= last . (take (n + 1))
<=> elementAt_w'pf n xs = (last . (take (n + 1))) xs
= last (take (n + 1) xs)
All I've used above is the precedence of the function composition
operator, and the fact that (f . g) x = f (g x).
------------------------------
Message: 3
Date: Wed, 30 Apr 2014 11:25:58 -0600
From: Dimitri DeFigueiredo <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] A little explanation! follow up
Message-ID: <[email protected]>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
I have a follow up question on this one. Is code like this:
elementAt_w'pf = (last .) . take . (+ 1)
considered good haskell? If so, will this ever become easy to read? What
do I do to practice reading this?
More specifically, how would I know in mind head, before asking the type
checker in GHCi to write this pipeline with the section (last .) instead
of simply making the (wrong) pipe:
elementAt_w'pf = last . take . (+ 1)
I am used to using pipes in the unix shell, but in the shell there is
always only one input and one output. There is no currying going on and
so it is very clear what we are dealing with. How do I learn this?
Thanks,
Dimitri
Em 30/04/14 09:53, Kyle Murphy escreveu:
> Near as I can tell, this is basically having to do with partial
> application and the order of precedence for the (.) operator. A great
> way to look at this stuff is using the :t command in GHCi and checking
> out the types involved.
>
> Prelude> :t (.)
> (.) :: (b -> c) -> (a -> b) -> a -> c
>
> Prelude> :t last
> last :: [a] -> a
>
> Prelude> :t (last .)
> (last .) :: (a -> [c]) -> a -> c
>
> Looking back at the type for (.) and plugging in "[a]" in place of "b"
> and "a" in place of "c" we get:
> ([b] -> b) -> (a -> [b]) -> a -> b
>
> and since "last" takes the place of the first function we can reduce
> that to:
> (a -> [b]) -> a -> b
>
> GHCi used "c" where we used "b" but you can clearly see the signatures
> are identical other than that detail.
>
> What we're left with is, (last .) is used to take a function from some
> type "a" that returns a list of type "b", and a "a" value, and then
> returns the last value from that list of "b" types.
>
>
> -R. Kyle Murphy
> --
> Curiosity was framed, Ignorance killed the cat.
>
>
> On Wed, Apr 30, 2014 at 11:29 AM, Gilberto Melfe
> <[email protected] <mailto:[email protected]>> wrote:
>
> Hello everybody !
>
> Could someone explain me exactly how this function works?
>
> elementAt_w'pf = (last .) . take . (+ 1)
>
> It's a posted solution to: 99 Haskell problems, problem 3.
>
> I'm having trouble with the "(last .) ." thing!
>
> Hope someone can help!
>
> Thank You!
>
> Gilberto
>
> _______________________________________________
> Beginners mailing list
> [email protected] <mailto:[email protected]>
> http://www.haskell.org/mailman/listinfo/beginners
>
>
>
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://www.haskell.org/mailman/listinfo/beginners
------------------------------
Message: 4
Date: Wed, 30 Apr 2014 19:40:10 +0200
From: Norbert Melzer <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] How to Update Cabal?
Message-ID:
<ca+bcvsumbcwbiszqjuhn4xonxdbejzsaskvfpwrsmfulpj+...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Probably one needs to ensure that ~/.cabal/bin is in the PATH-variable
before the path that cabal normally lives in. Alternatively you have to use
sudo and the --global switch for cabal (not recommended).
Am 30.04.2014 18:22 schrieb "Daniel Trstenjak" <[email protected]>:
>
> Hi Ari,
>
> On Wed, Apr 30, 2014 at 12:18:08PM -0400, Ari King wrote:
> > Is it best to clone the cabal git repo and install as described here? In
> which
> > case, should I remove the existing cabal installation from "/usr/bin"?
>
> It's just:
>
> cabal install cabal-install
>
>
> Greetings,
> Daniel
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://www.haskell.org/mailman/listinfo/beginners
>
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Message: 5
Date: Wed, 30 Apr 2014 14:15:57 -0400
From: Brent Yorgey <[email protected]>
To: [email protected]
Subject: Re: [Haskell-beginners] A little explanation! follow up
Message-ID: <[email protected]>
Content-Type: text/plain; charset=us-ascii
On Wed, Apr 30, 2014 at 11:25:58AM -0600, Dimitri DeFigueiredo wrote:
> I have a follow up question on this one. Is code like this:
>
> elementAt_w'pf = (last .) . take . (+ 1)
>
> considered good haskell?
Uses of composition sections like (last .) is not very common; I would
not consider it good haskell (though tastes may differ). On the other
hand, point-free notation, used in moderation, is considered good
Haskell style. For example, instead of
foo x = bar (baz (quux x))
you should write
foo = bar . baz . quux
In this case I would certainly consider the second definition better
style than the first.
> If so, will this ever become easy to read?
> What do I do to practice reading this?
Just read and write lots of code. Try transforming things into and
out of point-free notation as an exercise.
> More specifically, how would I know in mind head, before asking the
> type checker in GHCi to write this pipeline with the section (last .)
> instead of simply making the (wrong) pipe:
>
> elementAt_w'pf = last . take . (+ 1)
Simply by working out the types in your head. This becomes much
easier with practice.
> I am used to using pipes in the unix shell, but in the shell there is
> always only one input and one output. There is no currying going on
> and so it is very clear what we are dealing with. How do I learn
> this?
There is always only one input and one output in Haskell as well. The
thing that makes it more complicated is that unlike in the shell,
those inputs and outputs can themselves be functions.
-Brent
>
>
> Em 30/04/14 09:53, Kyle Murphy escreveu:
> >Near as I can tell, this is basically having to do with partial
> >application and the order of precedence for the (.) operator. A
> >great way to look at this stuff is using the :t command in GHCi and
> >checking out the types involved.
> >
> >Prelude> :t (.)
> >(.) :: (b -> c) -> (a -> b) -> a -> c
> >
> >Prelude> :t last
> >last :: [a] -> a
> >
> >Prelude> :t (last .)
> >(last .) :: (a -> [c]) -> a -> c
> >
> >Looking back at the type for (.) and plugging in "[a]" in place of
> >"b" and "a" in place of "c" we get:
> >([b] -> b) -> (a -> [b]) -> a -> b
> >
> >and since "last" takes the place of the first function we can
> >reduce that to:
> >(a -> [b]) -> a -> b
> >
> >GHCi used "c" where we used "b" but you can clearly see the
> >signatures are identical other than that detail.
> >
> >What we're left with is, (last .) is used to take a function from
> >some type "a" that returns a list of type "b", and a "a" value, and
> >then returns the last value from that list of "b" types.
> >
> >
> >-R. Kyle Murphy
> >--
> >Curiosity was framed, Ignorance killed the cat.
> >
> >
> >On Wed, Apr 30, 2014 at 11:29 AM, Gilberto Melfe
> ><[email protected] <mailto:[email protected]>> wrote:
> >
> > Hello everybody !
> >
> > Could someone explain me exactly how this function works?
> >
> > elementAt_w'pf = (last .) . take . (+ 1)
> >
> > It's a posted solution to: 99 Haskell problems, problem 3.
> >
> > I'm having trouble with the "(last .) ." thing!
> >
> > Hope someone can help!
> >
> > Thank You!
> >
> > Gilberto
> >
> > _______________________________________________
> > Beginners mailing list
> > [email protected] <mailto:[email protected]>
> > http://www.haskell.org/mailman/listinfo/beginners
> >
> >
> >
> >
> >_______________________________________________
> >Beginners mailing list
> >[email protected]
> >http://www.haskell.org/mailman/listinfo/beginners
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://www.haskell.org/mailman/listinfo/beginners
------------------------------
Message: 6
Date: Wed, 30 Apr 2014 14:16:45 -0400
From: Kyle Murphy <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] A little explanation!
Message-ID:
<CA+y6JcxTnRFZd=9bgfpkgdrw5mypfnrsavh1vxiuc8clewe...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
On the point of it being "retarded" there's some merit to that in that it's
written in point-free (or as some joking call it, pointless) style. The
idea is to remove the explicit variables and so the functions are
re-arranged and composed such that all the variables come at the end of the
statement and you're left with nothing but a chain of functions at the
start (which can sometimes be easier to understand since it's just function
composition). The downside, as you've seen is that it often hurts
readability and so there are some that are strongly opposed to point-free
style. If used carefully, and things are broken into smaller easier to
understand chunks (via for instance let or where clauses) it can still be
readable even using point-free style, but it's really dependent on both the
skill of the person writing the statement, and the familiarity and skill of
the person reading the statement. When in doubt about legibility it's
better to avoid point-free style, in particular if you start having to do
strange things like apply the (.) operator to itself (E.G. "(f .) . g" or
similar) you're probably being to clever for your own good. I wouldn't go
so far as to say you should *never* use point-free style, but you should
seriously consider if you're trying to just be clever and show off, or if
it really is making the statement simpler to understand via more clearly
showing how the functions are composed.
-R. Kyle Murphy
--
Curiosity was framed, Ignorance killed the cat.
On Wed, Apr 30, 2014 at 1:19 PM, Michael Orlitzky <[email protected]>wrote:
> On 04/30/2014 11:29 AM, Gilberto Melfe wrote:
> > Hello everybody !
> >
> > Could someone explain me exactly how this function works?
> >
> > elementAt_w'pf = (last .) . take . (+ 1)
> >
> > It's a posted solution to: 99 Haskell problems, problem 3.
> >
> > I'm having trouble with the "(last .) ." thing!
> >
>
> It's not your fault, that solution is retarded. Once you remove all of
> the intentional obfuscation, it looks like,
>
> elementAt_w'pf n xs = last (take (n + 1) xs)
>
> which is easy to understand. You can un-retard it step-by-step:
>
> elementAt_w'pf = (last .) . take . (+ 1)
>
> <=> elementAt_w'pf n = ((last .) . take . (+ 1)) n
>
> = (last .) ((take . (+ 1)) n)
>
> = (last .) (take (n + 1))
>
> = last . (take (n + 1))
>
> <=> elementAt_w'pf n xs = (last . (take (n + 1))) xs
>
> = last (take (n + 1) xs)
>
> All I've used above is the precedence of the function composition
> operator, and the fact that (f . g) x = f (g x).
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://www.haskell.org/mailman/listinfo/beginners
>
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