Le Tue, 01 Jan 2013 14:24:04 -0900,
Christopher Howard christopher.how...@frigidcode.com a écrit :
I'm working through a video lecture describing how to prove programs
correct, by first translating the program into a control flow
representation and then using propositional logic. In the
Le Sun, 25 Nov 2012 21:41:47 +,
Gytis Žilinskas gytis.zilins...@gmail.com a écrit :
Greetings,
I'm only taking my very first steps learning Haskell, but I believe
that this mailing list might be appropriate for my question.
How difficult would it be to study category theory and
Le Tue, 20 Nov 2012 06:54:25 -0500 (EST),
c...@lavabit.com a écrit :
Hello,
I know nothing about compilers and interpreters. I checked several
books, but none of them explained why we have to translate a
high-level language into a small (core) language. Is it impossible
(very hard) to
Le Tue, 20 Nov 2012 10:49:01 -0500 (EST),
c...@lavabit.com a écrit :
What would be the point in doing so?
Well, I don't know. Would it save some time? Why bother with a core
language?
The compilation process might be slightly faster, but I guess it
wouldn't be much noticeable.
Also I
Le Wed, 24 Oct 2012 12:36:52 +0700,
Kim-Ee Yeoh k...@atamo.com a écrit :
On Tue, Oct 16, 2012 at 9:37 PM, AUGER Cédric sedri...@gmail.com
wrote:
As I said, from the mathematical point of view, join (often noted μ
in category theory) is the (natural) transformation which with
return (η
composition is the holy grail, why not encourage the monadic
code, too, to be compositional? Nicety can wait; some syntax sugar
might take care of it.
And as you have pointed out, arrows make a superior choice in this
regard, but they are rather newer to monads.
@ AUGER Cédric
It would
Le Tue, 16 Oct 2012 09:51:29 -0400,
Jake McArthur jake.mcart...@gmail.com a écrit :
On Mon, Oct 15, 2012 at 11:29 PM, Dan Doel dan.d...@gmail.com wrote:
I'd be down with putting join in the class, but that tends to not be
terribly important for most cases, either.
Join is not the most
…
On Tue, Oct 16, 2012 at 10:37 AM, AUGER Cédric sedri...@gmail.com
wrote:
Le Tue, 16 Oct 2012 09:51:29 -0400,
Jake McArthur jake.mcart...@gmail.com a écrit :
On Mon, Oct 15, 2012 at 11:29 PM, Dan Doel dan.d...@gmail.com
wrote:
I'd be down with putting join in the class
Le Tue, 16 Oct 2012 11:58:44 -0400,
Dan Doel dan.d...@gmail.com a écrit :
On Tue, Oct 16, 2012 at 10:37 AM, AUGER Cédric sedri...@gmail.com
wrote:
join IS the most important from the categorical point of view.
In a way it is natural to define 'bind' from 'join', but in
Haskell
Le Mon, 15 Oct 2012 15:12:28 +0200,
Benjamin Franksen benjamin.frank...@helmholtz-berlin.de a écrit :
Ertugrul Söylemez wrote:
damodar kulkarni kdamodar2...@gmail.com wrote:
The Monad class makes us define bind (=) and unit (return) for
our monads.
Why the Kleisli composition (=) or
Le Sat, 8 Sep 2012 09:20:29 +0100,
Ramana Kumar ramana.ku...@cl.cam.ac.uk a écrit :
On Fri, Sep 7, 2012 at 5:56 PM, Edward Z. Yang ezy...@mit.edu wrote:
Excerpts from David Feuer's message of Fri Sep 07 12:06:00 -0400
2012:
They're not *usually* desirable, but when the code has been
Le Fri, 6 Jan 2012 10:59:29 +0100,
Yves Parès limestr...@gmail.com a écrit :
2012/1/6 AUGER Cédric sedri...@gmail.com
when you write forall a. exists b. a - b - a, then you allow the
caller to have access to b to produce a (didn't you write
a-b-a?)
Yes, sorry, I always assumed
Le Tue, 3 Jan 2012 20:46:15 +0100,
Yves Parès limestr...@gmail.com a écrit :
Actually, my question is why the different type can't be unified
with the
inferred type?
Because without ScopedTypeVariable, both types got expanded to :
legSome :: *forall nt* t s. LegGram nt t s - nt -
So is there anyway to force the scoping of variables, so that
f :: a - a
f x = x :: a
becomes valid?
Do we need to do it the Ocaml way, that is not declaring the first
line, activate XScopedVariables and do:
f :: a - a = \x - x :: a
or
f (x :: a) = x :: a
or
some other thing
?
I don't see the
Le Wed, 4 Jan 2012 14:41:21 +0100,
Yves Parès limestr...@gmail.com a écrit :
I expected the type of 'x' to be universally quantified, and thus
can be unified with 'forall a. a' with no problem
As I get it. 'x' is not universally quantified. f is. [1]
x would be universally quantified if
Le Wed, 4 Jan 2012 16:22:31 +0100,
Yves Parès limestr...@gmail.com a écrit :
Oleg explained why those work in his last post. It's the exact same
logic for each one.
f :: a - a
f x = x :: a
We explained that too: it's converted (alpha-converted, but I don't
exactly know what 'alpha'
doesn't
appear in the introduced type).
2012/1/4 AUGER Cédric sedri...@gmail.com
Le Wed, 4 Jan 2012 14:41:21 +0100,
Yves Parès limestr...@gmail.com a écrit :
I expected the type of 'x' to be universally quantified, and
thus can be unified with 'forall a. a' with no problem
Hi all, I am an Haskell newbie; can someone explain me why there is
no reported error in @legSome@ but there is one in @legSomeb@
(I used leksah as an IDE, and my compiler is:
$ ghc -v
Glasgow Haskell Compiler, Version 7.2.1, stage 2 booted by GHC version
6.12.3 )
What I do not understand is
Thanks all,
I finally give up to put Ord in the LegGram type.
What was annoying me was that @legsome@ was in fact an instance of a
class I defined. So I changed its signature to make it depend on Ord.
That is not very nice, since at first glance, there could be
implementations which does not
Le Sun, 1 Jan 2012 16:31:51 -0500,
Dan Doel dan.d...@gmail.com a écrit :
On Sun, Jan 1, 2012 at 3:26 PM, Ketil Malde ke...@malde.org wrote:
Chris Smith cdsm...@gmail.com writes:
I wonder: can writing to memory be called a “computational
effect”? If yes, then every computation is impure.
struggling with anything else today. Maybe a third try...
On 28/12/2011 19:38, AUGER Cédric wrote:
Le Wed, 28 Dec 2011 17:39:52 +,
Steve Hornesh006d3...@blueyonder.co.uk a écrit :
This is just my view on whether Haskell is pure, being offered up
for criticism. I haven't seen this view
Le Mon, 26 Dec 2011 19:30:20 -0800,
Alexander Solla alex.so...@gmail.com a écrit :
So we give meaning to syntax through our semantics. That is what this
whole conversation is all about. I am proposing we give Haskell
bottoms semantics that bring it in line with the bottoms from various
Le Wed, 28 Dec 2011 17:39:52 +,
Steve Horne sh006d3...@blueyonder.co.uk a écrit :
This is just my view on whether Haskell is pure, being offered up for
criticism. I haven't seen this view explicitly articulated anywhere
before, but it does seem to be implicit in a lot of explanations -
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