dainichi wrote:
Now to the point: Wouldn't it be great if I had a visual tool that
visually
showed me the graph while the above evaluation unfolded? I could use it to
show some of my co-workers to whom laziness is a mystery, what it's all
about.
Check out
Call-by-name lambda-calculus is strictly more expressive (in Felleisen
sense) than call-by-value lambda-calculus, and the call-by-need (aka, lazy)
lambda-calculus is observationally equivalent to the call-by-name.
One can add shift/reset to any of these calculi (CBV shift/reset is
most known;
About 7 years ago such a tool existed:
http://www.cs.kent.ac.uk/people/staff/cr3/toolbox/haskell/GHood/
I don't know if Claus is around. Perhaps he could give you more information.
Dominic.
___
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
Hi,
I'm developing toy application to learn HDBC. I have problem with
doing relations mapping. Actually I don't know is it a problem or
feature. ;)
Anyway, I have two tables with relation between them:
(this example is simplified to the whole structure of database, you can imagine)
CREATE
On 23 Feb 2008, at 3:30 AM, Radosław Grzanka wrote:
Hi,
I'm developing toy application to learn HDBC. I have problem with
doing relations mapping. Actually I don't know is it a problem or
feature. ;)
Anyway, I have two tables with relation between them:
(this example is simplified to the
Dear All,
banging my head against Haskell, but liking the feeling of hurting
brains. Just a simple question:
If
fmap (^4) [1,2,3] = \i - shows i
gives
1 16 81
then why does
let i = fmap (^4) [1,2,3] in shows i
give
[1,16,81]
Probably very simple, but there must be a delicate
There's a link on the HackageDB Introduction page that gets you the
latest versions of all packages (30MB).
Ross,
Thanks for the archive URL. I parsed through all the hackagedb
modules. I also added the display of repository source (ghc, hdb) and
package source. HackageDB is 3-4 times bigger
fmap (^4) [1,2,3] = \i - shows i
gives
1 16 81
You are in the list comprehension in a monadic expression. shows is
called three times (i is int).
then why does
let i = fmap (^4) [1,2,3] in shows i
give
[1,16,81]
shows is called once (i is a list).
On Sat, Feb 23, 2008 at 8:00 AM, Harri Kiiskinen [EMAIL PROTECTED] wrote:
then why does
let i = fmap (^4) [1,2,3] in shows i
give
[1,16,81]
I'll probably mess this up somewhere, but if I do, reset assured that
someone else here will correct me ;)
fmap (^4) [1,2,3] == [1,16,81] so
2008/2/23, Harri Kiiskinen [EMAIL PROTECTED]:
Dear All,
banging my head against Haskell, but liking the feeling of hurting
brains. Just a simple question:
If
fmap (^4) [1,2,3] = \i - shows i
gives
1 16 81
In the List Monad, (=) is defined as concatMap, so this code can be
Peter Verswyvelen wrote:
PS: This job offer was already placed in the Haskell Café a while ago,
but I was advised there to place it in the main Haskell list. I hope
this is not considered as spam.
I certainly don't see it as spam. One day there will be a dozen jobs
for Haskell programmers
Harri Kiiskinen wrote:
fmap (^4) [1,2,3] = \i - shows i
let i = fmap (^4) [1,2,3] in shows i
Probably very simple, but there must be a delicate difference between
these two expressions. I just don't get it.
First, let's simplify these expressions using the following equation:
fmap
About 7 years ago such a tool existed:
http://www.cs.kent.ac.uk/people/staff/cr3/toolbox/haskell/GHood/
GHood was never intended to visualize graph reduction
directly (*). instead, it visualized observations - ie, you
could see if and when which parts of an observed data
structure was
On Sat, Feb 23, 2008 at 1:05 AM, [EMAIL PROTECTED] wrote:
Adding control effects (shift/reset) changes the expressivity
results. Now all three calculi are distinct and none subsumes the
other. For example, the expression
reset( (\x - 1) (abort 2))
evaluates to 1 in call-by-name
On Sat, Feb 23, 2008 at 1:05 AM, [EMAIL PROTECTED] wrote:
reset ((\x - x + x) (shift f f))
This one doesn't typecheck, since you can't unify the types (a - r) and r.
--
Taral [EMAIL PROTECTED]
Please let me know if there's any further trouble I can give you.
-- Unknown
Hi, all, I am continuing to mess with my little scheme interpreter,
and I decided that it would be nice to be able to hit control-C in the
middle of a long-running scheme computation to interrupt that and
return to the lisp prompt; hitting control-C and getting back to the
shell prompt works, but
---
Haskell Weekly News
http://sequence.complete.org/hwn/20080223
Issue 70 - February 23, 2008
---
Welcome to issue 70 of HWN, a newsletter covering
Henning Thielemann wrote:
It seems that algorithms on graphs can be implemented particularly
efficient in low-level languages with pointers and in-place updates. E.g.
topological sort needs only linear time, provided that dereferencing
pointers requires constant time. I could simulate pointer
On Sat, Feb 23, 2008 at 10:59:40AM -0500, Steve Lihn wrote:
I parsed through all the hackagedb
modules. I also added the display of repository source (ghc, hdb) and
package source. HackageDB is 3-4 times bigger than GHC core. The
result is interesting, looking at the most used modules move up
Taral [EMAIL PROTECTED] wrote in article [EMAIL PROTECTED] in
gmane.comp.lang.haskell.cafe:
On Sat, Feb 23, 2008 at 1:05 AM, [EMAIL PROTECTED] wrote:
reset ((\x - x + x) (shift f f))
This one doesn't typecheck, since you can't unify the types (a - r) and r.
Some type systems for
Hi all,
I try not to be too rude, although I'm rather disgusted.
I know there are several sites out on the web where solutions to PE problems
are given. That is of course absolutely against the sporting spirit of
Project Euler, but hey, not all people are sporting.
I've found
Daniel Fischer wrote:
Hi all,
I try not to be too rude, although I'm rather disgusted.
I know there are several sites out on the web where solutions to PE problems
are given. That is of course absolutely against the sporting spirit of
Project Euler, but hey, not all people are sporting.
I've
Hello Uwe,
Saturday, February 23, 2008, 11:35:35 PM, you wrote:
mysighandler =
Catch (do hPutStrLn stderr caught a signal!
fail Interrupt!)
scheme calculation doesn't get interrupted at all! I see in the
System.Posix.Signals documentation that the signal handler gets
invoked
Thanks, Bulat, I'll look into this!
On 2/23/08, Bulat Ziganshin [EMAIL PROTECTED] wrote:
Hello Uwe,
Saturday, February 23, 2008, 11:35:35 PM, you wrote:
mysighandler =
Catch (do hPutStrLn stderr caught a signal!
fail Interrupt!)
scheme calculation doesn't get
You're going the right way about having the answers published in more
ways than just the Haskell wiki. I'm only making a prediction, not a
threat.
Might be. And I've been over-angered.
Having the Haskell code for the solutions in the wiki might be legitimate, but
a) no code should be put
On 2/23/08, Bulat Ziganshin [EMAIL PROTECTED] wrote:
[about my question about keyboard interrupts]
you should store thread id of thread running interpreter and send
async exception to it. control.concurrent is probably contains all
required functions
Most splendid! Here's what I did
Shame on you Haskell wiki! :)
Perhaps it's a conpiracy to avoid wasting too much of the community effort
on PE, and direct that energy to darcs :)
- Quan
___
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
27 matches
Mail list logo
