| Unless lightning strikes and tomorrow morning I figure out what's the
| deal with the spurious Mac OSX crashes, I think this might be my last
| network app in Haskell. I should really be spending time on the
| business end of the app intead of figuring out platform differences
| and the like.
On Nov 18, 2005, at 10:17 AM, Simon Peyton-Jones wrote:
I hope you don't abandon Haskell altogether. Without steady, friendly
pressure from applications-end folk like you, things won't improve.
Nah, I'm just having a very frustrating Friday. I think I need some
direction in which to dig
This would be a good new thread to discuss it ;-)
On Nov 18, 2005, at 10:42 AM, Jan Stoklasa (gmail) wrote:
Hi,
so sad, so true...
At least haskell ideas sneak into mainstream languages under
disguise (LINQ
anyone?). C-Java-C# syntax that business developers and their
bosses love
so much
Sebastian Sylvan wrote:
How about (¤)? It looks like a ring to me, I'm not sure where that's
located on a EN keyboard, but it's not terribly inconvenient on my SE
keyboard. f ¤ g looks better than f . g for function composition, if
you ask me.
That symbol actually does look
On 11/18/05, Tomasz Zielonka [EMAIL PROTECTED] wrote:
On Thu, Nov 17, 2005 at 06:56:09PM +0100, Sebastian Sylvan wrote:
Some people do use it more often than I do, but I find that in most
cases except simple pipelined functions it only makes the code
harder to read.
But this case is
I always fancied () as a synonym for 'mappend'
John
--
John Meacham - ⑆repetae.net⑆john⑈
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Fraser Wilson [EMAIL PROTECTED] writes:
Isn't there a potential for confusion with function
composition (f . g)?
Perhaps, but I always have spaces on either side when it's function
composition.
Good for you. Syntax that changes depending on spacing is my number
one
On Fri, Nov 18, 2005 at 12:21:09PM +0100, Sebastian Sylvan wrote:
On 11/18/05, Tomasz Zielonka [EMAIL PROTECTED] wrote:
On Thu, Nov 17, 2005 at 06:56:09PM +0100, Sebastian Sylvan wrote:
Some people do use it more often than I do, but I find that in most
cases except simple pipelined
On 18 November 2005 10:48, Joel Reymont wrote:
On Nov 18, 2005, at 10:17 AM, Simon Peyton-Jones wrote:
I hope you don't abandon Haskell altogether. Without steady,
friendly pressure from applications-end folk like you, things won't
improve.
Nah, I'm just having a very frustrating
On Nov 18, 2005, at 1:55 PM, Simon Marlow wrote:
You can get debugging output by compiling your program with -debug,
and
then running it with some of the -Dsomething options (use +RTS -?
for
a list, +RTS -Ds is a good one to start with).
I'm still working on a repro case but here's what I
On 18 November 2005 14:42, Joel Reymont wrote:
On Nov 18, 2005, at 1:55 PM, Simon Marlow wrote:
You can get debugging output by compiling your program with -debug,
and then running it with some of the -Dsomething options (use +RTS
-? for a list, +RTS -Ds is a good one to start with).
I'm
On 11/18/05, Tomasz Zielonka [EMAIL PROTECTED] wrote:
On Fri, Nov 18, 2005 at 12:21:09PM +0100, Sebastian Sylvan wrote:
On 11/18/05, Tomasz Zielonka [EMAIL PROTECTED] wrote:
On Thu, Nov 17, 2005 at 06:56:09PM +0100, Sebastian Sylvan wrote:
Some people do use it more often than I do,
On 18/11/05, Wolfgang Jeltsch [EMAIL PROTECTED] wrote:
Am Mittwoch, 16. November 2005 20:02 schrieb Cale Gibbard:
[...]
It's unfortunate, but if you don't put a little bit of effort into
defending your forms, they will eventually get quite a lot of spam.
Cleaning up 600+ pages by hand
On Nov 18, 2005, at 2:17 AM, Simon Peyton-Jones wrote:
I hope you don't abandon Haskell altogether. Without steady, friendly
pressure from applications-end folk like you, things won't improve.
It's incredibly valuable feedback. But I can see that when you
have to
deliver something next
This is a very late response to an old thread...
Tom Hawkins wrote:
In a pure language, is it possible to detect cycles in recursive
data structures? For example, is it possible to determine that
cyclic has a loop? ...
data Expr = Constant Int | Addition Expr Expr
cyclic :: Expr
cyclic
On Fri, 18 Nov 2005, Paul Hudak wrote:
For example:
fe1,fe2 :: Fix Expr
fe1 e = Add (Const 1) (Const 1) -- non-recursive
fe2 e = Add (Const 1) e -- recursive
Do you mean
fe1 _ = Add (Const 1) Loop
?
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test.wagerlabs.com seems really slow for me right now. I've mirrored
the repo on my own machine (might not be 100% reliable, but should
stay up nearly all of the time). The mirror address is
http://vx.hn.org/postmortem/
- Cale
On 18/11/05, Joel Reymont [EMAIL PROTECTED] wrote:
Folks,
This is
Am Freitag, 18. November 2005 17:16 schrieb Jason Dagit:
[...]
I was playing with one of the Haskell OpenGL libraries (actually it's
a refined FFI) over the summer and some things about it rubbed me the
wrong way. I wanted to try fixing them but I really couldn't figure
out how to get ahold
I'm still trying to settle on a feel for good programming style in
Haskell. One thing I've been making some use of lately is
(\|) = flip ($)
infixl 0 \|
Then expressions like
f4 $ f3 $ f2 $ f1 $ x
become
x \|
f1 \|
f2 \|
f3 \|
f4
I've seen something like this on haWiki using (#), but I
--- Paul Hudak [EMAIL PROTECTED] wrote:
This is a very late response to an old thread...
Tom Hawkins wrote:
In a pure language, is it possible to detect cycles in recursive
data structures? For example, is it possible to determine that
cyclic has a loop? ...
data Expr =
The exception is actually from withTimeOut. Removing calls to that
lets the handshake proceed. The server is using a client handshake,
though, so the handshake of client vs. client goes on indefinitely.
I'm fixing the server side and once that is done will clean up SSL at
the end of the
Maybe this is old hat, but the question about detecting loops in data
structures got me thinking about this. I know you can encode the cons
operator (and ordinary lists) in pure lambda calculus, but how could
you possibly represent something like [0, 1..]? One thought that
occurss to me is to
What do you mean by represent?
It's easy enough to write down the lambda term that is the
encoding of [0..].
-- Lennart
Greg Woodhouse wrote:
Maybe this is old hat, but the question about detecting loops in data
structures got me thinking about this. I know you can encode the cons
Henning Thielemann wrote:
On Fri, 18 Nov 2005, Paul Hudak wrote:
For example:
fe1,fe2 :: Fix Expr
fe1 e = Add (Const 1) (Const 1) -- non-recursive
fe2 e = Add (Const 1) e -- recursive
Do you mean
fe1 _ = Add (Const 1) Loop
?
No, I really meant it as written. I included this
--- Lennart Augustsson [EMAIL PROTECTED] wrote:
What do you mean by represent?
It's easy enough to write down the lambda term that is the
encoding of [0..].
-- Lennart
You mean like \x - x ? If I apply it to the Church numeral i, I get i
in return. But that hardly seems
Greg Woodhouse wrote:
--- Paul Hudak [EMAIL PROTECTED] wrote:
Tom Hawkins wrote:
In a pure language, is it possible to detect cycles in recursive
data structures? For example, is it possible to determine that
cyclic has a loop? ...
data Expr = Constant Int | Addition Expr Expr
How about:
nil = \ n c . n
cons x xs = \ n c . c x xs
zero = \ z s . z
suc n = \ z s . s n
listFromZero = Y ( \ from n . cons n (from (suc n))) zero
(Untested, so I might have some mistake.)
-- Lennart
Greg Woodhouse wrote:
--- Lennart Augustsson [EMAIL PROTECTED] wrote:
What
--- Paul Hudak [EMAIL PROTECTED] wrote:
I suspect from your other post that you haven't seen the standard
trick of encoding infinite data structures as fixpoints. Suppose you
have a lambda calculus term for cons, as well as for the numeral 1.
Then the infinite list of ones is just:
--- Lennart Augustsson [EMAIL PROTECTED] wrote:
How about:
nil = \ n c . n
cons x xs = \ n c . c x xs
zero = \ z s . z
suc n = \ z s . s n
listFromZero = Y ( \ from n . cons n (from (suc n))) zero
(Untested, so I might have some mistake.)
-- Lennart
Okay, I see what
Greg Woodhouse wrote:
--- Paul Hudak [EMAIL PROTECTED] wrote:
Y (\ones. cons 1 ones)
where Y (aka the paradoxical combinator or fixed point
combinator) is defined as:
\f. (\x. f (x x)) (\x. f (x x))
Now, this is I have seen, but it frankly strikes me as a formal
trick. I've never felt that
On Friday 18 November 2005 02:59, you wrote:
On Nov 17, 2005, at 1:52 PM, Benjamin Franksen wrote:
...
Yes, yes, yes. I'd rather use a different operator for record
selection.
For instance the colon (:). Yes, I know it is the 'cons' operator
for a
certain concrete data type that
--- Paul Hudak [EMAIL PROTECTED] wrote:
The important property of Y is this:
Y f = f (Y f)
Right. This is just a formal statement of the property thaat f fixex Y
f. I'm with you so far.
In this way you can see it as unwinding the function, one step at a
time. If we define f as
Greg Woodhouse wrote:
--- Paul Hudak [EMAIL PROTECTED] wrote:
The important property of Y is this:
Y f = f (Y f)
Right. This is just a formal statement of the property thaat f fixex Y
f. I'm with you so far.
In this way you can see it as unwinding the function, one step at a
time. If
On Fri, Nov 18, 2005 at 04:22:59PM +0100, Sebastian Sylvan wrote:
Yes. I just don't think it's used enough to warrant giving it one of
the best symbols.
grep -o ' [-+.*/[EMAIL PROTECTED] ' GenUtil.hs | sort | uniq -c | sort -n
1 $!
1 *
8 +
10 ==
12 -
Greg Woodhouse wrote:
--- Lennart Augustsson [EMAIL PROTECTED] wrote:
How about:
nil = \ n c . n
cons x xs = \ n c . c x xs
zero = \ z s . z
suc n = \ z s . s n
listFromZero = Y ( \ from n . cons n (from (suc n))) zero
(Untested, so I might have some mistake.)
-- Lennart
Greg Woodhouse wrote:
Perhaps the issue is that the manipulations below are purely syntactic,
But all the computation rules of the lambda calculus are syntactic
in that sense. When you can prove things by symbol pushing it's
the easiest way.
But as Paul Hudak mentioned, there definitions that
--- Lennart Augustsson [EMAIL PROTECTED] wrote:
Unfolding Y is indeed part of the algorithm to generate the list.
The lambda calculus is just another programming language, so
why does this disturb you?
Well...think about this way. The function
f i = [1, 1 ..]!!i
is just a constant
--- Lennart Augustsson [EMAIL PROTECTED] wrote:
It computes the fix point which you can also define as
oo
fix f = lub f^i(_|_)
i=0
where f^i is f iterated i times. Is that a definition
of fixpoint that makes you happier?
Believe it or not, yes.
My simple-mindness and naïveté begin to bother me. I begin to get lost,
and I don't know anymore what is the problem...
Greg Woodhouse a eacute;crit:
--- Paul Hudak [EMAIL PROTECTED] wrote:
...
The important property of Y is this:
Y f = f (Y f)
Right. This is just a formal statement of
I'm happy to report that the problem can be reproduced by running the
code from my darcs repo at http://test.wagerlabs.com/postmortem. See
the README file. I'm on Mac OSX 10.4.3.
The server just sits there, goes through the SSL handshake and...
does nothing else. The clients go through the
On 18/11/05, Greg Woodhouse [EMAIL PROTECTED] wrote:
--- Lennart Augustsson [EMAIL PROTECTED] wrote:
I guess I'm not doing a very good job of expressing myself. I see that
if you define Y as you do, then the various functions you list have the
property that Y f = f (Y f).
I don't want to
On Saturday 19 November 2005 02:16, Greg Woodhouse wrote:
--- [EMAIL PROTECTED] wrote:
fix f = f (fix f) -- Here you have your Y. No typeless lambda.
g f n = n : f n-- This is a generic *non-recursive* `repeat`
ones = fix g 1 -- Guess what.
Very nice! I honestly would not have
Cale Gibbard wrote:
Y = (\f. (\x. f (x x)) (\x. f (x x)))
In a sense, the real definition of Y is Y f = f (Y f), this lambda
term just happens to have that property, but such functions aren't
rare.
Actually no, the real definition is the top one, because the other one
isn't even a valid
On 18/11/05, Paul Hudak [EMAIL PROTECTED] wrote:
Cale Gibbard wrote:
Y = (\f. (\x. f (x x)) (\x. f (x x)))
In a sense, the real definition of Y is Y f = f (Y f), this lambda
term just happens to have that property, but such functions aren't
rare.
Actually no, the real definition is the
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