Hi Stephan,
S. Günther wrote:
Is it possible to change a particular node of the
doubly linked list? That is to say, that would like
to have a function:
update :: DList a - a - DList a
where
update node newValue
returns a list where only the value at the node
which is passed in is set to the new
hi all, not sure if there is someone still working during holiday like me :
)
I got a little problem in implementing some operations on tree.
suppose we have a tree date type defined:
data Tree a = Leaf a | Branch (Tree a) (Tree a)
I want to do a concatenation on these tree just like the
I'm not working, but still checking mail.
If you don't care about balancing the tree or the order of elements, you can
just use
Branch :: Tree a - Tree a - Tree a
as a concatenation operator. Check with GHCi to see that the Branch constructor
actually has the above type.
/ Emil
Max
On 31 Dec 2008, at 16:02, Max cs wrote:
hi all, not sure if there is someone still working during holiday
like me : )
I got a little problem in implementing some operations on tree.
suppose we have a tree date type defined:
data Tree a = Leaf a | Branch (Tree a) (Tree a)
I want to do a
Aside:
lookPairs :: RqData [(String,String)]
lookPairs = asks fst = return . map (\(n,vbs)-(n,L.unpack $ inputValue
vbs))
Looks like an opportunity for semantic editor combinators [1]. Something
like
lookPairs = (fmap.fmap.fmap) (L.unpack . inputValue) (asks fst)
Or specialize the edit
Forgot to send this to the list.
On Wed, 31 Dec 2008 16:05:10 +0100, Max cs max.cs.2...@googlemail.com
wrote:
hi all, not sure if there is someone still working during holiday like
me :
)
I got a little problem in implementing some operations on tree.
suppose we have a tree date type
Ticket #635 Replace use of select() in the I/O manager with
epoll/kqueue/etc. (http://hackage.haskell.org/trac/ghc/ticket/635)
dates back from 2005. Now its 2009 and GHC can handle hundreds of
thousands of threads, yet having more than 1024 file descriptors open
is still impossible. This
On Wed, 31 Dec 2008 17:19:09 +0100, Max cs max.cs.2...@googlemail.com
wrote:
Hi Henk-Jan van Tuyl,
Thank you very much for your reply!
I think the concatenation should be different to thhe
treeConcat :: Tree a - Tree a - Tree a
the above is a combination of two trees instead of a
On 31 Dec 2008, at 21:18, Henk-Jan van Tuyl wrote:
On Wed, 31 Dec 2008 17:19:09 +0100, Max cs max.cs.
2...@googlemail.com wrote:
Hi Henk-Jan van Tuyl,
Thank you very much for your reply!
I think the concatenation should be different to thhe
treeConcat :: Tree a - Tree a - Tree a
the
As someone suggested me, I can read the logs from Writer and WriterT
as computation goes by,
if the monoid for the Writer is lazy readable.
This has been true until I tried to put the IO inside WriterT
{-# LANGUAGE FlexibleContexts #-}
import Control.Monad.Writer
k :: (MonadWriter [Int] m) =
On Wed, 31 Dec 2008 21:25:02 +0100, Thomas Davie tom.da...@gmail.com
wrote:
On 31 Dec 2008, at 21:18, Henk-Jan van Tuyl wrote:
On Wed, 31 Dec 2008 17:19:09 +0100, Max cs max.cs.2...@googlemail.com
wrote:
Hi Henk-Jan van Tuyl,
Thank you very much for your reply!
I think the
On Wed, 2008-12-31 at 21:48 +0100, Paolino wrote:
As someone suggested me, I can read the logs from Writer and WriterT as
computation goes by,
if the monoid for the Writer is lazy readable.
This has been true until I tried to put the IO inside WriterT
{-# LANGUAGE FlexibleContexts #-}
IO is not lazy; you never make it to print.
Consider this program:
k = f 0 where
f n = do
lift (print n)
tell [n]
f (n+1)
weird :: IO [Int]
weird = do
(_, ns) - runWriterT k
return (take 20 ns)
What should weird print? According to k, it prints every
Also, it's actually really hard to tie the knot in the update; without
some kind of distinguished node that allows you to know that it is the
beginning/end of the list.
For example, in this DList:
1,1,1, lots of times, 1, 2, 1, 1, ... lots of times, 1, (loop)
If you change the 3rd 1, how
Dear all,
Happy New Year!
I am learning the Induction Proof over Haskell, I saw some proofs for the
equivalence of two functions will have a case called 'bottom' but some of them
do no have. What kind of situation we should also include the bottom case to
the proof? How about the functions
2008/12/31 ra...@msn.com
Dear all,
Happy New Year!
I am learning the Induction Proof over Haskell, I saw some proofs for the
equivalence of two functions will have a case called 'bottom' but some of
them do no have. What kind of situation we should also include the bottom
case to the
I've released a new version of the monte-carlo packages for haskell.
Here are the highlights for monte-carlo:
Changes in 0.2:
* More general type class, MonadMC, which allows all the functions to
work
in both MC and MCT monads.
* Functions to sample from discrete distributions.
*
Luke Palmer wrote:
First, by simple definition, id _|_ = _|_. Now let's consider foo _|_.
The Haskell semantics say that pattern matching on _|_ yields _|_, so
foo _|_ = _|_. So they are equivalent on _|_ also. Thus foo and id are
exactly the same function.
Would it in general also be
On Thu, 2009-01-01 at 02:16 +0100, Martijn van Steenbergen wrote:
Luke Palmer wrote:
First, by simple definition, id _|_ = _|_. Now let's consider foo _|_.
The Haskell semantics say that pattern matching on _|_ yields _|_, so
foo _|_ = _|_. So they are equivalent on _|_ also. Thus foo
Thanks for the answers to all.
Untying the knot was (and still is) exactly the problem I was facing.
I knew that the whole list had to be rebuild and wasn't concerned
with performance since at that point I just wanted to know how to
do it and if it is possible at all. After I realized that it
I am afraid I am still confused.
foo [] = ...
foo (x:xs) = ...
There is an implied:
foo _|_ = _|_
The right side cannot be anything but _|_. If it could, then that would
imply we could solve the halting problem:
in a proof, how I could say the right side must be _|_ without defining foo
On Thu, 2009-01-01 at 03:50 +, ra...@msn.com wrote:
I am afraid I am still confused.
foo [] = ...
foo (x:xs) = ...
There is an implied:
foo _|_ = _|_
The right side cannot be anything but _|_. If it could, then that
would imply we could solve the halting problem:
in a
Am Donnerstag, 1. Januar 2009 04:50 schrieb ra...@msn.com:
I am afraid I am still confused.
foo [] = ...
foo (x:xs) = ...
There is an implied:
foo _|_ = _|_
The right side cannot be anything but _|_. If it could, then that would
imply we could solve the halting problem:
in a
On Wed, 2008-12-31 at 22:08 -0600, Jonathan Cast wrote:
On Thu, 2009-01-01 at 03:50 +, ra...@msn.com wrote:
I am afraid I am still confused.
foo [] = ...
foo (x:xs) = ...
There is an implied:
foo _|_ = _|_
The right side cannot be anything but _|_. If it could, then that
I'm pleased to announce the release of gitit-0.4.1, which I've just
uploaded to HackageDB. Gitit is a wiki program that stores pages in
a git repostory.
Gitit now has support for (optional) captchas, using the reCAPTCHA
service. I've packaged up the reCAPTCHA code as a separate library on
I must ask why runWriterT k :: State s (a,[Int]) is working.
Looks like I could runIO the same way I evalState there.
In that case I wouldn't wait for the State s action to finish.
Thanks
2008/12/31 Derek Elkins derek.a.elk...@gmail.com
On Wed, 2008-12-31 at 21:48 +0100, Paolino wrote:
As
How do I read IO is not lazy ?
Is IO (=) forcing the evaluation of its arguments, causing the unwanted
neverending loop?
And, this happens even in (MonadTrans t = t IO) (=) ?
Thanks
paolino
2008/12/31 Ryan Ingram ryani.s...@gmail.com
IO is not lazy; you never make it to print.
Consider
hi all, I want to define a data type Tree a that can either be a or Branch
(Tree a) (Tree a)?
I tried
data Tree a = a | Branch (Tree a) (Tree a) deriving Show
but it seems not accpetable in haskell ?
any way I could achieve this ?
Thanks
On 2009 Jan 1, at 2:32, Max.cs wrote:
data Tree a = a | Branch (Tree a) (Tree a) deriving Show
but it seems not accpetable in haskell ?
You need a constructor in both legs of the type:
data Tree a = Leaf a | Branch (Tree a) (Tree a) deriving Show
--
brandon s. allbery
You need some type constructor:
data Tree a = Leaf a | Branch (Tree a) (Tree a)
Am 01.01.2009 um 08:32 schrieb Max.cs:
hi all, I want to define a data type Tree a that can either be a
or Branch (Tree a) (Tree a)?
I tried
data Tree a = a | Branch (Tree a) (Tree a) deriving Show
but it
30 matches
Mail list logo
