On Sat, Dec 08, 2007 at 02:59:16PM -0200, Felipe Lessa wrote:
Hello!
I see from
http://www.haskell.org/haskellwiki/Monads_as_computation#Do_notation
that
do { v - x ; stmts }
= x = \v - do { stmts }
However, look at this GHCi session:
Prelude let return' = return :: a - Maybe a
On Sat, Dec 08, 2007 at 03:28:58PM -0200, Felipe Lessa wrote:
On Dec 8, 2007 3:12 PM, Ilya Tsindlekht [EMAIL PROTECTED] wrote:
On Sat, Dec 08, 2007 at 02:59:16PM -0200, Felipe Lessa wrote:
Prelude do {1 - return 3; return' ok}
Nothing
Prelude return 3 = \1 - return' ok
*** Exception
On Sun, Sep 16, 2007 at 06:11:03PM +0200, B K wrote:
Hello,
Are there any Haskell Hackers on this mailing list who live in Israel?
I am interested in starting an Israel Haskell User Group.
I am here, although I probably do not really count for Haskell Hacker.
On Thu, Jul 05, 2007 at 12:58:06AM +1000, Alexis Hazell wrote:
On Tuesday 03 July 2007 09:51, Arie Peterson wrote:
No, there is a 'State s' monad provided (for arbitrary state type 's'),
which implements the 'get' and 'put' methods. In other words, 'State s' is
an instance of the
On Tue, Jul 03, 2007 at 10:53:33AM +, Peter Verswyvelen wrote:
Ah, thanks for the correction. So if I understand it correctly, this is
currying:
when
f :: (a,b) - c
then
g :: a - (b,c)
g :: a-b-c
is the curried form of f? So currying has to do with tuples?
And partial
On Wed, Jun 13, 2007 at 02:37:37PM +0100, PR Stanley wrote:
Hi
Can you think of a fourth way of redefining disjunct using pattern matching?
vee :: Bool - Bool - Bool
vee _ True = True
vee True _ = True
vee _ _ = False
ve :: Bool - Bool - Bool
ve True True = True
ve True False = True
ve
On Wed, Jun 06, 2007 at 12:23:03PM +1200, Daniel McAllansmith wrote:
Hello.
I've got a system of linear inequalities representing half-spaces. The
half-spaces may or may not form a convex hull.
I need to find the integral coordinates that are contained within the convex
hull, if there
On Wed, Jun 06, 2007 at 01:39:32PM +0400, Dmitri O.Kondratiev wrote:
Monad class contains declaration
*fail* :: String - m a
and provides default implementation for 'fail' as:
fail s = error s
On the other hand Prelude defines:
*
error* :: String - a
which stops execution and
On Wed, Jun 06, 2007 at 03:48:18PM +0200, Steffen Mazanek wrote:
Hello,
is there a function f::[a-b]-a-[b] in the libraries? Couldn't find one
using
hoogle although this seems to be quite a common thing...
Steffen
Just to add to what others have said, yet another way to implement it
On Mon, Jun 04, 2007 at 05:42:35PM +0300, Juozas Gaigalas wrote:
Hello,
I am a somewhat experienced programmer and a complete Haskell newbie, so I
hope this is the correct ML for my question.
I have decided to learn Haskell and started with Graham Hutton's book.
Everything was going
On Tue, Jun 05, 2007 at 10:41:44AM +0100, Rodrigo Queiro wrote:
To back him up, it seems that the lhaskell.vim syntax highlighter is broken
with Vim 7.1. Here, it is definitely using lhaskell.vim, but doesn't seem to
be parsing the code in between \begin{code} and \end{code} as Haskell.
Works
On Sun, May 27, 2007 at 05:34:33PM -0400, Brandon S. Allbery KF8NH wrote:
On May 27, 2007, at 17:23 , Andrew Coppin wrote:
Personally, I try to avoid ever using list comprehensions. But
every now and then I discover an expression which is apparently not
expressible without them - which
On Sat, May 19, 2007 at 04:10:29PM +0100, Jon Fairbairn wrote:
[...]
foldl f z l = case l of
(x:xs) - foldl f (f z x) xs
[] - z
which reveals that foldl in the RHS isn't really the
leftmost outermost function, but case is -- the tail call is
to case. It
On Sat, May 19, 2007 at 09:16:46PM +0100, Jon Fairbairn wrote:
Ilya Tsindlekht [EMAIL PROTECTED] writes:
By definition, tail recursive function is recursive function in which
the value returned from a recursive call is immediately returned without
modification as value of the top-level
On Sat, May 05, 2007 at 12:09:03AM -0700, [EMAIL PROTECTED] wrote:
Ilya Tsindlekht wrote:
Does the definition of monad silently assume that if f and f' are equal
in the sense that they return the same value for any argument o correct
type then m = f = m = f'
Of course NOT! Here's
Does the definition of monad silently assume that if f and f' are equal
in the sense that they return the same value for any argument o correct
type then m = f = m = f'
More specifically, the definition says x = return = x. How does one
justify from this that x = (return . id) = x?
Are values of
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