Hi,
I'm a poor lonesome newbie in Haskell world, and I would like to add a string
typed on the prompt to a list of strings which is already defined.
It would look like something like :
type Path = [String]
currentPath::Path
currentPath = []
getpiece ::IO String
getpiece = do c -getLine
On Sunday 17 February 2002 08:20, christophe certain wrote:
Hi,
I'm a poor lonesome newbie in Haskell world, and I would like to add a
string typed on the prompt to a list of strings which is already defined.
It would look like something like :
type Path = [String]
currentPath::Path
You seem to expect currentPath to be updated by putpiece? This won't
happen
in Haskell. Once you've declared
currentPath=[]
it will always be [].
Values never change. If you want the functional equivalent of accumulator
variables they have to be an argument of a recursive
On Sun, 17 Feb 2002, Cagdas Ozgenc wrote:
Hi Adrian,
How can I add a function that sorts this list that I read from the user and
accumulate using the function that you described? I am not asking for a sort
algorithm of course, I am just wondering how to feed the IO Path as an input
to a
On Sun, 17 Feb 2002, Jay Cox wrote:
(snip)
PS: Anybody got any other suggestions for IO monad entry-level docs?
(snip)
Simon's Tackling the Awkward Squad paper was a revelation for me.
-- Mark
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Claus Reinke [EMAIL PROTECTED] writes:
Haskell definitely supports abstraction and composition, so we can
factor out application aspects (not just text) that need localisation,
and link them (dynamically?) with the main parts of our applications.
Some systematic approach would be useful,
Title: Message
hello,
below is the
code that i wrote as an excercise for myself (I am still learning
haskell).
it
implements a straighforward way to simplify boolean expressions, and should be
self-explanatory.
my question
is, if i have an expression such as ((Const False) :: subexp),
konst writes:
my question is, if i have an expression such as ((Const False) ::
subexp), will subexp be reduced first (according to the definition
'simplify (x :: y) = simplify' ((simplify x) :: (simplify y))') or
will laziness do the right thing, and emit (Const False) without looking
into
