Hi,
I'm a bit puzzled by this observatio that I made. I have a function
that, pseudocoded, lookes somewhat like
f i as bs cs = ins i (f (i+1) as) ++ ins i (f (i+1) bs) ++ ins i (f (i+1) cs)
where ins i = manipulates the first element of the list
Now, without the ins'es, the
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
I've got a debian install (2.2.18kernel, 2.95.4 gnu c compiler) with ghc
installed using hmake. I also found DtdToHaskell was installed on my
computer. I run it on any Dtd and get the following:
spg@Further:~/clean$ DtdToHaskell clean.dtd
module
On 18 Feb 2002, Ketil Z. Malde wrote:
Hi,
I'm a bit puzzled by this observatio that I made. I have a function
that, pseudocoded, lookes somewhat like
f i as bs cs = ins i (f (i+1) as) ++ ins i (f (i+1) bs) ++ ins i (f (i+1) cs)
where ins i = manipulates the first element of the
On Mon, 18 Feb 2002, Jay Cox wrote:
On 18 Feb 2002, Ketil Z. Malde wrote:
Hi,
I'm a bit puzzled by this observatio that I made. I have a function
that, pseudocoded, lookes somewhat like
f i as bs cs = ins i (f (i+1) as) ++ ins i (f (i+1) bs) ++ ins i (f (i+1) cs)
where
Artie Gold writes:
One way to think of it is to look at a program as a partially ordered
set of calculations; some calculations need to occur before others,
other groups can occur in any order. In an imperative language you
specify a total ordering (which is overkill).
This is a weak argument.
Jay Cox [EMAIL PROTECTED] writes:
where ins i = manipulates the first element of the list
if you mean that (ins i) :: [a] - [a] manipulates the first element of
the list it takes then of course it is strict. because in
It is strict in the head of the list, yes. I.e. it is defined
