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 d
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 argumen
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)
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
-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 DT
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 function
