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On Wednesday 26 April 2006 04:36 pm, Will Entriken wrote:
> i think that looks find, unless you know the first value of d, before
> you enter the loop, just make it that.
>
> good luck
> -WE
>
> On 4/26/06, Dmitry Suzdalev <dimsuz@gmail.com> wrote:
> > On Thursday 27 April 2006 00:12, Will Entriken wrote:
> > > std::numeric_limits<double>::min() return the minimum positive value.
> >
> > Hmm... yes? :)
> > Bahhh, I should read docs more carefully, thanks! :)
> >
> > > you could try maxd =3D -1 if d will be non-negative. that will be pre=
tty
> > > obvious what is going on.
> >
> > Nope, d might be negative - that's why I didn't use -1 for maxd.
> > So, then another question:
> > Maybe I'm wrong when I say that
> > "-std::numeric_limits<double>::max()" is messy-looking?
> >
> > Now that I know (thanks, Will), that min() return minimum _positive_
> > value, I think that using -max() is all right. Is this so?
> >
> > Thanks!
> > Dmitry.
> >
> > >> Visit http://mail.kde.org/mailman/listinfo/kde-devel#unsub to
> > >> unsubscribe <<
> >>
> >> Visit http://mail.kde.org/mailman/listinfo/kde-devel#unsub to
> >> unsubscribe <<

Will can you please not top post? It breaks quoting and my train of though =
on=20
the matter....

=2D-=20
Gary L. Greene, Jr.
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