Re: [boost] [odeint] Iterator semantics

Subject: Re: [boost] [odeint] Iterator semantics
From: Jeffrey Lee Hellrung, Jr. (jeffrey.hellrung_at_[hidden])
Date: 2012-07-12 02:50:36

On Wed, Jul 11, 2012 at 11:21 PM, Karsten Ahnert <karsten.ahnert_at_[hidden]
> wrote:

> On 07/12/2012 12:39 AM, Jeffrey Lee Hellrung, Jr. wrote:
> > On Wed, Jul 11, 2012 at 2:57 PM, Karsten Ahnert
> > <karsten.ahnert_at_[hidden]>wrote:
> >
> >> On 07/11/2012 07:44 PM, Jeffrey Lee Hellrung, Jr. wrote:
> >>> On Wed, Jul 11, 2012 at 9:04 AM, Karsten Ahnert
> >>> <karsten.ahnert_at_[hidden]>wrote:
>
[...]

> >>> This is problematic since equality here means that the time
> >>>> (t) of the begin iterator is smaller then the time of the end
> iterator.
> >>>>
> >>>
> >>> Can you elaborate? I don't understand why this is what equality means.
> >>
> >> One needs to implement the != operator (or equal if you use
> >> Boost.Iterator). A naive implementation of operator!= (or the equal() if
> >> you use boost.iterator) could check that the current time of the ODE is
> >> smaller then the time of other iterator:
> >>
> >> bool equal( iterator other )
> >> {
> >>     return this->t < other->t;
> >> }
> >>
> >
> > :: confused :: I would think iterator1 op iterator2 is equivalent to
> time1
> > op time2.
>
> No, this is not the case iterator != iterator2 is equivalaent to time1 <
> time2 . The first operation is commutative the second one not.
>

Well, you can define operator== and operator!= however you want, and some
definitions will make sense and some won't (e.g., non-/anti-symmetry).
*All* I've been trying to do here is coax out of you *why* you're defining
operator== and operator!= this way :)

(Especially, with my limited understanding of the application, when it
seems time1 op time2 would do what you want, or perhaps a fuzzy version
taking into account the time step.)

[...]

- Jeff