CGAL users discussion list

["Bart Janssens" <>]

On 5/7/08, samuel kim 
<>
 wrote:
> If I did as the above, the handle of facet was that of triangle#1. So I want
> to ask you a question, when I do this,
>       hf_new = pHalfedge->prev()->opposite()->facet(); // expecting to get 
> the
> handle of facet of triangle#2
>       pHalfedge = hf_new->facet_begin(); // expecting to get the handle of
> half_edge2 in the triangle#2
> Which edge handle should be returned in triangle#2? 

Hi Samuel,

I don't think you need to get the facets when changing between triangles. 
Instead, you can just do
facet_2_edge_2 = pHalfedge->prev()->opposite()
facet_3_edge_2 = facet_2_edge_2 ->next()->opposite()

If you are sure you get the right edge in facet 1, that should work.

> And what is the base rule to get to know the exact pointer of the handle of
> half_edge data structure whenever travel triangles? Because I have tried
> many times, and found that the pointer of handle of half_edge data structure
> was varying at each triangle which was not expectable.

Not sure what you want to do here. I think the CGAL way is to stick with the 
handles, and avoid pointers.

Cheers,

-- 
Bart Janssens