> Hi WanMil,
>
> I am hitting a problem in BoundaryUtil.splitToElements()
>
> I create a boundary object with
> Boundary boundary = new Boundary(nodeElem.area,
> nodeElem.boundary.getTags());
>
> Now I want to verify it with
> List<BoundaryElement> boundaryElements = boundary.getBoundaryElements();
> if (boundaryElements.size() == 0)
> continue;
>
> My problem: I am hitting this assertion in BoundaryUtil.splitToElements():
>
> assert bElements.get(0).isOuter() : log.threadTag()+" first element is not
> outer. "+ bElements;
>
> The area looks like this:
> [Curve[0, (448156.0, 2442187.0), (448156.0, 2442187.0), D], Curve[1,
> (448156.0, 2442187.0), (448177.0, 2442274.0), D], Curve[1, (448177.0,
> 2442274.0), (448176.4834437087, 2442270.543046358), U], Curve[1,
> (448176.48344370874, 2442270.543046358), (448156.1621621622, 2442187.0), U]]
>
> Can you tell me what's wrong with that? The area is split into one list of
> coords, and that describes a way that is considered to be anti-clockwise:
> [2442187/448156, 2442274/448177, 2442271/448176, 2442187/448156]
>
> I am not sure where I should search the error.
>
> Ciao,
> Gerd
Hi Gerd,
it is possible that a boundary has a hole created by an inner way in a
multipolygon. The
List<List<Coord>> areaElements = Java2DConverter.areaToShapes(area);
call converts the area into distinct polygons where clockwise polygons
describe outer ways and counterclockwise polygons describe inner way
(holes). As it is mentioned in the comment the areaElements list starts
with the most inner polygon and continues with more and more outer
polygons. Therefore the last polygon is the most outer polygon (and also
the biggest one). As the biggest polygon cannot be an inner polygon it
must be an outer one. This is the check.
I haven't played through your testcase but either the Java2DConverter
has a bug, or the code
if (singleElement.size() <= 3) {
// need at least 4 items to describe a polygon
continue;
}
removes the last outer element or the input area is defect (I don't know
if that's ever possible).
WanMil
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