Yes, which is what I wrote:
"What is the idea behind having a color in ShapeDrawable?"
I cant see any reason why there would be a color associated to a shape?
Create a triangle, and by the way, its red. That was my question.
So if you WOULD like to control the color by disabling light and use
diffuse, it will not work.
It will be white (due to the call in ShapeDrawable).
Anyway, there are ways around this. I just thought it was very inconsistent
to associate a color to a ShapeDrawable.
/Anders
On Fri, Dec 2, 2011 at 9:53 AM, Filip Arlet <fili...@seznam.cz> wrote:
> Hi,
>
> in OpenGL if GL_LIGHTING is disabled, the final color of polygon is
> determined by glColor.
>
> If you see ShapeDrawable::drawImplemenation(), there is glColor call.
>
>
> Code:
>
> osg::State& state = *renderInfo.getState();
> GLBeginEndAdapter& gl = state.getGLBeginEndAdapter();
>
> if (_shape.valid())
> {
> gl.Color4fv(_color.ptr()); ///// <---------
>
> DrawShapeVisitor dsv(state,_tessellationHints.get());
>
> _shape->accept(dsv);
> }
>
>
>
>
> Cheers,
> Filip[/code]
>
> ------------------
> Read this topic online here:
> http://forum.openscenegraph.org/viewtopic.php?p=44181#44181
>
>
>
>
>
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> osg-users@lists.openscenegraph.org
> http://lists.openscenegraph.org/listinfo.cgi/osg-users-openscenegraph.org
>
--
__________________________________________
Anders Backman, HPC2N
90187 Umeå University, Sweden
and...@cs.umu.se http://www.hpc2n.umu.se
Cell: +46-70-392 64 67
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