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I want to render a scene from multiple positions and orientations. I think the geometry shader is a good choice to not use the Vertex and tessellation shaders for each camera position again. So I want to do layered rendering. I already did that once, but this time I simultaneously want to use additional clip planes. Each camera position has its own clip planes, which should not be visible from other cameras. My question: is it possible to define clipPlanes (glClipPlane) for a specific layer? Or is it better to do that during the fragment shader step, by doing dot-product, check the sign and discard if necessary? Does I have other options (except: rendering the scene in a loop for each position / orientation)?

Thanks a lot!

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The GS is responsible for assigning a particular primitive to a particular layer. The GS is also responsible for writing values for gl_ClipDistance for a particular primitive. So the GS is perfectly capable of doing both of those things: assigning a primitive to a layer and doing the clip-plane computations for that layer.

And if you need to "deactivate" a clip plane for a particular primitive assigned to a particular layer, just assign the gl_ClipDistance[i] for that to 1.0 for all of its vertices.

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  • $\begingroup$ only to be sure... do you meen gl_ClipDistance[] instead of gl_ClipPlane[]? Does the distance value be interpolated for the triangle during the rasterization step? $\endgroup$
    – Thomas
    Feb 4, 2021 at 9:02
  • $\begingroup$ @Thomas: I fixed the variable. And all primitive vertex outputs are interpolated (unless you declare them to not be interpolated with the flat qualifier). $\endgroup$
    – Nicol Bolas
    Feb 4, 2021 at 14:45
  • $\begingroup$ Thanks a lot! Then I'll try to implement it that way =) $\endgroup$
    – Thomas
    Feb 4, 2021 at 15:53

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