How are vertices transformed in WebGL in indexed and non-indexed geometries?

Question

I am trying to digest these two links:

https://www.khronos.org/opengl/wiki/Rendering_Pipeline_Overview https://www.khronos.org/opengl/wiki/Vertex_Shader

The pipeline overview says that vertex shader runs before the primitive assembly.

The second one mentions this:

A vertex shader is (usually) invariant with its input. That is, within a single Drawing Command, two vertex shader invocations that get the exact same input attributes will return binary identical results. Because of this, if OpenGL can detect that a vertex shader invocation is being given the same inputs as a previous invocation, it is allowed to reuse the results of the previous invocation, instead of wasting valuable time executing something that it already knows the answer to.

OpenGL implementations generally do not do this by actually comparing the input values (that would take far too long). Instead, this optimization typically only happens when using indexed rendering functions. If a particular index is specified more than once (within the same Instanced Rendering), then this vertex is guaranteed to result in the exact same input data.

Therefore, implementations employ a cache on the results of vertex shaders. If an index/instance pair comes up again, and the result is still in the cache, then the vertex shader is not executed again. Thus, there can be fewer vertex shader invocations than there are vertices specified.

So if i have two quads with two triangles each:

indexed:

verts: { 0 1 2 3 }
tris:  { 0 1 2 }
         { 1 2 3 }

soup:

verts: { 0 1 2 3 4 5 }
tris:  { 0 1 2 } 
             { 3 4 5 }

and perhaps a vertex shader that looks like this:

uniform mat4 mvm;
uniform mat4 pm;

attribute vec3 position;

void main (){
  vec4 res;
  for ( int i = 0; i < 256; i++ ){
     res = pm * mvm * vec4(position,1.);
  }
gl_Position = res;

Should I care that one has 4 vertices while the other one has 6? Is this even true from gpu to gpu, will one invoke the vertex shader 4 times vs 6? How is this affected by the cache:

If an index/instance pair comes up again, and the result is still in the cache...

How is the primitive number related to performance here? In both cases i have the same amount of primitives.

In the case of a very simple fragment shader, but an expensive vertex shader:

void main(){
  gl_FragColor = vec4(1.);
}

And a tessellated quad (100x100 segments) can i say that the indexed version will run faster, or can run faster, or maybe say nothing?

gman gman · Accepted Answer · 2018-02-15T00:47:24

Like everything in GPUs according to the spec you can say nothing. It's up to the driver and GPU. In reality though in your example 4 vertices will run faster than 6 pretty much everywhere?

Search for vertex order optimization and lots of articles come up

Linear-Speed Vertex Cache Optimisation

Triangle Order Optimization

AMD Triangle Order Optimization Tool

Triangle Order Optimization for Graphics Hardware Computation Culling

unrelated but another example of the spec vs realtiy is that according to the spec depth testing happens AFTER the fragment shader runs (otherwise you couldn't set gl_FragDepth in the fragment shader. In reality though as long as the results are the same the driver/GPU can do whatever it wants so fragment shaders that don't set gl_FragDepth or discard certain fragments are depth tested first and only run if the test passes.

How are vertices transformed in WebGL in indexed and non-indexed geometries?

1 Answers