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#Actualsamoth

Posted 18 March 2013 - 11:18 AM

Rendering is almost always fragment shader or ROP bound (or geometry shader bound, if there's enough output), and only very rarely vertex shader bound. Fragment shader work, as pointed out by mhagain, does not apply.

Therefore, you can pretty much consider the cost "zero".

This is true even more so as degenerate triangles reuse vertex indices of adjacent non-degenerate triangles. Which means they will be transformed, go into the post-transform-cache, and be reused. You transform them once, and you would need to transform them once anyway.

 

So the only real cost is 3 extra indices in the index buffer, which is neglegible both memory and bandwidth wise.


#4samoth

Posted 18 March 2013 - 11:17 AM

Rendering is almost always fragment shader or ROP bound (or geometry shader bound, if there's enough output), and only very rarely vertex shader bound. Fragment shader work, as pointed out by mhagain, does not apply.

Therefore, you can pretty much consider the cost "zero".

This is true even more so as degenerate triangles reuse vertex indices of adjacent non-degenerate triangles. Which means they will be transformed, go into the post-transform-cache, and be reused. You transform them once, and you would need to transform them once anyway. So the only real cost is 3 extra indices in the index buffer, which is neglegible both memory and bandwidth wise.

#3samoth

Posted 18 March 2013 - 11:16 AM

Rendering is almost always fragment shader or ROP bound (or geometry shader bound, if there's enough output), and only very rarely vertex shader bound.

Therefore, you can pretty much consider the cost "zero".

This is true even more so as degenerate triangles reuse vertex indices of adjacent non-degenerate triangles. Which means they will be transformed, go into the post-transform-cache, and be reused. You transform them once, and you would need to transform them once anyway. So the only real cost is 3 extra indices in the index buffer, which is neglegible both memory and bandwidth wise.

#2samoth

Posted 18 March 2013 - 11:15 AM

Rendering is almost always fragment shader or ROP bound (or geometry shader bound, if there's enough output), and only very rarely vertex shader bound.

Therefore, you can pretty much consider the cost "zero".

This is true even more so as degenerate triangles reuse vertex indices of adjacent non-degenerate triangles. Which means they will be transformed, go into the post-transform-cache, and be reused. You transform them once, and you would need to transform them once anyway.

#1samoth

Posted 18 March 2013 - 11:14 AM

Rendering is almost always fragment shader or ROP bound (or geometry shader bound, if there's enough output), and only very rarely vertex shader bound. Therefore, you can pretty much consider the cost "zero". This is true even more so as degenerate triangles reuse vertex indices of adjacent non-degenerate triangles. Which means they will be transformed, go into the post-transform-cache, and be reused. You transform them once, and you would need to transform them once anyway.

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