Im trying to understand how a fragment shader is invoked during a multisampled rendering, Ive carefully read the specs but Im a bit confused..
I setup a multisampled FBO with a color and a depth/stencil renderbuffers.. fine
I setup a simple 1-line vertex shader just for transform and another 1-line fragment shader which just returns a black color.. and its fine
I draw a line - the line is displayed smoothed.. fine
now, If in the fragment shader, instead of dumbly returning a black color, I texelFetch from a multisampled texture, always reading sample number 0 (that is, black), I get a jagged line..
whats the difference between a simple myoutcolor=vec4(0.0) and a myoutcolor=texelFetch(mysampler2DMS,...,0) ?
the specs state that a fragment shader is executed at sample-level only if it statically uses gl_SampleID, gl_Samplemask[], or has a "sample" storage qualifiers.. (in all these cases this means GLSL #version 400 or above), otherwise it gets executed at fragment level..
but how can that be? in the above example, I got a smoothed line even if I always returned the same color, so I tought the fragment shader was ALWAYS executed at sample level in a multisampled framebuffer.. on the other hand, if it was not the case, I should get smooth results even if returning always the same sample from a MS texture..
can somebody help me figure this out?
PS: Im currently using #version 330, but I found that every multisampling-aware statement Id need to correctly handle MS (gl_SampleID, gl_Samplemask[], sample..) is only supported in GLSL 400 or above.. so, whats exactly the support GLSL 330 offer for correctly handling multisampling? how can I handle MS textures in GLSL 330?
Because that's what multisampling is.
Multisampling is all about invoking the FS once per pixel and distributing that value over the samples within that pixel area that the primitive covers. So if the FS is executed for a pixel on the edge of the primitive, then the result will be distributed to only some of the samples of that pixel.