Can a GLSL fragment shader run without a framebuffer and similar inconveniences? - c

Question

Reapeating the above: Can a GLSL fragment shader run without a framebuffer and any rasterization stage?
This perfect answer gives an insight about where to start with SSBO's. The answer has a link to OpenGL ARB extension that has a boilerplate code. The code works for me if made with some changes to work with OpenGL compute programs. But, I really does not get it, how to do with a fragment program? And without any other buffers than SSBO.
The code clearly has fragment source code without any pixel operations, only SSBO ones.
in vec4 color;
void main()
{
uint fragmentNumber = atomicCounterIncrement(fragmentCounter);
if (fragmentNumber < maxFragmentCount) {
fragments[fragmentNumber].position = ivec2(gl_FragCoord.xy);
fragments[fragmentNumber].color = color;
}
}
And later in the C program file:
// Generate, bind, and specify the data store for the atomic counter.
glGenBuffers(1, &counterBuffer);
glBindBufferBase(GL_ATOMIC_COUNTER_BUFFER, 0, counterBuffer);
glBufferData(GL_ATOMIC_COUNTER_BUFFER, sizeof(GLuint), NULL,
GL_DYNAMIC_DRAW);
// Reset the atomic counter to zero, then draw stuff. This will record
// values into the shader storage buffer as fragments are generated.
GLuint zero = 0;
glBufferSubData(GL_ATOMIC_COUNTER_BUFFER, 0, sizeof(GLuint), &zero);
glUseProgram(program);
glDrawElements(GL_TRIANGLES, ...);
As per my setup, I do not have any output with the means of OpenGL pixels. I wish it to stay so. Is it possible, or am I missing something?
P.S The above setup gives me error invalid framebuffer operation after glDrawElements immediately followed by glFinish.
Update 21.03.2021
There is a Framebuffers with no attachments. The only thing you should set in its state is its width and height. And that is somewhat at the course that anyone's heading, if one wish to minimize setup.
The minus of the aformentioned, is that it is still requires some geometry to be fed to rasterization stage. To start the shader stages, you know. But, as a plus, one gets geometry rasterization, wish it or not.
If I have time, I leave some code as a reminder for miself.

Answer 1

Can a GLSL fragment shader run without a framebuffer and similar inconveniences?
No. The fragment shaders need the step that invokes them. The stage that produce fragments called rasterization.
From the khronos wiki:
A Fragment Shader is the Shader stage that
will process a Fragment generated by the Rasterization
into a set of colors and a single depth value.
The fragment shader is the OpenGL pipeline stage after a primitive is rasterized.
And the rasterization needs a render step to produce fragments. The rendering is done to somewhere.
In OpenGL, it is done to framebuffer. So without a framebuffer, you can not render, hence OpenGL
can not produce fragments.
Setup of a framebuffer can be minimized by
Framebuffers with no attachments.
But one needs to supply geometry and render it, to invoke fragment shaders.
Fragment shaders can read and write to arbitrary SSBO. But the usage is not similar to compute shaders.
Fragment shaders invoke on each produced fragment, and compute shaders can be invoked, as I may say, arbitrary.
Many thanks to all commenters who had pointed me to the obvious, by now, reason why the fragment shaders need a render operation.