Opengl double overflow

Question

I have double 'radius' = 2.0E-45, when i set it to ~2.0E-46 calculation collapse resulting in white screen. So seems like issue is overflow. I wrote the same algorithm but using nubma cuda and f64 (double precision) 'radius'. And everything works fine. I am using f32 texture buffer for 'depth_array' (there is no float64 dtype for this), but numba implementation works fine with f32, and opengl implementation also works fine until 'radius' bigger than ~2.0E-46. Why numba implementation works, while opengl not? I want to stick with opengl. Is there any possibility to fix it?

I only put in parts that use the 'radius'. All other variables are double type. (code is messy and just a scratch)

#version 150
#extension GL_ARB_gpu_shader_fp64 : enable

double radius = 2.0E-45;

...

dvec2 pixel = dvec2(gl_FragCoord.xy) + dvec2(-0.5+(double(x)+0.5)/double(AA),-0.5+(double(y)+0.5)/double(AA));
dvec2 c = pixel/dvec2(width, height) * dvec2(radius, radius) + dvec2(-radius/2, -radius/2);
color.rgb += sample(c);

...

vec3 sample(dvec2 dn)
    {
        vec3 color = vec3(0.0,0.0,0.0);
        dvec2 d0 = dn;

        double zn_size = 0.0;
        int i = 0;
        while (i < depth)
        {
            int x = i % depth;
            dvec2 value = dvec2(texelFetch(depth_array, x).rg);
            dn = complex_mul(dn, value + dn);
            dn = dn + d0;

            i++;
            x = i % depth;
            value = dvec2(texelFetch(depth_array, x).rg);
            dvec2 zn = value * 0.5 + dn;

            zn_size = dot(zn, zn);
            if (zn_size > r)
            {
                double fraciter = (zn_size-r)/(r2-r);
                double iter = double(i) - fraciter;
                double m = sqrt(iter)*mul*2.0;

                color = sin(vec3(.1, .15, .2)*float(m)*0.5)*.5+0.5;
                break;
            }
        }
        return color;
    }

Nicol Bolas · Accepted Answer · 2019-10-06 13:38:15Z

3

In GLSL, the literal value 2.0E-45 has the type float. That means the value will be squashed into the valid range of a float before it gets assigned to a value.

If you want a literal to be a double, then it needs to use the proper suffix: 2.0E-45lf.

answered Oct 6, 2019 at 13:38

Nicol Bolas

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1 Comment

Егор Коханов Over a year ago

Yeah, thx that worked. I forgot to mention that i tried to do double radius = 0.00...(a lot of zeros)...0002 and it didn't work.

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