c++ - OpenGL: How to render perfect rectangular gradient?

Question

I can render triangular gradient with simply just one triangle and using glColor for each corner.

But how to render perfect rectangular gradient? I tried with one quad, but the middle will get ugly seam. I also tried with texture of 2x2 size, it was like it should be done: proper blending from each corner, but the texture sampling precision becomes unprecise when stretched too much (i started to see pixels bigger than 1x1 size).

Is there some way of calculating this in a shader perhaps?

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Edit: Link to images were broken(removed).

Answer 1

Indeed, the kind of gradient you want relies on 4 colors at each pixel, where OpenGL typically only interpolates input over triangles (so 3 inputs). Getting the perfect gradient is not possible just with the standard interpolants.

Now, as you mentioned, a 2x2 texture can do it. If you did see precision issues, I suggest switching the format of the texture to something that typically requires more precision (like a float texture).

Last, and as you mentioned also in your question, you can solve this with a shader. Say you pass an extra attribute per-vertex that corresponds to (u,v) = (0,0) (0,1) (1,0) (1,0) all the way to the pixel shader (with the vertex shader just doing a pass-through).

You can do the following in the pixel shader (note, the idea here is sound, but I did not test the code):

Vertex shader snippet:

varying vec2 uv;
attribute vec2 uvIn;

uv = uvIn;

Fragment shader:

uniform vec3 color0;
uniform vec3 color1;
varying vec2 uv;

// from wikipedia on bilinear interpolation on unit square:
// f(x,y) = f(0,0)(1-x)(1-y) + f(1,0)x(1-y) + f(0,1)(1-x)y + f(1,1) xy. 
// applied here:
// gl_FragColor = color0 * ((1-x)*(1-y) + x*y) + color1*(x*(1-y) + (1-x)*y)
// gl_FragColor = color0 * (1 - x - y + 2 * x * y) + color1 * (x + y - 2 * x * y)
// after simplification:
// float temp = (x + y - 2 * x * y);
// gl_FragColor = color0 * (1-temp) + color1 * temp;
gl_FragColor = mix(color0, color1, uv.u + uv.v - 2 * uv.u * uv.v);