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ac/nir: optimize out multiplications in small line culling

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Reviewed-by: Timur Kristóf <timur.kristof@gmail.com>
Acked-by: Pierre-Eric Pelloux-Prayer <pierre-eric.pelloux-prayer@amd.com>
Part-of: <https://gitlab.freedesktop.org/mesa/mesa/-/merge_requests/26917>
This commit is contained in:
Marek Olšák 2024-01-03 09:49:07 -05:00 • committed by Marge Bot
parent 0d74123633
commit 4747bd6ba8
1 changed files with 24 additions and 18 deletions
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42
src/amd/common/ac_nir_cull.c
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@ -164,19 +164,31 @@ ac_nir_cull_triangle(nir_builder *b,
static void
rotate_45degrees(nir_builder *b, nir_def *v[2])
{
/* sin(45) == cos(45) */
nir_def *sincos45 = nir_imm_float(b, 0.707106781);
/* x2 = x*cos45 - y*sin45 = x*sincos45 - y*sincos45
* y2 = x*sin45 + y*cos45 = x*sincos45 + y*sincos45
/* Rotating a triangle by 45 degrees:
*
* x2 = x*cos(45) - y*sin(45)
* y2 = x*sin(45) + y*cos(45)
*
* Since sin(45) == cos(45), we can write:
*
* x2 = x*cos(45) - y*cos(45) = (x - y) * cos(45)
* y2 = x*cos(45) + y*cos(45) = (x + y) * cos(45)
*
* The width of each square (rotated diamond) is sqrt(0.5), so we have to scale it to 1
* by multiplying by 1/sqrt(0.5) = sqrt(2) because we want round() to give us the position
* of the closest center of the square (rotated diamond). After scaling, we get:
*
* x2 = (x - y) * cos(45) * sqrt(2)
* y2 = (x + y) * cos(45) * sqrt(2)
*
* Since cos(45) * sqrt(2) = 1, we get:
*
* x2 = x - y
* y2 = x + y
*/
nir_def *first = nir_fmul(b, v[0], sincos45);
/* Doing 2x ffma while duplicating the multiplication is 33% faster than fmul+fadd+fadd. */
nir_def *result[2] = {
nir_ffma(b, nir_fneg(b, v[1]), sincos45, first),
nir_ffma(b, v[1], sincos45, first),
};
nir_def *result[2];
result[0] = nir_fsub(b, v[0], v[1]);
result[1] = nir_fadd(b, v[0], v[1]);
memcpy(v, result, sizeof(result));
}
@ -254,12 +266,6 @@ cull_small_primitive_line(nir_builder *b, nir_def *pos[3][4],
nir_def *rounded_to_eq[2];
for (unsigned chan = 0; chan < 2; chan++) {
/* The width of each square is sqrt(0.5), so scale it to 1 because we want
* round() to give us the position of the closest center of a square (diamond).
*/
v0[chan] = nir_fmul_imm(b, v0[chan], 1.414213562);
v1[chan] = nir_fmul_imm(b, v1[chan], 1.414213562);
/* Compute the bounding box around both vertices. We do this because we must
* enlarge the line area by the precision of the rasterizer.
*/