ccpr: Remove constant scale when solving the cubic midtangent
We only care about the direction of the tangent vector, not the the
magnitude, so a 3x scale is irrelevant.
Bug: skia:
Change-Id: Icd2d82faf2c700fc794f8d4a59f21b32398b64f0
Reviewed-on: https://skia-review.googlesource.com/142245
Reviewed-by: Greg Daniel <egdaniel@google.com>
Commit-Queue: Chris Dalton <csmartdalton@google.com>
diff --git a/src/gpu/ccpr/GrCCGeometry.cpp b/src/gpu/ccpr/GrCCGeometry.cpp
index 30d93ad..17a54af 100644
--- a/src/gpu/ccpr/GrCCGeometry.cpp
+++ b/src/gpu/ccpr/GrCCGeometry.cpp
@@ -638,9 +638,7 @@
// This function finds the T value whose tangent angle is halfway between the tangents at T=0 and
// T=1 (tan0 and tan1).
static inline float find_midtangent(const Sk2f& tan0, const Sk2f& tan1,
- float scale2, const Sk2f& C2,
- float scale1, const Sk2f& C1,
- float scale0, const Sk2f& C0) {
+ const Sk2f& C2, const Sk2f& C1, const Sk2f& C0) {
// Tangents point in the direction of increasing T, so tan0 and -tan1 both point toward the
// midtangent. 'n' will therefore bisect tan0 and -tan1, giving us the normal to the midtangent.
//
@@ -660,9 +658,6 @@
Sk4f C[2];
Sk2f::Store4(C, C2, C1, C0, 0);
Sk4f coeffs = C[0]*n[0] + C[1]*n[1];
- if (1 != scale2 || 1 != scale1 || 1 != scale0) {
- coeffs *= Sk4f(scale2, scale1, scale0, 0);
- }
// Now solve the quadratic.
float a = coeffs[0], b = coeffs[1], c = coeffs[2];
@@ -682,9 +677,9 @@
const Sk2f& p3, const Sk2f& tan0,
const Sk2f& tan1,
int maxFutureSubdivisions) {
- float midT = find_midtangent(tan0, tan1, 3, p3 + (p1 - p2)*3 - p0,
- 6, p0 - p1*2 + p2,
- 3, p1 - p0);
+ float midT = find_midtangent(tan0, tan1, p3 + (p1 - p2)*3 - p0,
+ (p0 - p1*2 + p2)*2,
+ p1 - p0);
// Use positive logic since NaN fails comparisons. (However midT should not be NaN since we cull
// near-flat cubics in cubicTo().)
if (!(midT > 0 && midT < 1)) {
@@ -715,9 +710,9 @@
// tangent line. Since the denominator scales dx and dy uniformly, we can throw it out
// completely after evaluating the derivative with the standard quotient rule. This leaves
// us with a simpler quadratic function that we use to find the midtangent.
- float midT = find_midtangent(tan0, tan1, 1, (w - 1) * (p2 - p0),
- 1, (p2 - p0) - 2*w*(p1 - p0),
- 1, w*(p1 - p0));
+ float midT = find_midtangent(tan0, tan1, (w - 1) * (p2 - p0),
+ (p2 - p0) - 2*w*(p1 - p0),
+ w*(p1 - p0));
// Use positive logic since NaN fails comparisons. (However midT should not be NaN since we
// cull near-linear conics above. And while w=0 is flat, it's not a line and has valid
// midtangents.)