SkDCubicToQuads.cpp 1.3 KB

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  1. /*
  2. * Copyright 2015 Google Inc.
  3. *
  4. * Use of this source code is governed by a BSD-style license that can be
  5. * found in the LICENSE file.
  6. */
  7. /*
  8. http://stackoverflow.com/questions/2009160/how-do-i-convert-the-2-control-points-of-a-cubic-curve-to-the-single-control-poi
  9. */
  10. /*
  11. Let's call the control points of the cubic Q0..Q3 and the control points of the quadratic P0..P2.
  12. Then for degree elevation, the equations are:
  13. Q0 = P0
  14. Q1 = 1/3 P0 + 2/3 P1
  15. Q2 = 2/3 P1 + 1/3 P2
  16. Q3 = P2
  17. In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways to compute P1 from
  18. the equations above:
  19. P1 = 3/2 Q1 - 1/2 Q0
  20. P1 = 3/2 Q2 - 1/2 Q3
  21. If this is a degree-elevated cubic, then both equations will give the same answer for P1. Since
  22. it's likely not, your best bet is to average them. So,
  23. P1 = -1/4 Q0 + 3/4 Q1 + 3/4 Q2 - 1/4 Q3
  24. */
  25. #include "src/pathops/SkPathOpsCubic.h"
  26. #include "src/pathops/SkPathOpsQuad.h"
  27. // used for testing only
  28. SkDQuad SkDCubic::toQuad() const {
  29. SkDQuad quad;
  30. quad[0] = fPts[0];
  31. const SkDPoint fromC1 = {(3 * fPts[1].fX - fPts[0].fX) / 2, (3 * fPts[1].fY - fPts[0].fY) / 2};
  32. const SkDPoint fromC2 = {(3 * fPts[2].fX - fPts[3].fX) / 2, (3 * fPts[2].fY - fPts[3].fY) / 2};
  33. quad[1].fX = (fromC1.fX + fromC2.fX) / 2;
  34. quad[1].fY = (fromC1.fY + fromC2.fY) / 2;
  35. quad[2] = fPts[3];
  36. return quad;
  37. }