cobalt / cobalt / c22907dbc00f1b681ca5a66f4cbc333d84fae0a5 / . / src / third_party / skia / src / pathops / SkDCubicToQuads.cpp

/* | |

* Copyright 2015 Google Inc. | |

* | |

* Use of this source code is governed by a BSD-style license that can be | |

* found in the LICENSE file. | |

*/ | |

/* | |

http://stackoverflow.com/questions/2009160/how-do-i-convert-the-2-control-points-of-a-cubic-curve-to-the-single-control-poi | |

*/ | |

/* | |

Let's call the control points of the cubic Q0..Q3 and the control points of the quadratic P0..P2. | |

Then for degree elevation, the equations are: | |

Q0 = P0 | |

Q1 = 1/3 P0 + 2/3 P1 | |

Q2 = 2/3 P1 + 1/3 P2 | |

Q3 = P2 | |

In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways to compute P1 from | |

the equations above: | |

P1 = 3/2 Q1 - 1/2 Q0 | |

P1 = 3/2 Q2 - 1/2 Q3 | |

If this is a degree-elevated cubic, then both equations will give the same answer for P1. Since | |

it's likely not, your best bet is to average them. So, | |

P1 = -1/4 Q0 + 3/4 Q1 + 3/4 Q2 - 1/4 Q3 | |

*/ | |

#include "SkPathOpsCubic.h" | |

#include "SkPathOpsQuad.h" | |

// used for testing only | |

SkDQuad SkDCubic::toQuad() const { | |

SkDQuad quad; | |

quad[0] = fPts[0]; | |

const SkDPoint fromC1 = {(3 * fPts[1].fX - fPts[0].fX) / 2, (3 * fPts[1].fY - fPts[0].fY) / 2}; | |

const SkDPoint fromC2 = {(3 * fPts[2].fX - fPts[3].fX) / 2, (3 * fPts[2].fY - fPts[3].fY) / 2}; | |

quad[1].fX = (fromC1.fX + fromC2.fX) / 2; | |

quad[1].fY = (fromC1.fY + fromC2.fY) / 2; | |

quad[2] = fPts[3]; | |

return quad; | |

} |