本文整理汇总了C++中SkOpAngle::lengthen方法的典型用法代码示例。如果您正苦于以下问题：C++ SkOpAngle::lengthen方法的具体用法？C++ SkOpAngle::lengthen怎么用？C++ SkOpAngle::lengthen使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类SkOpAngle的用法示例。
在下文中一共展示了SkOpAngle::lengthen方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: line
/*(
for quads and cubics, set up a parameterized line (e.g. LineParameters )
for points [0] to [1]. See if point [2] is on that line, or on one side
or the other. If it both quads' end points are on the same side, choose
the shorter tangent. If the tangents are equal, choose the better second
tangent angle
maybe I could set up LineParameters lazily
*/
bool SkOpAngle::operator<(const SkOpAngle& rh) const {
double y = dy();
double ry = rh.dy();
if ((y < 0) ^ (ry < 0)) { // OPTIMIZATION: better to use y * ry < 0 ?
return y < 0;
}
double x = dx();
double rx = rh.dx();
if (y == 0 && ry == 0 && x * rx < 0) {
return x < rx;
}
double x_ry = x * ry;
double rx_y = rx * y;
double cmp = x_ry - rx_y;
if (!approximately_zero(cmp)) {
return cmp < 0;
}
if (approximately_zero(x_ry) && approximately_zero(rx_y)
&& !approximately_zero_squared(cmp)) {
return cmp < 0;
}
// at this point, the initial tangent line is coincident
// see if edges curl away from each other
if (fSide * rh.fSide <= 0 && (!approximately_zero(fSide)
|| !approximately_zero(rh.fSide))) {
// FIXME: running demo will trigger this assertion
// (don't know if commenting out will trigger further assertion or not)
// commenting it out allows demo to run in release, though
return fSide < rh.fSide;
}
// see if either curve can be lengthened and try the tangent compare again
if (/* cmp && */ (*fSpans)[fEnd].fOther != rh.fSegment // tangents not absolutely identical
&& (*rh.fSpans)[rh.fEnd].fOther != fSegment) { // and not intersecting
SkOpAngle longer = *this;
SkOpAngle rhLonger = rh;
if (longer.lengthen() | rhLonger.lengthen()) {
return longer < rhLonger;
}
}
if ((fVerb == SkPath::kLine_Verb && approximately_zero(x) && approximately_zero(y))
|| (rh.fVerb == SkPath::kLine_Verb
&& approximately_zero(rx) && approximately_zero(ry))) {
// See general unsortable comment below. This case can happen when
// one line has a non-zero change in t but no change in x and y.
fUnsortable = true;
rh.fUnsortable = true;
return this < &rh; // even with no solution, return a stable sort
}
if ((*rh.fSpans)[SkMin32(rh.fStart, rh.fEnd)].fTiny
|| (*fSpans)[SkMin32(fStart, fEnd)].fTiny) {
fUnsortable = true;
rh.fUnsortable = true;
return this < &rh; // even with no solution, return a stable sort
}
SkASSERT(fVerb >= SkPath::kQuad_Verb);
SkASSERT(rh.fVerb >= SkPath::kQuad_Verb);
// FIXME: until I can think of something better, project a ray from the
// end of the shorter tangent to midway between the end points
// through both curves and use the resulting angle to sort
// FIXME: some of this setup can be moved to set() if it works, or cached if it's expensive
double len = fTangent1.normalSquared();
double rlen = rh.fTangent1.normalSquared();
SkDLine ray;
SkIntersections i, ri;
int roots, rroots;
bool flip = false;
do {
bool useThis = (len < rlen) ^ flip;
const SkDCubic& part = useThis ? fCurvePart : rh.fCurvePart;
SkPath::Verb partVerb = useThis ? fVerb : rh.fVerb;
ray[0] = partVerb == SkPath::kCubic_Verb && part[0].approximatelyEqual(part[1]) ?
part[2] : part[1];
ray[1].fX = (part[0].fX + part[partVerb].fX) / 2;
ray[1].fY = (part[0].fY + part[partVerb].fY) / 2;
SkASSERT(ray[0] != ray[1]);
roots = (i.*CurveRay[fVerb])(fPts, ray);
rroots = (ri.*CurveRay[rh.fVerb])(rh.fPts, ray);
} while ((roots == 0 || rroots == 0) && (flip ^= true));
if (roots == 0 || rroots == 0) {
// FIXME: we don't have a solution in this case. The interim solution
// is to mark the edges as unsortable, exclude them from this and
// future computations, and allow the returned path to be fragmented
fUnsortable = true;
rh.fUnsortable = true;
return this < &rh; // even with no solution, return a stable sort
}
SkDPoint loc;
double best = SK_ScalarInfinity;
SkDVector dxy;
double dist;
int index;
//.........这里部分代码省略.........