int intersect() {
addEndPoints();
double rootVals[2];
int roots = intersectRay(rootVals);
for (int index = 0; index < roots; ++index) {
double quadT = rootVals[index];
double lineT = findLineT(quadT);
if (pinTs(quadT, lineT)) {
intersections.insert(quadT, lineT);
}
}
return intersections.fUsed;
}
int intersect() {
addEndPoints();
double rootVals[2];
int roots = intersectRay(rootVals);
for (int index = 0; index < roots; ++index) {
double quadT = rootVals[index];
double lineT = findLineT(quadT);
if (PinTs(&quadT, &lineT)) {
SkDPoint pt = line.xyAtT(lineT);
intersections->insert(quadT, lineT, pt);
}
}
return intersections->used();
}
int intersect() {
addExactEndPoints();
if (fAllowNear) {
addNearEndPoints();
}
double rootVals[3];
int roots = intersectRay(rootVals);
for (int index = 0; index < roots; ++index) {
double cubicT = rootVals[index];
double lineT = findLineT(cubicT);
SkDPoint pt;
if (pinTs(&cubicT, &lineT, &pt, kPointUninitialized) && uniqueAnswer(cubicT, pt)) {
fIntersections->insert(cubicT, lineT, pt);
}
}
checkCoincident();
return fIntersections->used();
}
int intersect() {
addExactEndPoints();
if (fAllowNear) {
addNearEndPoints();
}
if (fIntersections->used() == 2) {
// FIXME : need sharable code that turns spans into coincident if middle point is on
} else {
double rootVals[2];
int roots = intersectRay(rootVals);
for (int index = 0; index < roots; ++index) {
double quadT = rootVals[index];
double lineT = findLineT(quadT);
SkDPoint pt;
if (pinTs(&quadT, &lineT, &pt, kPointUninitialized)) {
fIntersections->insert(quadT, lineT, pt);
}
}
}
return fIntersections->used();
}
int intersect() {
/*
solve by rotating line+quad so line is horizontal, then finding the roots
set up matrix to rotate quad to x-axis
|cos(a) -sin(a)|
|sin(a) cos(a)|
note that cos(a) = A(djacent) / Hypoteneuse
sin(a) = O(pposite) / Hypoteneuse
since we are computing Ts, we can ignore hypoteneuse, the scale factor:
| A -O |
| O A |
A = line[1].x - line[0].x (adjacent side of the right triangle)
O = line[1].y - line[0].y (opposite side of the right triangle)
for each of the three points (e.g. n = 0 to 2)
quad[n].y' = (quad[n].y - line[0].y) * A - (quad[n].x - line[0].x) * O
*/
double adj = line[1].x - line[0].x;
double opp = line[1].y - line[0].y;
double r[3];
for (int n = 0; n < 3; ++n) {
r[n] = (quad[n].y - line[0].y) * adj - (quad[n].x - line[0].x) * opp;
}
double A = r[2];
double B = r[1];
double C = r[0];
A += C - 2 * B; // A = a - 2*b + c
B -= C; // B = -(b - c)
int roots = quadraticRoots(A, B, C, intersections.fT[0]);
for (int index = 0; index < roots; ) {
double lineT = findLineT(intersections.fT[0][index]);
if (lineIntersects(lineT, index, roots)) {
++index;
}
}
return roots;
}
