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提交 009f710e 编写于 作者: D dlila
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4493128: CubicCurve2D intersects method fails 

Summary: Now using subdivision code in sun.awt.geom.Curve.
Reviewed-by: flar
上级 ee1a5ea8
隐藏空白更改
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Showing with 7 addition and 197 deletion
src/share/classes/java/awt/geom/CubicCurve2D.java
浏览文件 @ 009f710e
......@@ -1387,203 +1387,13 @@ public abstract class CubicCurve2D implements Shape, Cloneable {
return false;
}
// Trivially accept if either endpoint is inside the rectangle
// (not on its border since it may end there and not go inside)
// Record where they lie with respect to the rectangle.
// -1 => left, 0 => inside, 1 => right
double x1 = getX1();
double y1 = getY1();
int x1tag = getTag(x1, x, x+w);
int y1tag = getTag(y1, y, y+h);
if (x1tag == INSIDE && y1tag == INSIDE) {
return true;
}
double x2 = getX2();
double y2 = getY2();
int x2tag = getTag(x2, x, x+w);
int y2tag = getTag(y2, y, y+h);
if (x2tag == INSIDE && y2tag == INSIDE) {
return true;
}
double ctrlx1 = getCtrlX1();
double ctrly1 = getCtrlY1();
double ctrlx2 = getCtrlX2();
double ctrly2 = getCtrlY2();
int ctrlx1tag = getTag(ctrlx1, x, x+w);
int ctrly1tag = getTag(ctrly1, y, y+h);
int ctrlx2tag = getTag(ctrlx2, x, x+w);
int ctrly2tag = getTag(ctrly2, y, y+h);
// Trivially reject if all points are entirely to one side of
// the rectangle.
if (x1tag < INSIDE && x2tag < INSIDE &&
ctrlx1tag < INSIDE && ctrlx2tag < INSIDE)
{
return false; // All points left
}
if (y1tag < INSIDE && y2tag < INSIDE &&
ctrly1tag < INSIDE && ctrly2tag < INSIDE)
{
return false; // All points above
}
if (x1tag > INSIDE && x2tag > INSIDE &&
ctrlx1tag > INSIDE && ctrlx2tag > INSIDE)
{
return false; // All points right
}
if (y1tag > INSIDE && y2tag > INSIDE &&
ctrly1tag > INSIDE && ctrly2tag > INSIDE)
{
return false; // All points below
}
// Test for endpoints on the edge where either the segment
// or the curve is headed "inwards" from them
// Note: These tests are a superset of the fast endpoint tests
// above and thus repeat those tests, but take more time
// and cover more cases
if (inwards(x1tag, x2tag, ctrlx1tag) &&
inwards(y1tag, y2tag, ctrly1tag))
{
// First endpoint on border with either edge moving inside
return true;
}
if (inwards(x2tag, x1tag, ctrlx2tag) &&
inwards(y2tag, y1tag, ctrly2tag))
{
// Second endpoint on border with either edge moving inside
return true;
}
// Trivially accept if endpoints span directly across the rectangle
boolean xoverlap = (x1tag * x2tag <= 0);
boolean yoverlap = (y1tag * y2tag <= 0);
if (x1tag == INSIDE && x2tag == INSIDE && yoverlap) {
return true;
}
if (y1tag == INSIDE && y2tag == INSIDE && xoverlap) {
return true;
}
// We now know that both endpoints are outside the rectangle
// but the 4 points are not all on one side of the rectangle.
// Therefore the curve cannot be contained inside the rectangle,
// but the rectangle might be contained inside the curve, or
// the curve might intersect the boundary of the rectangle.
double[] eqn = new double[4];
double[] res = new double[4];
if (!yoverlap) {
// Both y coordinates for the closing segment are above or
// below the rectangle which means that we can only intersect
// if the curve crosses the top (or bottom) of the rectangle
// in more than one place and if those crossing locations
// span the horizontal range of the rectangle.
fillEqn(eqn, (y1tag < INSIDE ? y : y+h), y1, ctrly1, ctrly2, y2);
int num = solveCubic(eqn, res);
num = evalCubic(res, num, true, true, null,
x1, ctrlx1, ctrlx2, x2);
// odd counts imply the crossing was out of [0,1] bounds
// otherwise there is no way for that part of the curve to
// "return" to meet its endpoint
return (num == 2 &&
getTag(res[0], x, x+w) * getTag(res[1], x, x+w) <= 0);
}
// Y ranges overlap. Now we examine the X ranges
if (!xoverlap) {
// Both x coordinates for the closing segment are left of
// or right of the rectangle which means that we can only
// intersect if the curve crosses the left (or right) edge
// of the rectangle in more than one place and if those
// crossing locations span the vertical range of the rectangle.
fillEqn(eqn, (x1tag < INSIDE ? x : x+w), x1, ctrlx1, ctrlx2, x2);
int num = solveCubic(eqn, res);
num = evalCubic(res, num, true, true, null,
y1, ctrly1, ctrly2, y2);
// odd counts imply the crossing was out of [0,1] bounds
// otherwise there is no way for that part of the curve to
// "return" to meet its endpoint
return (num == 2 &&
getTag(res[0], y, y+h) * getTag(res[1], y, y+h) <= 0);
}
// The X and Y ranges of the endpoints overlap the X and Y
// ranges of the rectangle, now find out how the endpoint
// line segment intersects the Y range of the rectangle
double dx = x2 - x1;
double dy = y2 - y1;
double k = y2 * x1 - x2 * y1;
int c1tag, c2tag;
if (y1tag == INSIDE) {
c1tag = x1tag;
} else {
c1tag = getTag((k + dx * (y1tag < INSIDE ? y : y+h)) / dy, x, x+w);
}
if (y2tag == INSIDE) {
c2tag = x2tag;
} else {
c2tag = getTag((k + dx * (y2tag < INSIDE ? y : y+h)) / dy, x, x+w);
}
// If the part of the line segment that intersects the Y range
// of the rectangle crosses it horizontally - trivially accept
if (c1tag * c2tag <= 0) {
return true;
}
// Now we know that both the X and Y ranges intersect and that
// the endpoint line segment does not directly cross the rectangle.
//
// We can almost treat this case like one of the cases above
// where both endpoints are to one side, except that we may
// get one or three intersections of the curve with the vertical
// side of the rectangle. This is because the endpoint segment
// accounts for the other intersection in an even pairing. Thus,
// with the endpoint crossing we end up with 2 or 4 total crossings.
//
// (Remember there is overlap in both the X and Y ranges which
// means that the segment itself must cross at least one vertical
// edge of the rectangle - in particular, the "near vertical side"
// - leaving an odd number of intersections for the curve.)
//
// Now we calculate the y tags of all the intersections on the
// "near vertical side" of the rectangle. We will have one with
// the endpoint segment, and one or three with the curve. If
// any pair of those vertical intersections overlap the Y range
// of the rectangle, we have an intersection. Otherwise, we don't.
// c1tag = vertical intersection class of the endpoint segment
//
// Choose the y tag of the endpoint that was not on the same
// side of the rectangle as the subsegment calculated above.
// Note that we can "steal" the existing Y tag of that endpoint
// since it will be provably the same as the vertical intersection.
c1tag = ((c1tag * x1tag <= 0) ? y1tag : y2tag);
// Now we have to calculate an array of solutions of the curve
// with the "near vertical side" of the rectangle. Then we
// need to sort the tags and do a pairwise range test to see
// if either of the pairs of crossings spans the Y range of
// the rectangle.
//
// Note that the c2tag can still tell us which vertical edge
// to test against.
fillEqn(eqn, (c2tag < INSIDE ? x : x+w), x1, ctrlx1, ctrlx2, x2);
int num = solveCubic(eqn, res);
num = evalCubic(res, num, true, true, null, y1, ctrly1, ctrly2, y2);
// Now put all of the tags into a bucket and sort them. There
// is an intersection iff one of the pairs of tags "spans" the
// Y range of the rectangle.
int tags[] = new int[num+1];
for (int i = 0; i < num; i++) {
tags[i] = getTag(res[i], y, y+h);
}
tags[num] = c1tag;
Arrays.sort(tags);
return ((num >= 1 && tags[0] * tags[1] <= 0) ||
(num >= 3 && tags[2] * tags[3] <= 0));
int numCrossings = rectCrossings(x, y, w, h);
// the intended return value is
// numCrossings != 0 || numCrossings == Curve.RECT_INTERSECTS
// but if (numCrossings != 0) numCrossings == INTERSECTS won't matter
// and if !(numCrossings != 0) then numCrossings == 0, so
// numCrossings != RECT_INTERSECT
return numCrossings != 0;
}
/**
......
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