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*
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* particular file as subject to the "Classpath" exception as provided
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*
* This code is distributed in the hope that it will be useful, but WITHOUT
* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
* FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
* version 2 for more details (a copy is included in the LICENSE file that
* accompanied this code).
*
* You should have received a copy of the GNU General Public License version
* 2 along with this work; if not, write to the Free Software Foundation,
* Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
*
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package com.sun.javafx.geom;
import java.util.NoSuchElementException;
import com.sun.javafx.geom.transform.BaseTransform;
A utility class to iterate over the path segments of an arc
through the PathIterator interface.
Version: 10 Feb 1997
/**
* A utility class to iterate over the path segments of an arc
* through the PathIterator interface.
*
* @version 10 Feb 1997
*/
class ArcIterator implements PathIterator {
double x, y, w, h, angStRad, increment, cv;
BaseTransform transform;
int index;
int arcSegs;
int lineSegs;
ArcIterator(Arc2D a, BaseTransform at) {
this.w = a.width / 2;
this.h = a.height / 2;
this.x = a.x + w;
this.y = a.y + h;
this.angStRad = -Math.toRadians(a.start);
this.transform = at;
double ext = -a.extent;
if (ext >= 360.0 || ext <= -360) {
arcSegs = 4;
this.increment = Math.PI / 2;
// btan(Math.PI / 2);
this.cv = 0.5522847498307933;
if (ext < 0) {
increment = -increment;
cv = -cv;
}
} else {
arcSegs = (int) Math.ceil(Math.abs(ext) / 90.0);
this.increment = Math.toRadians(ext / arcSegs);
this.cv = btan(increment);
if (cv == 0) {
arcSegs = 0;
}
}
switch (a.getArcType()) {
case Arc2D.OPEN:
lineSegs = 0;
break;
case Arc2D.CHORD:
lineSegs = 1;
break;
case Arc2D.PIE:
lineSegs = 2;
break;
}
if (w < 0 || h < 0) {
arcSegs = lineSegs = -1;
}
}
Return the winding rule for determining the insideness of the
path.
See Also: - WIND_EVEN_ODD
- PathIterator.WIND_NON_ZERO
/**
* Return the winding rule for determining the insideness of the
* path.
* @see #WIND_EVEN_ODD
* @see #WIND_NON_ZERO
*/
public int getWindingRule() {
return WIND_NON_ZERO;
}
Tests if there are more points to read.
Returns: true if there are more points to read
/**
* Tests if there are more points to read.
* @return true if there are more points to read
*/
public boolean isDone() {
return index > arcSegs + lineSegs;
}
Moves the iterator to the next segment of the path forwards
along the primary direction of traversal as long as there are
more points in that direction.
/**
* Moves the iterator to the next segment of the path forwards
* along the primary direction of traversal as long as there are
* more points in that direction.
*/
public void next() {
++index;
}
/*
* btan computes the length (k) of the control segments at
* the beginning and end of a cubic bezier that approximates
* a segment of an arc with extent less than or equal to
* 90 degrees. This length (k) will be used to generate the
* 2 bezier control points for such a segment.
*
* Assumptions:
* a) arc is centered on 0,0 with radius of 1.0
* b) arc extent is less than 90 degrees
* c) control points should preserve tangent
* d) control segments should have equal length
*
* Initial data:
* start angle: ang1
* end angle: ang2 = ang1 + extent
* start point: P1 = (x1, y1) = (cos(ang1), sin(ang1))
* end point: P4 = (x4, y4) = (cos(ang2), sin(ang2))
*
* Control points:
* P2 = (x2, y2)
* | x2 = x1 - k * sin(ang1) = cos(ang1) - k * sin(ang1)
* | y2 = y1 + k * cos(ang1) = sin(ang1) + k * cos(ang1)
*
* P3 = (x3, y3)
* | x3 = x4 + k * sin(ang2) = cos(ang2) + k * sin(ang2)
* | y3 = y4 - k * cos(ang2) = sin(ang2) - k * cos(ang2)
*
* The formula for this length (k) can be found using the
* following derivations:
*
* Midpoints:
* a) bezier (t = 1/2)
* bPm = P1 * (1-t)^3 +
* 3 * P2 * t * (1-t)^2 +
* 3 * P3 * t^2 * (1-t) +
* P4 * t^3 =
* = (P1 + 3P2 + 3P3 + P4)/8
*
* b) arc
* aPm = (cos((ang1 + ang2)/2), sin((ang1 + ang2)/2))
*
* Let angb = (ang2 - ang1)/2; angb is half of the angle
* between ang1 and ang2.
*
* Solve the equation bPm == aPm
*
* a) For xm coord:
* x1 + 3*x2 + 3*x3 + x4 = 8*cos((ang1 + ang2)/2)
*
* cos(ang1) + 3*cos(ang1) - 3*k*sin(ang1) +
* 3*cos(ang2) + 3*k*sin(ang2) + cos(ang2) =
* = 8*cos((ang1 + ang2)/2)
*
* 4*cos(ang1) + 4*cos(ang2) + 3*k*(sin(ang2) - sin(ang1)) =
* = 8*cos((ang1 + ang2)/2)
*
* 8*cos((ang1 + ang2)/2)*cos((ang2 - ang1)/2) +
* 6*k*sin((ang2 - ang1)/2)*cos((ang1 + ang2)/2) =
* = 8*cos((ang1 + ang2)/2)
*
* 4*cos(angb) + 3*k*sin(angb) = 4
*
* k = 4 / 3 * (1 - cos(angb)) / sin(angb)
*
* b) For ym coord we derive the same formula.
*
* Since this formula can generate "NaN" values for small
* angles, we will derive a safer form that does not involve
* dividing by very small values:
* (1 - cos(angb)) / sin(angb) =
* = (1 - cos(angb))*(1 + cos(angb)) / sin(angb)*(1 + cos(angb)) =
* = (1 - cos(angb)^2) / sin(angb)*(1 + cos(angb)) =
* = sin(angb)^2 / sin(angb)*(1 + cos(angb)) =
* = sin(angb) / (1 + cos(angb))
*
*/
private static double btan(double increment) {
increment /= 2.0;
return 4.0 / 3.0 * Math.sin(increment) / (1.0 + Math.cos(increment));
}
Returns the coordinates and type of the current path segment in
the iteration.
The return value is the path segment type:
SEG_MOVETO, SEG_LINETO, SEG_QUADTO, SEG_CUBICTO, or SEG_CLOSE.
A float array of length 6 must be passed in and may be used to
store the coordinates of the point(s).
Each point is stored as a pair of float x,y coordinates.
SEG_MOVETO and SEG_LINETO types will return one point,
SEG_QUADTO will return two points,
SEG_CUBICTO will return 3 points
and SEG_CLOSE will not return any points.
See Also:
/**
* Returns the coordinates and type of the current path segment in
* the iteration.
* The return value is the path segment type:
* SEG_MOVETO, SEG_LINETO, SEG_QUADTO, SEG_CUBICTO, or SEG_CLOSE.
* A float array of length 6 must be passed in and may be used to
* store the coordinates of the point(s).
* Each point is stored as a pair of float x,y coordinates.
* SEG_MOVETO and SEG_LINETO types will return one point,
* SEG_QUADTO will return two points,
* SEG_CUBICTO will return 3 points
* and SEG_CLOSE will not return any points.
* @see #SEG_MOVETO
* @see #SEG_LINETO
* @see #SEG_QUADTO
* @see #SEG_CUBICTO
* @see #SEG_CLOSE
*/
public int currentSegment(float[] coords) {
if (isDone()) {
throw new NoSuchElementException("arc iterator out of bounds");
}
double angle = angStRad;
if (index == 0) {
coords[0] = (float) (x + Math.cos(angle) * w);
coords[1] = (float) (y + Math.sin(angle) * h);
if (transform != null) {
transform.transform(coords, 0, coords, 0, 1);
}
return SEG_MOVETO;
}
if (index > arcSegs) {
if (index == arcSegs + lineSegs) {
return SEG_CLOSE;
}
coords[0] = (float) x;
coords[1] = (float) y;
if (transform != null) {
transform.transform(coords, 0, coords, 0, 1);
}
return SEG_LINETO;
}
angle += increment * (index - 1);
double relx = Math.cos(angle);
double rely = Math.sin(angle);
coords[0] = (float) (x + (relx - cv * rely) * w);
coords[1] = (float) (y + (rely + cv * relx) * h);
angle += increment;
relx = Math.cos(angle);
rely = Math.sin(angle);
coords[2] = (float) (x + (relx + cv * rely) * w);
coords[3] = (float) (y + (rely - cv * relx) * h);
coords[4] = (float) (x + relx * w);
coords[5] = (float) (y + rely * h);
if (transform != null) {
transform.transform(coords, 0, coords, 0, 3);
}
return SEG_CUBICTO;
}
}