/*
+ *
+ * Copyright © 2000 Keith Packard, member of The XFree86 Project, Inc.
* Copyright © 2000 SuSE, Inc.
+ * 2005 Lars Knoll & Zack Rusin, Trolltech
* Copyright © 2007 Red Hat, Inc.
*
+ *
* Permission to use, copy, modify, distribute, and sell this software and its
* documentation for any purpose is hereby granted without fee, provided that
* the above copyright notice appear in all copies and that both that
* copyright notice and this permission notice appear in supporting
- * documentation, and that the name of SuSE not be used in advertising or
- * publicity pertaining to distribution of the software without specific,
- * written prior permission. SuSE makes no representations about the
- * suitability of this software for any purpose. It is provided "as is"
- * without express or implied warranty.
+ * documentation, and that the name of Keith Packard not be used in
+ * advertising or publicity pertaining to distribution of the software without
+ * specific, written prior permission. Keith Packard makes no
+ * representations about the suitability of this software for any purpose. It
+ * is provided "as is" without express or implied warranty.
*
- * SuSE DISCLAIMS ALL WARRANTIES WITH REGARD TO THIS SOFTWARE, INCLUDING ALL
- * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS, IN NO EVENT SHALL SuSE
- * BE LIABLE FOR ANY SPECIAL, INDIRECT OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES
- * WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN AN ACTION
- * OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING OUT OF OR IN
- * CONNECTION WITH THE USE OR PERFORMANCE OF THIS SOFTWARE.
+ * THE COPYRIGHT HOLDERS DISCLAIM ALL WARRANTIES WITH REGARD TO THIS
+ * SOFTWARE, INCLUDING ALL IMPLIED WARRANTIES OF MERCHANTABILITY AND
+ * FITNESS, IN NO EVENT SHALL THE COPYRIGHT HOLDERS BE LIABLE FOR ANY
+ * SPECIAL, INDIRECT OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES
+ * WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN
+ * AN ACTION OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING
+ * OUT OF OR IN CONNECTION WITH THE USE OR PERFORMANCE OF THIS
+ * SOFTWARE.
*/
#include <config.h>
#include <stdlib.h>
+#include <math.h>
#include "pixman-private.h"
static void
-radial_gradient_property_changed (pixman_image_t *image)
+radial_gradient_get_scanline_32 (pixman_image_t *image, int x, int y, int width,
+ uint32_t *buffer, uint32_t *mask, uint32_t maskBits)
{
- image->common.get_scanline_32 = (scanFetchProc)pixmanFetchGradient;
- image->common.get_scanline_64 = (scanFetchProc)_pixman_image_get_scanline_64_generic;
+ /*
+ * In the radial gradient problem we are given two circles (c₁,r₁) and
+ * (c₂,r₂) that define the gradient itself. Then, for any point p, we
+ * must compute the value(s) of t within [0.0, 1.0] representing the
+ * circle(s) that would color the point.
+ *
+ * There are potentially two values of t since the point p can be
+ * colored by both sides of the circle, (which happens whenever one
+ * circle is not entirely contained within the other).
+ *
+ * If we solve for a value of t that is outside of [0.0, 1.0] then we
+ * use the extend mode (NONE, REPEAT, REFLECT, or PAD) to map to a
+ * value within [0.0, 1.0].
+ *
+ * Here is an illustration of the problem:
+ *
+ * p₂
+ * p •
+ * • ╲
+ * · ╲r₂
+ * p₁ · ╲
+ * • θ╲
+ * ╲ ╌╌•
+ * ╲r₁ · c₂
+ * θ╲ ·
+ * ╌╌•
+ * c₁
+ *
+ * Given (c₁,r₁), (c₂,r₂) and p, we must find an angle θ such that two
+ * points p₁ and p₂ on the two circles are collinear with p. Then, the
+ * desired value of t is the ratio of the length of p₁p to the length
+ * of p₁p₂.
+ *
+ * So, we have six unknown values: (p₁x, p₁y), (p₂x, p₂y), θ and t.
+ * We can also write six equations that constrain the problem:
+ *
+ * Point p₁ is a distance r₁ from c₁ at an angle of θ:
+ *
+ * 1. p₁x = c₁x + r₁·cos θ
+ * 2. p₁y = c₁y + r₁·sin θ
+ *
+ * Point p₂ is a distance r₂ from c₂ at an angle of θ:
+ *
+ * 3. p₂x = c₂x + r2·cos θ
+ * 4. p₂y = c₂y + r2·sin θ
+ *
+ * Point p lies at a fraction t along the line segment p₁p₂:
+ *
+ * 5. px = t·p₂x + (1-t)·p₁x
+ * 6. py = t·p₂y + (1-t)·p₁y
+ *
+ * To solve, first subtitute 1-4 into 5 and 6:
+ *
+ * px = t·(c₂x + r₂·cos θ) + (1-t)·(c₁x + r₁·cos θ)
+ * py = t·(c₂y + r₂·sin θ) + (1-t)·(c₁y + r₁·sin θ)
+ *
+ * Then solve each for cos θ and sin θ expressed as a function of t:
+ *
+ * cos θ = (-(c₂x - c₁x)·t + (px - c₁x)) / ((r₂-r₁)·t + r₁)
+ * sin θ = (-(c₂y - c₁y)·t + (py - c₁y)) / ((r₂-r₁)·t + r₁)
+ *
+ * To simplify this a bit, we define new variables for several of the
+ * common terms as shown below:
+ *
+ * p₂
+ * p •
+ * • ╲
+ * · ┆ ╲r₂
+ * p₁ · ┆ ╲
+ * • pdy┆ ╲
+ * ╲ ┆ •c₂
+ * ╲r₁ ┆ · ┆
+ * ╲ ·┆ ┆cdy
+ * •╌╌╌╌┴╌╌╌╌╌╌╌┘
+ * c₁ pdx cdx
+ *
+ * cdx = (c₂x - c₁x)
+ * cdy = (c₂y - c₁y)
+ * dr = r₂-r₁
+ * pdx = px - c₁x
+ * pdy = py - c₁y
+ *
+ * Note that cdx, cdy, and dr do not depend on point p at all, so can
+ * be pre-computed for the entire gradient. The simplifed equations
+ * are now:
+ *
+ * cos θ = (-cdx·t + pdx) / (dr·t + r₁)
+ * sin θ = (-cdy·t + pdy) / (dr·t + r₁)
+ *
+ * Finally, to get a single function of t and eliminate the last
+ * unknown θ, we use the identity sin²θ + cos²θ = 1. First, square
+ * each equation, (we knew a quadratic was coming since it must be
+ * possible to obtain two solutions in some cases):
+ *
+ * cos²θ = (cdx²t² - 2·cdx·pdx·t + pdx²) / (dr²·t² + 2·r₁·dr·t + r₁²)
+ * sin²θ = (cdy²t² - 2·cdy·pdy·t + pdy²) / (dr²·t² + 2·r₁·dr·t + r₁²)
+ *
+ * Then add both together, set the result equal to 1, and express as a
+ * standard quadratic equation in t of the form At² + Bt + C = 0
+ *
+ * (cdx² + cdy² - dr²)·t² - 2·(cdx·pdx + cdy·pdy + r₁·dr)·t + (pdx² + pdy² - r₁²) = 0
+ *
+ * In other words:
+ *
+ * A = cdx² + cdy² - dr²
+ * B = -2·(pdx·cdx + pdy·cdy + r₁·dr)
+ * C = pdx² + pdy² - r₁²
+ *
+ * And again, notice that A does not depend on p, so can be
+ * precomputed. From here we just use the quadratic formula to solve
+ * for t:
+ *
+ * t = (-2·B ± ⎷(B² - 4·A·C)) / 2·A
+ */
+
+ gradient_t *gradient = (gradient_t *)image;
+ source_image_t *source = (source_image_t *)image;
+ radial_gradient_t *radial = (radial_gradient_t *)image;
+ uint32_t *end = buffer + width;
+ GradientWalker walker;
+ pixman_bool_t affine = TRUE;
+ double cx = 1.;
+ double cy = 0.;
+ double cz = 0.;
+ double rx = x + 0.5;
+ double ry = y + 0.5;
+ double rz = 1.;
+
+ _pixman_gradient_walker_init (&walker, gradient, source->common.repeat);
+
+ if (source->common.transform) {
+ pixman_vector_t v;
+ /* reference point is the center of the pixel */
+ v.vector[0] = pixman_int_to_fixed(x) + pixman_fixed_1/2;
+ v.vector[1] = pixman_int_to_fixed(y) + pixman_fixed_1/2;
+ v.vector[2] = pixman_fixed_1;
+ if (!pixman_transform_point_3d (source->common.transform, &v))
+ return;
+
+ cx = source->common.transform->matrix[0][0]/65536.;
+ cy = source->common.transform->matrix[1][0]/65536.;
+ cz = source->common.transform->matrix[2][0]/65536.;
+ rx = v.vector[0]/65536.;
+ ry = v.vector[1]/65536.;
+ rz = v.vector[2]/65536.;
+ affine = source->common.transform->matrix[2][0] == 0 && v.vector[2] == pixman_fixed_1;
+ }
+
+ if (affine) {
+ while (buffer < end) {
+ if (!mask || *mask++ & maskBits)
+ {
+ double pdx, pdy;
+ double B, C;
+ double det;
+ double c1x = radial->c1.x / 65536.0;
+ double c1y = radial->c1.y / 65536.0;
+ double r1 = radial->c1.radius / 65536.0;
+ pixman_fixed_48_16_t t;
+
+ pdx = rx - c1x;
+ pdy = ry - c1y;
+
+ B = -2 * ( pdx * radial->cdx
+ + pdy * radial->cdy
+ + r1 * radial->dr);
+ C = (pdx * pdx + pdy * pdy - r1 * r1);
+
+ det = (B * B) - (4 * radial->A * C);
+ if (det < 0.0)
+ det = 0.0;
+
+ if (radial->A < 0)
+ t = (pixman_fixed_48_16_t) ((- B - sqrt(det)) / (2.0 * radial->A) * 65536);
+ else
+ t = (pixman_fixed_48_16_t) ((- B + sqrt(det)) / (2.0 * radial->A) * 65536);
+
+ *(buffer) = _pixman_gradient_walker_pixel (&walker, t);
+ }
+ ++buffer;
+
+ rx += cx;
+ ry += cy;
+ }
+ } else {
+ /* projective */
+ while (buffer < end) {
+ if (!mask || *mask++ & maskBits)
+ {
+ double pdx, pdy;
+ double B, C;
+ double det;
+ double c1x = radial->c1.x / 65536.0;
+ double c1y = radial->c1.y / 65536.0;
+ double r1 = radial->c1.radius / 65536.0;
+ pixman_fixed_48_16_t t;
+ double x, y;
+
+ if (rz != 0) {
+ x = rx/rz;
+ y = ry/rz;
+ } else {
+ x = y = 0.;
+ }
+
+ pdx = x - c1x;
+ pdy = y - c1y;
+
+ B = -2 * ( pdx * radial->cdx
+ + pdy * radial->cdy
+ + r1 * radial->dr);
+ C = (pdx * pdx + pdy * pdy - r1 * r1);
+
+ det = (B * B) - (4 * radial->A * C);
+ if (det < 0.0)
+ det = 0.0;
+
+ if (radial->A < 0)
+ t = (pixman_fixed_48_16_t) ((- B - sqrt(det)) / (2.0 * radial->A) * 65536);
+ else
+ t = (pixman_fixed_48_16_t) ((- B + sqrt(det)) / (2.0 * radial->A) * 65536);
+
+ *(buffer) = _pixman_gradient_walker_pixel (&walker, t);
+ }
+ ++buffer;
+
+ rx += cx;
+ ry += cy;
+ rz += cz;
+ }
+ }
+
}
static void
-radial_gradient_get_scanline_32 (pixman_image_t *image, int x, int y, int width,
- uint32_t *buffer, uint32_t *mask, uint32_t maskBits)
+radial_gradient_property_changed (pixman_image_t *image)
{
-
+ image->common.get_scanline_32 = (scanFetchProc)radial_gradient_get_scanline_32;
+ image->common.get_scanline_64 = (scanFetchProc)_pixman_image_get_scanline_64_generic;
}
PIXMAN_EXPORT pixman_image_t *
{
pixman_image_t *image;
radial_gradient_t *radial;
-
+
return_val_if_fail (n_stops >= 2, NULL);
-
+
image = _pixman_image_allocate();
-
+
if (!image)
return NULL;
-
+
radial = &image->radial;
-
+
if (!_pixman_init_gradient (&radial->common, stops, n_stops))
{
free (image);
return NULL;
}
-
+
image->type = RADIAL;
-
+
radial->c1.x = inner->x;
radial->c1.y = inner->y;
radial->c1.radius = inner_radius;
radial->A = (radial->cdx * radial->cdx
+ radial->cdy * radial->cdy
- radial->dr * radial->dr);
-
+
image->common.property_changed = radial_gradient_property_changed;
-
+
radial_gradient_property_changed (image);
return image;
if (pict->common.type == LINEAR) {
assert (0);
} else {
-/*
- * In the radial gradient problem we are given two circles (c₁,r₁) and
- * (c₂,r₂) that define the gradient itself. Then, for any point p, we
- * must compute the value(s) of t within [0.0, 1.0] representing the
- * circle(s) that would color the point.
- *
- * There are potentially two values of t since the point p can be
- * colored by both sides of the circle, (which happens whenever one
- * circle is not entirely contained within the other).
- *
- * If we solve for a value of t that is outside of [0.0, 1.0] then we
- * use the extend mode (NONE, REPEAT, REFLECT, or PAD) to map to a
- * value within [0.0, 1.0].
- *
- * Here is an illustration of the problem:
- *
- * p₂
- * p •
- * • ╲
- * · ╲r₂
- * p₁ · ╲
- * • θ╲
- * ╲ ╌╌•
- * ╲r₁ · c₂
- * θ╲ ·
- * ╌╌•
- * c₁
- *
- * Given (c₁,r₁), (c₂,r₂) and p, we must find an angle θ such that two
- * points p₁ and p₂ on the two circles are collinear with p. Then, the
- * desired value of t is the ratio of the length of p₁p to the length
- * of p₁p₂.
- *
- * So, we have six unknown values: (p₁x, p₁y), (p₂x, p₂y), θ and t.
- * We can also write six equations that constrain the problem:
- *
- * Point p₁ is a distance r₁ from c₁ at an angle of θ:
- *
- * 1. p₁x = c₁x + r₁·cos θ
- * 2. p₁y = c₁y + r₁·sin θ
- *
- * Point p₂ is a distance r₂ from c₂ at an angle of θ:
- *
- * 3. p₂x = c₂x + r2·cos θ
- * 4. p₂y = c₂y + r2·sin θ
- *
- * Point p lies at a fraction t along the line segment p₁p₂:
- *
- * 5. px = t·p₂x + (1-t)·p₁x
- * 6. py = t·p₂y + (1-t)·p₁y
- *
- * To solve, first subtitute 1-4 into 5 and 6:
- *
- * px = t·(c₂x + r₂·cos θ) + (1-t)·(c₁x + r₁·cos θ)
- * py = t·(c₂y + r₂·sin θ) + (1-t)·(c₁y + r₁·sin θ)
- *
- * Then solve each for cos θ and sin θ expressed as a function of t:
- *
- * cos θ = (-(c₂x - c₁x)·t + (px - c₁x)) / ((r₂-r₁)·t + r₁)
- * sin θ = (-(c₂y - c₁y)·t + (py - c₁y)) / ((r₂-r₁)·t + r₁)
- *
- * To simplify this a bit, we define new variables for several of the
- * common terms as shown below:
- *
- * p₂
- * p •
- * • ╲
- * · ┆ ╲r₂
- * p₁ · ┆ ╲
- * • pdy┆ ╲
- * ╲ ┆ •c₂
- * ╲r₁ ┆ · ┆
- * ╲ ·┆ ┆cdy
- * •╌╌╌╌┴╌╌╌╌╌╌╌┘
- * c₁ pdx cdx
- *
- * cdx = (c₂x - c₁x)
- * cdy = (c₂y - c₁y)
- * dr = r₂-r₁
- * pdx = px - c₁x
- * pdy = py - c₁y
- *
- * Note that cdx, cdy, and dr do not depend on point p at all, so can
- * be pre-computed for the entire gradient. The simplifed equations
- * are now:
- *
- * cos θ = (-cdx·t + pdx) / (dr·t + r₁)
- * sin θ = (-cdy·t + pdy) / (dr·t + r₁)
- *
- * Finally, to get a single function of t and eliminate the last
- * unknown θ, we use the identity sin²θ + cos²θ = 1. First, square
- * each equation, (we knew a quadratic was coming since it must be
- * possible to obtain two solutions in some cases):
- *
- * cos²θ = (cdx²t² - 2·cdx·pdx·t + pdx²) / (dr²·t² + 2·r₁·dr·t + r₁²)
- * sin²θ = (cdy²t² - 2·cdy·pdy·t + pdy²) / (dr²·t² + 2·r₁·dr·t + r₁²)
- *
- * Then add both together, set the result equal to 1, and express as a
- * standard quadratic equation in t of the form At² + Bt + C = 0
- *
- * (cdx² + cdy² - dr²)·t² - 2·(cdx·pdx + cdy·pdy + r₁·dr)·t + (pdx² + pdy² - r₁²) = 0
- *
- * In other words:
- *
- * A = cdx² + cdy² - dr²
- * B = -2·(pdx·cdx + pdy·cdy + r₁·dr)
- * C = pdx² + pdy² - r₁²
- *
- * And again, notice that A does not depend on p, so can be
- * precomputed. From here we just use the quadratic formula to solve
- * for t:
- *
- * t = (-2·B ± ⎷(B² - 4·A·C)) / 2·A
- */
if (pict->common.type == RADIAL) {
- /* radial or conical */
- pixman_bool_t affine = TRUE;
- double cx = 1.;
- double cy = 0.;
- double cz = 0.;
- double rx = x + 0.5;
- double ry = y + 0.5;
- double rz = 1.;
-
- if (pict->common.transform) {
- pixman_vector_t v;
- /* reference point is the center of the pixel */
- v.vector[0] = pixman_int_to_fixed(x) + pixman_fixed_1/2;
- v.vector[1] = pixman_int_to_fixed(y) + pixman_fixed_1/2;
- v.vector[2] = pixman_fixed_1;
- if (!pixman_transform_point_3d (pict->common.transform, &v))
- return;
-
- cx = pict->common.transform->matrix[0][0]/65536.;
- cy = pict->common.transform->matrix[1][0]/65536.;
- cz = pict->common.transform->matrix[2][0]/65536.;
- rx = v.vector[0]/65536.;
- ry = v.vector[1]/65536.;
- rz = v.vector[2]/65536.;
- affine = pict->common.transform->matrix[2][0] == 0 && v.vector[2] == pixman_fixed_1;
- }
-
- radial_gradient_t *radial = (radial_gradient_t *)pict;
- if (affine) {
- while (buffer < end) {
- if (!mask || *mask++ & maskBits)
- {
- double pdx, pdy;
- double B, C;
- double det;
- double c1x = radial->c1.x / 65536.0;
- double c1y = radial->c1.y / 65536.0;
- double r1 = radial->c1.radius / 65536.0;
- pixman_fixed_48_16_t t;
-
- pdx = rx - c1x;
- pdy = ry - c1y;
-
- B = -2 * ( pdx * radial->cdx
- + pdy * radial->cdy
- + r1 * radial->dr);
- C = (pdx * pdx + pdy * pdy - r1 * r1);
-
- det = (B * B) - (4 * radial->A * C);
- if (det < 0.0)
- det = 0.0;
-
- if (radial->A < 0)
- t = (pixman_fixed_48_16_t) ((- B - sqrt(det)) / (2.0 * radial->A) * 65536);
- else
- t = (pixman_fixed_48_16_t) ((- B + sqrt(det)) / (2.0 * radial->A) * 65536);
-
- *(buffer) = _pixman_gradient_walker_pixel (&walker, t);
- }
- ++buffer;
-
- rx += cx;
- ry += cy;
- }
- } else {
- /* projective */
- while (buffer < end) {
- if (!mask || *mask++ & maskBits)
- {
- double pdx, pdy;
- double B, C;
- double det;
- double c1x = radial->c1.x / 65536.0;
- double c1y = radial->c1.y / 65536.0;
- double r1 = radial->c1.radius / 65536.0;
- pixman_fixed_48_16_t t;
- double x, y;
-
- if (rz != 0) {
- x = rx/rz;
- y = ry/rz;
- } else {
- x = y = 0.;
- }
-
- pdx = x - c1x;
- pdy = y - c1y;
-
- B = -2 * ( pdx * radial->cdx
- + pdy * radial->cdy
- + r1 * radial->dr);
- C = (pdx * pdx + pdy * pdy - r1 * r1);
-
- det = (B * B) - (4 * radial->A * C);
- if (det < 0.0)
- det = 0.0;
-
- if (radial->A < 0)
- t = (pixman_fixed_48_16_t) ((- B - sqrt(det)) / (2.0 * radial->A) * 65536);
- else
- t = (pixman_fixed_48_16_t) ((- B + sqrt(det)) / (2.0 * radial->A) * 65536);
-
- *(buffer) = _pixman_gradient_walker_pixel (&walker, t);
- }
- ++buffer;
-
- rx += cx;
- ry += cy;
- rz += cz;
- }
- }
+ assert (0);
} else /* SourcePictTypeConical */ {
/* radial or conical */
pixman_bool_t affine = TRUE;
projects / profile / ivi / pixman.git / commitdiff