3 * Copyright © 2000 Keith Packard, member of The XFree86 Project, Inc.
4 * Copyright © 2000 SuSE, Inc.
5 * 2005 Lars Knoll & Zack Rusin, Trolltech
6 * Copyright © 2007 Red Hat, Inc.
9 * Permission to use, copy, modify, distribute, and sell this software and its
10 * documentation for any purpose is hereby granted without fee, provided that
11 * the above copyright notice appear in all copies and that both that
12 * copyright notice and this permission notice appear in supporting
13 * documentation, and that the name of Keith Packard not be used in
14 * advertising or publicity pertaining to distribution of the software without
15 * specific, written prior permission. Keith Packard makes no
16 * representations about the suitability of this software for any purpose. It
17 * is provided "as is" without express or implied warranty.
19 * THE COPYRIGHT HOLDERS DISCLAIM ALL WARRANTIES WITH REGARD TO THIS
20 * SOFTWARE, INCLUDING ALL IMPLIED WARRANTIES OF MERCHANTABILITY AND
21 * FITNESS, IN NO EVENT SHALL THE COPYRIGHT HOLDERS BE LIABLE FOR ANY
22 * SPECIAL, INDIRECT OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES
23 * WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN
24 * AN ACTION OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING
25 * OUT OF OR IN CONNECTION WITH THE USE OR PERFORMANCE OF THIS
32 #include "pixman-private.h"
35 radial_gradient_get_scanline_32 (pixman_image_t *image,
44 * In the radial gradient problem we are given two circles (c₁,r₁) and
45 * (c₂,r₂) that define the gradient itself. Then, for any point p, we
46 * must compute the value(s) of t within [0.0, 1.0] representing the
47 * circle(s) that would color the point.
49 * There are potentially two values of t since the point p can be
50 * colored by both sides of the circle, (which happens whenever one
51 * circle is not entirely contained within the other).
53 * If we solve for a value of t that is outside of [0.0, 1.0] then we
54 * use the extend mode (NONE, REPEAT, REFLECT, or PAD) to map to a
55 * value within [0.0, 1.0].
57 * Here is an illustration of the problem:
71 * Given (c₁,r₁), (c₂,r₂) and p, we must find an angle θ such that two
72 * points p₁ and p₂ on the two circles are collinear with p. Then, the
73 * desired value of t is the ratio of the length of p₁p to the length
76 * So, we have six unknown values: (p₁x, p₁y), (p₂x, p₂y), θ and t.
77 * We can also write six equations that constrain the problem:
79 * Point p₁ is a distance r₁ from c₁ at an angle of θ:
81 * 1. p₁x = c₁x + r₁·cos θ
82 * 2. p₁y = c₁y + r₁·sin θ
84 * Point p₂ is a distance r₂ from c₂ at an angle of θ:
86 * 3. p₂x = c₂x + r2·cos θ
87 * 4. p₂y = c₂y + r2·sin θ
89 * Point p lies at a fraction t along the line segment p₁p₂:
91 * 5. px = t·p₂x + (1-t)·p₁x
92 * 6. py = t·p₂y + (1-t)·p₁y
94 * To solve, first subtitute 1-4 into 5 and 6:
96 * px = t·(c₂x + r₂·cos θ) + (1-t)·(c₁x + r₁·cos θ)
97 * py = t·(c₂y + r₂·sin θ) + (1-t)·(c₁y + r₁·sin θ)
99 * Then solve each for cos θ and sin θ expressed as a function of t:
101 * cos θ = (-(c₂x - c₁x)·t + (px - c₁x)) / ((r₂-r₁)·t + r₁)
102 * sin θ = (-(c₂y - c₁y)·t + (py - c₁y)) / ((r₂-r₁)·t + r₁)
104 * To simplify this a bit, we define new variables for several of the
105 * common terms as shown below:
125 * Note that cdx, cdy, and dr do not depend on point p at all, so can
126 * be pre-computed for the entire gradient. The simplifed equations
129 * cos θ = (-cdx·t + pdx) / (dr·t + r₁)
130 * sin θ = (-cdy·t + pdy) / (dr·t + r₁)
132 * Finally, to get a single function of t and eliminate the last
133 * unknown θ, we use the identity sin²θ + cos²θ = 1. First, square
134 * each equation, (we knew a quadratic was coming since it must be
135 * possible to obtain two solutions in some cases):
137 * cos²θ = (cdx²t² - 2·cdx·pdx·t + pdx²) / (dr²·t² + 2·r₁·dr·t + r₁²)
138 * sin²θ = (cdy²t² - 2·cdy·pdy·t + pdy²) / (dr²·t² + 2·r₁·dr·t + r₁²)
140 * Then add both together, set the result equal to 1, and express as a
141 * standard quadratic equation in t of the form At² + Bt + C = 0
143 * (cdx² + cdy² - dr²)·t² - 2·(cdx·pdx + cdy·pdy + r₁·dr)·t + (pdx² + pdy² - r₁²) = 0
147 * A = cdx² + cdy² - dr²
148 * B = -2·(pdx·cdx + pdy·cdy + r₁·dr)
149 * C = pdx² + pdy² - r₁²
151 * And again, notice that A does not depend on p, so can be
152 * precomputed. From here we just use the quadratic formula to solve
155 * t = (-2·B ± ⎷(B² - 4·A·C)) / 2·A
158 gradient_t *gradient = (gradient_t *)image;
159 source_image_t *source = (source_image_t *)image;
160 radial_gradient_t *radial = (radial_gradient_t *)image;
161 uint32_t *end = buffer + width;
162 pixman_gradient_walker_t walker;
163 pixman_bool_t affine = TRUE;
171 _pixman_gradient_walker_init (&walker, gradient, source->common.repeat);
173 if (source->common.transform)
176 /* reference point is the center of the pixel */
177 v.vector[0] = pixman_int_to_fixed (x) + pixman_fixed_1 / 2;
178 v.vector[1] = pixman_int_to_fixed (y) + pixman_fixed_1 / 2;
179 v.vector[2] = pixman_fixed_1;
181 if (!pixman_transform_point_3d (source->common.transform, &v))
184 cx = source->common.transform->matrix[0][0] / 65536.;
185 cy = source->common.transform->matrix[1][0] / 65536.;
186 cz = source->common.transform->matrix[2][0] / 65536.;
188 rx = v.vector[0] / 65536.;
189 ry = v.vector[1] / 65536.;
190 rz = v.vector[2] / 65536.;
193 source->common.transform->matrix[2][0] == 0 &&
194 v.vector[2] == pixman_fixed_1;
201 if (!mask || *mask++ & mask_bits)
206 double c1x = radial->c1.x / 65536.0;
207 double c1y = radial->c1.y / 65536.0;
208 double r1 = radial->c1.radius / 65536.0;
209 pixman_fixed_48_16_t t;
214 B = -2 * (pdx * radial->cdx +
217 C = pdx * pdx + pdy * pdy - r1 * r1;
219 det = (B * B) - (4 * radial->A * C);
224 t = (pixman_fixed_48_16_t) ((-B - sqrt (det)) / (2.0 * radial->A) * 65536);
226 t = (pixman_fixed_48_16_t) ((-B + sqrt (det)) / (2.0 * radial->A) * 65536);
228 *buffer = _pixman_gradient_walker_pixel (&walker, t);
241 if (!mask || *mask++ & mask_bits)
246 double c1x = radial->c1.x / 65536.0;
247 double c1y = radial->c1.y / 65536.0;
248 double r1 = radial->c1.radius / 65536.0;
249 pixman_fixed_48_16_t t;
265 B = -2 * (pdx * radial->cdx +
268 C = (pdx * pdx + pdy * pdy - r1 * r1);
270 det = (B * B) - (4 * radial->A * C);
275 t = (pixman_fixed_48_16_t) ((-B - sqrt (det)) / (2.0 * radial->A) * 65536);
277 t = (pixman_fixed_48_16_t) ((-B + sqrt (det)) / (2.0 * radial->A) * 65536);
279 *buffer = _pixman_gradient_walker_pixel (&walker, t);
292 radial_gradient_property_changed (pixman_image_t *image)
294 image->common.get_scanline_32 = radial_gradient_get_scanline_32;
295 image->common.get_scanline_64 = _pixman_image_get_scanline_generic_64;
298 PIXMAN_EXPORT pixman_image_t *
299 pixman_image_create_radial_gradient (pixman_point_fixed_t * inner,
300 pixman_point_fixed_t * outer,
301 pixman_fixed_t inner_radius,
302 pixman_fixed_t outer_radius,
303 const pixman_gradient_stop_t *stops,
306 pixman_image_t *image;
307 radial_gradient_t *radial;
309 return_val_if_fail (n_stops >= 2, NULL);
311 image = _pixman_image_allocate ();
316 radial = &image->radial;
318 if (!_pixman_init_gradient (&radial->common, stops, n_stops))
324 image->type = RADIAL;
326 radial->c1.x = inner->x;
327 radial->c1.y = inner->y;
328 radial->c1.radius = inner_radius;
329 radial->c2.x = outer->x;
330 radial->c2.y = outer->y;
331 radial->c2.radius = outer_radius;
332 radial->cdx = pixman_fixed_to_double (radial->c2.x - radial->c1.x);
333 radial->cdy = pixman_fixed_to_double (radial->c2.y - radial->c1.y);
334 radial->dr = pixman_fixed_to_double (radial->c2.radius - radial->c1.radius);
335 radial->A = (radial->cdx * radial->cdx +
336 radial->cdy * radial->cdy -
337 radial->dr * radial->dr);
339 image->common.property_changed = radial_gradient_property_changed;
341 radial_gradient_property_changed (image);