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, int x, int y, int width,
36 uint32_t *buffer, uint32_t *mask, uint32_t maskBits)
39 * In the radial gradient problem we are given two circles (c₁,r₁) and
40 * (c₂,r₂) that define the gradient itself. Then, for any point p, we
41 * must compute the value(s) of t within [0.0, 1.0] representing the
42 * circle(s) that would color the point.
44 * There are potentially two values of t since the point p can be
45 * colored by both sides of the circle, (which happens whenever one
46 * circle is not entirely contained within the other).
48 * If we solve for a value of t that is outside of [0.0, 1.0] then we
49 * use the extend mode (NONE, REPEAT, REFLECT, or PAD) to map to a
50 * value within [0.0, 1.0].
52 * Here is an illustration of the problem:
66 * Given (c₁,r₁), (c₂,r₂) and p, we must find an angle θ such that two
67 * points p₁ and p₂ on the two circles are collinear with p. Then, the
68 * desired value of t is the ratio of the length of p₁p to the length
71 * So, we have six unknown values: (p₁x, p₁y), (p₂x, p₂y), θ and t.
72 * We can also write six equations that constrain the problem:
74 * Point p₁ is a distance r₁ from c₁ at an angle of θ:
76 * 1. p₁x = c₁x + r₁·cos θ
77 * 2. p₁y = c₁y + r₁·sin θ
79 * Point p₂ is a distance r₂ from c₂ at an angle of θ:
81 * 3. p₂x = c₂x + r2·cos θ
82 * 4. p₂y = c₂y + r2·sin θ
84 * Point p lies at a fraction t along the line segment p₁p₂:
86 * 5. px = t·p₂x + (1-t)·p₁x
87 * 6. py = t·p₂y + (1-t)·p₁y
89 * To solve, first subtitute 1-4 into 5 and 6:
91 * px = t·(c₂x + r₂·cos θ) + (1-t)·(c₁x + r₁·cos θ)
92 * py = t·(c₂y + r₂·sin θ) + (1-t)·(c₁y + r₁·sin θ)
94 * Then solve each for cos θ and sin θ expressed as a function of t:
96 * cos θ = (-(c₂x - c₁x)·t + (px - c₁x)) / ((r₂-r₁)·t + r₁)
97 * sin θ = (-(c₂y - c₁y)·t + (py - c₁y)) / ((r₂-r₁)·t + r₁)
99 * To simplify this a bit, we define new variables for several of the
100 * common terms as shown below:
120 * Note that cdx, cdy, and dr do not depend on point p at all, so can
121 * be pre-computed for the entire gradient. The simplifed equations
124 * cos θ = (-cdx·t + pdx) / (dr·t + r₁)
125 * sin θ = (-cdy·t + pdy) / (dr·t + r₁)
127 * Finally, to get a single function of t and eliminate the last
128 * unknown θ, we use the identity sin²θ + cos²θ = 1. First, square
129 * each equation, (we knew a quadratic was coming since it must be
130 * possible to obtain two solutions in some cases):
132 * cos²θ = (cdx²t² - 2·cdx·pdx·t + pdx²) / (dr²·t² + 2·r₁·dr·t + r₁²)
133 * sin²θ = (cdy²t² - 2·cdy·pdy·t + pdy²) / (dr²·t² + 2·r₁·dr·t + r₁²)
135 * Then add both together, set the result equal to 1, and express as a
136 * standard quadratic equation in t of the form At² + Bt + C = 0
138 * (cdx² + cdy² - dr²)·t² - 2·(cdx·pdx + cdy·pdy + r₁·dr)·t + (pdx² + pdy² - r₁²) = 0
142 * A = cdx² + cdy² - dr²
143 * B = -2·(pdx·cdx + pdy·cdy + r₁·dr)
144 * C = pdx² + pdy² - r₁²
146 * And again, notice that A does not depend on p, so can be
147 * precomputed. From here we just use the quadratic formula to solve
150 * t = (-2·B ± ⎷(B² - 4·A·C)) / 2·A
153 gradient_t *gradient = (gradient_t *)image;
154 source_image_t *source = (source_image_t *)image;
155 radial_gradient_t *radial = (radial_gradient_t *)image;
156 uint32_t *end = buffer + width;
157 pixman_gradient_walker_t walker;
158 pixman_bool_t affine = TRUE;
166 _pixman_gradient_walker_init (&walker, gradient, source->common.repeat);
168 if (source->common.transform) {
170 /* reference point is the center of the pixel */
171 v.vector[0] = pixman_int_to_fixed(x) + pixman_fixed_1/2;
172 v.vector[1] = pixman_int_to_fixed(y) + pixman_fixed_1/2;
173 v.vector[2] = pixman_fixed_1;
174 if (!pixman_transform_point_3d (source->common.transform, &v))
177 cx = source->common.transform->matrix[0][0]/65536.;
178 cy = source->common.transform->matrix[1][0]/65536.;
179 cz = source->common.transform->matrix[2][0]/65536.;
180 rx = v.vector[0]/65536.;
181 ry = v.vector[1]/65536.;
182 rz = v.vector[2]/65536.;
183 affine = source->common.transform->matrix[2][0] == 0 && v.vector[2] == pixman_fixed_1;
187 while (buffer < end) {
188 if (!mask || *mask++ & maskBits)
193 double c1x = radial->c1.x / 65536.0;
194 double c1y = radial->c1.y / 65536.0;
195 double r1 = radial->c1.radius / 65536.0;
196 pixman_fixed_48_16_t t;
201 B = -2 * ( pdx * radial->cdx
204 C = (pdx * pdx + pdy * pdy - r1 * r1);
206 det = (B * B) - (4 * radial->A * C);
211 t = (pixman_fixed_48_16_t) ((- B - sqrt(det)) / (2.0 * radial->A) * 65536);
213 t = (pixman_fixed_48_16_t) ((- B + sqrt(det)) / (2.0 * radial->A) * 65536);
215 *(buffer) = _pixman_gradient_walker_pixel (&walker, t);
224 while (buffer < end) {
225 if (!mask || *mask++ & maskBits)
230 double c1x = radial->c1.x / 65536.0;
231 double c1y = radial->c1.y / 65536.0;
232 double r1 = radial->c1.radius / 65536.0;
233 pixman_fixed_48_16_t t;
246 B = -2 * ( pdx * radial->cdx
249 C = (pdx * pdx + pdy * pdy - r1 * r1);
251 det = (B * B) - (4 * radial->A * C);
256 t = (pixman_fixed_48_16_t) ((- B - sqrt(det)) / (2.0 * radial->A) * 65536);
258 t = (pixman_fixed_48_16_t) ((- B + sqrt(det)) / (2.0 * radial->A) * 65536);
260 *(buffer) = _pixman_gradient_walker_pixel (&walker, t);
273 radial_gradient_property_changed (pixman_image_t *image)
275 image->common.get_scanline_32 = radial_gradient_get_scanline_32;
276 image->common.get_scanline_64 = _pixman_image_get_scanline_64_generic;
279 PIXMAN_EXPORT pixman_image_t *
280 pixman_image_create_radial_gradient (pixman_point_fixed_t *inner,
281 pixman_point_fixed_t *outer,
282 pixman_fixed_t inner_radius,
283 pixman_fixed_t outer_radius,
284 const pixman_gradient_stop_t *stops,
287 pixman_image_t *image;
288 radial_gradient_t *radial;
290 return_val_if_fail (n_stops >= 2, NULL);
292 image = _pixman_image_allocate();
297 radial = &image->radial;
299 if (!_pixman_init_gradient (&radial->common, stops, n_stops))
305 image->type = RADIAL;
307 radial->c1.x = inner->x;
308 radial->c1.y = inner->y;
309 radial->c1.radius = inner_radius;
310 radial->c2.x = outer->x;
311 radial->c2.y = outer->y;
312 radial->c2.radius = outer_radius;
313 radial->cdx = pixman_fixed_to_double (radial->c2.x - radial->c1.x);
314 radial->cdy = pixman_fixed_to_double (radial->c2.y - radial->c1.y);
315 radial->dr = pixman_fixed_to_double (radial->c2.radius - radial->c1.radius);
316 radial->A = (radial->cdx * radial->cdx
317 + radial->cdy * radial->cdy
318 - radial->dr * radial->dr);
320 image->common.property_changed = radial_gradient_property_changed;
322 radial_gradient_property_changed (image);