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* Copyright © 2000 Keith Packard, member of The XFree86 Project, Inc.
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* Copyright © 2000 SuSE, Inc.
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* 2005 Lars Knoll & Zack Rusin, Trolltech
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* Copyright © 2007 Red Hat, Inc.
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* Permission to use, copy, modify, distribute, and sell this software and its
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* documentation for any purpose is hereby granted without fee, provided that
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* the above copyright notice appear in all copies and that both that
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* copyright notice and this permission notice appear in supporting
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* documentation, and that the name of Keith Packard not be used in
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* advertising or publicity pertaining to distribution of the software without
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* specific, written prior permission. Keith Packard makes no
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* representations about the suitability of this software for any purpose. It
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* is provided "as is" without express or implied warranty.
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* THE COPYRIGHT HOLDERS DISCLAIM ALL WARRANTIES WITH REGARD TO THIS
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* SOFTWARE, INCLUDING ALL IMPLIED WARRANTIES OF MERCHANTABILITY AND
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* FITNESS, IN NO EVENT SHALL THE COPYRIGHT HOLDERS BE LIABLE FOR ANY
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* SPECIAL, INDIRECT OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES
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* WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN
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* AN ACTION OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING
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* OUT OF OR IN CONNECTION WITH THE USE OR PERFORMANCE OF THIS
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#include "pixman-private.h"
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radial_gradient_get_scanline_32 (pixman_image_t *image,
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* In the radial gradient problem we are given two circles (c₁,r₁) and
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* (c₂,r₂) that define the gradient itself. Then, for any point p, we
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* must compute the value(s) of t within [0.0, 1.0] representing the
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* circle(s) that would color the point.
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* There are potentially two values of t since the point p can be
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* colored by both sides of the circle, (which happens whenever one
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* circle is not entirely contained within the other).
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* If we solve for a value of t that is outside of [0.0, 1.0] then we
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* use the extend mode (NONE, REPEAT, REFLECT, or PAD) to map to a
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* value within [0.0, 1.0].
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* Here is an illustration of the problem:
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* Given (c₁,r₁), (c₂,r₂) and p, we must find an angle θ such that two
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* points p₁ and p₂ on the two circles are collinear with p. Then, the
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* desired value of t is the ratio of the length of p₁p to the length
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* So, we have six unknown values: (p₁x, p₁y), (p₂x, p₂y), θ and t.
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* We can also write six equations that constrain the problem:
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* Point p₁ is a distance r₁ from c₁ at an angle of θ:
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* 1. p₁x = c₁x + r₁·cos θ
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* 2. p₁y = c₁y + r₁·sin θ
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* Point p₂ is a distance r₂ from c₂ at an angle of θ:
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* 3. p₂x = c₂x + r2·cos θ
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* 4. p₂y = c₂y + r2·sin θ
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* Point p lies at a fraction t along the line segment p₁p₂:
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* 5. px = t·p₂x + (1-t)·p₁x
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* 6. py = t·p₂y + (1-t)·p₁y
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* To solve, first subtitute 1-4 into 5 and 6:
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* px = t·(c₂x + r₂·cos θ) + (1-t)·(c₁x + r₁·cos θ)
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* py = t·(c₂y + r₂·sin θ) + (1-t)·(c₁y + r₁·sin θ)
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* Then solve each for cos θ and sin θ expressed as a function of t:
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* cos θ = (-(c₂x - c₁x)·t + (px - c₁x)) / ((r₂-r₁)·t + r₁)
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* sin θ = (-(c₂y - c₁y)·t + (py - c₁y)) / ((r₂-r₁)·t + r₁)
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* To simplify this a bit, we define new variables for several of the
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* common terms as shown below:
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* Note that cdx, cdy, and dr do not depend on point p at all, so can
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* be pre-computed for the entire gradient. The simplifed equations
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* cos θ = (-cdx·t + pdx) / (dr·t + r₁)
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* sin θ = (-cdy·t + pdy) / (dr·t + r₁)
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* Finally, to get a single function of t and eliminate the last
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* unknown θ, we use the identity sin²θ + cos²θ = 1. First, square
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* each equation, (we knew a quadratic was coming since it must be
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* possible to obtain two solutions in some cases):
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* cos²θ = (cdx²t² - 2·cdx·pdx·t + pdx²) / (dr²·t² + 2·r₁·dr·t + r₁²)
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* sin²θ = (cdy²t² - 2·cdy·pdy·t + pdy²) / (dr²·t² + 2·r₁·dr·t + r₁²)
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* Then add both together, set the result equal to 1, and express as a
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* standard quadratic equation in t of the form At² + Bt + C = 0
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* (cdx² + cdy² - dr²)·t² - 2·(cdx·pdx + cdy·pdy + r₁·dr)·t + (pdx² + pdy² - r₁²) = 0
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* A = cdx² + cdy² - dr²
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* B = -2·(pdx·cdx + pdy·cdy + r₁·dr)
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* C = pdx² + pdy² - r₁²
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* And again, notice that A does not depend on p, so can be
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* precomputed. From here we just use the quadratic formula to solve
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* t = (-2·B ± ⎷(B² - 4·A·C)) / 2·A
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gradient_t *gradient = (gradient_t *)image;
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source_image_t *source = (source_image_t *)image;
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radial_gradient_t *radial = (radial_gradient_t *)image;
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uint32_t *end = buffer + width;
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pixman_gradient_walker_t walker;
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pixman_bool_t affine = TRUE;
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_pixman_gradient_walker_init (&walker, gradient, source->common.repeat);
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if (source->common.transform)
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/* reference point is the center of the pixel */
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v.vector[0] = pixman_int_to_fixed (x) + pixman_fixed_1 / 2;
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v.vector[1] = pixman_int_to_fixed (y) + pixman_fixed_1 / 2;
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v.vector[2] = pixman_fixed_1;
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if (!pixman_transform_point_3d (source->common.transform, &v))
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cx = source->common.transform->matrix[0][0] / 65536.;
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cy = source->common.transform->matrix[1][0] / 65536.;
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cz = source->common.transform->matrix[2][0] / 65536.;
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rx = v.vector[0] / 65536.;
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ry = v.vector[1] / 65536.;
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rz = v.vector[2] / 65536.;
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source->common.transform->matrix[2][0] == 0 &&
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v.vector[2] == pixman_fixed_1;
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/* When computing t over a scanline, we notice that some expressions
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* are constant so we can compute them just once. Given:
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* t = (-2·B ± ⎷(B² - 4·A·C)) / 2·A
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* A = cdx² + cdy² - dr² [precomputed as radial->A]
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* B = -2·(pdx·cdx + pdy·cdy + r₁·dr)
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* C = pdx² + pdy² - r₁²
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* Since we have an affine transformation, we know that (pdx, pdy)
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* increase linearly with each pixel,
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* we can then express B in terms of an linear increment along
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* B = B₀ + n·cB, with
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* B₀ = -2·(pdx₀·cdx + pdy₀·cdy + r₁·dr) and
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* cB = -2·(cx·cdx + cy·cdy)
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* Thus we can replace the full evaluation of B per-pixel (4 multiplies,
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* 2 additions) with a single addition.
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double r1 = radial->c1.radius / 65536.;
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double r1sq = r1 * r1;
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double pdx = rx - radial->c1.x / 65536.;
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double pdy = ry - radial->c1.y / 65536.;
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double A = radial->A;
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double invA = -65536. / (2. * A);
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double B = -2. * (pdx*radial->cdx + pdy*radial->cdy + r1*radial->dr);
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double cB = -2. * (cx*radial->cdx + cy*radial->cdy);
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pixman_bool_t invert = A * radial->dr < 0;
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if (!mask || *mask++ & mask_bits)
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pixman_fixed_48_16_t t;
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double det = B * B + A4 * (pdx * pdx + pdy * pdy - r1sq);
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t = (pixman_fixed_48_16_t) (B * invA);
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t = (pixman_fixed_48_16_t) ((B + sqrt (det)) * invA);
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t = (pixman_fixed_48_16_t) ((B - sqrt (det)) * invA);
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*buffer = _pixman_gradient_walker_pixel (&walker, t);
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if (!mask || *mask++ & mask_bits)
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double c1x = radial->c1.x / 65536.0;
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double c1y = radial->c1.y / 65536.0;
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double r1 = radial->c1.radius / 65536.0;
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pixman_fixed_48_16_t t;
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B = -2 * (pdx * radial->cdx +
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C = (pdx * pdx + pdy * pdy - r1 * r1);
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det = (B * B) - (4 * radial->A * C);
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if (radial->A * radial->dr < 0)
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t = (pixman_fixed_48_16_t) ((-B - sqrt (det)) / (2.0 * radial->A) * 65536);
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t = (pixman_fixed_48_16_t) ((-B + sqrt (det)) / (2.0 * radial->A) * 65536);
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*buffer = _pixman_gradient_walker_pixel (&walker, t);
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radial_gradient_property_changed (pixman_image_t *image)
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image->common.get_scanline_32 = radial_gradient_get_scanline_32;
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image->common.get_scanline_64 = _pixman_image_get_scanline_generic_64;
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PIXMAN_EXPORT pixman_image_t *
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pixman_image_create_radial_gradient (pixman_point_fixed_t * inner,
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pixman_point_fixed_t * outer,
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pixman_fixed_t inner_radius,
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pixman_fixed_t outer_radius,
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const pixman_gradient_stop_t *stops,
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pixman_image_t *image;
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radial_gradient_t *radial;
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return_val_if_fail (n_stops >= 2, NULL);
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image = _pixman_image_allocate ();
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radial = &image->radial;
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if (!_pixman_init_gradient (&radial->common, stops, n_stops))
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image->type = RADIAL;
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radial->c1.x = inner->x;
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radial->c1.y = inner->y;
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radial->c1.radius = inner_radius;
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radial->c2.x = outer->x;
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radial->c2.y = outer->y;
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radial->c2.radius = outer_radius;
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radial->cdx = pixman_fixed_to_double (radial->c2.x - radial->c1.x);
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radial->cdy = pixman_fixed_to_double (radial->c2.y - radial->c1.y);
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radial->dr = pixman_fixed_to_double (radial->c2.radius - radial->c1.radius);
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radial->A = (radial->cdx * radial->cdx +
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radial->cdy * radial->cdy -
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radial->dr * radial->dr);
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image->common.property_changed = radial_gradient_property_changed;