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/*
*
* 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 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.
*
* 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.
*/
#ifdef HAVE_CONFIG_H
#include <config.h>
#endif
#include <stdlib.h>
#include <math.h>
#include "pixman-private.h"
static void
radial_gradient_get_scanline_32 (pixman_image_t *image,
int x,
int y,
int width,
uint32_t * buffer,
const uint32_t *mask,
uint32_t mask_bits)
{
/*
* 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;
pixman_gradient_walker_t 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)
{
/* When computing t over a scanline, we notice that some expressions
* are constant so we can compute them just once. Given:
*
* t = (-2·B ± ⎷(B² - 4·A·C)) / 2·A
*
* where
*
* A = cdx² + cdy² - dr² [precomputed as radial->A]
* B = -2·(pdx·cdx + pdy·cdy + r₁·dr)
* C = pdx² + pdy² - r₁²
*
* Since we have an affine transformation, we know that (pdx, pdy)
* increase linearly with each pixel,
*
* pdx = pdx₀ + n·cx,
* pdy = pdy₀ + n·cy,
*
* we can then express B in terms of an linear increment along
* the scanline:
*
* B = B₀ + n·cB, with
* B₀ = -2·(pdx₀·cdx + pdy₀·cdy + r₁·dr) and
* cB = -2·(cx·cdx + cy·cdy)
*
* Thus we can replace the full evaluation of B per-pixel (4 multiplies,
* 2 additions) with a single addition.
*/
double r1 = radial->c1.radius / 65536.;
double r1sq = r1 * r1;
double pdx = rx - radial->c1.x / 65536.;
double pdy = ry - radial->c1.y / 65536.;
double A = radial->A;
double invA = -65536. / (2. * A);
double A4 = -4. * A;
double B = -2. * (pdx*radial->cdx + pdy*radial->cdy + r1*radial->dr);
double cB = -2. * (cx*radial->cdx + cy*radial->cdy);
pixman_bool_t invert = A * radial->dr < 0;
while (buffer < end)
{
if (!mask || *mask++ & mask_bits)
{
pixman_fixed_48_16_t t;
double det = B * B + A4 * (pdx * pdx + pdy * pdy - r1sq);
if (det <= 0.)
t = (pixman_fixed_48_16_t) (B * invA);
else if (invert)
t = (pixman_fixed_48_16_t) ((B + sqrt (det)) * invA);
else
t = (pixman_fixed_48_16_t) ((B - sqrt (det)) * invA);
*buffer = _pixman_gradient_walker_pixel (&walker, t);
}
++buffer;
pdx += cx;
pdy += cy;
B += cB;
}
}
else
{
/* projective */
while (buffer < end)
{
if (!mask || *mask++ & mask_bits)
{
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 * radial->dr < 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_property_changed (pixman_image_t *image)
{
image->common.get_scanline_32 = radial_gradient_get_scanline_32;
image->common.get_scanline_64 = _pixman_image_get_scanline_generic_64;
}
PIXMAN_EXPORT pixman_image_t *
pixman_image_create_radial_gradient (pixman_point_fixed_t * inner,
pixman_point_fixed_t * outer,
pixman_fixed_t inner_radius,
pixman_fixed_t outer_radius,
const pixman_gradient_stop_t *stops,
int n_stops)
{
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->c2.x = outer->x;
radial->c2.y = outer->y;
radial->c2.radius = outer_radius;
radial->cdx = pixman_fixed_to_double (radial->c2.x - radial->c1.x);
radial->cdy = pixman_fixed_to_double (radial->c2.y - radial->c1.y);
radial->dr = pixman_fixed_to_double (radial->c2.radius - radial->c1.radius);
radial->A = (radial->cdx * radial->cdx +
radial->cdy * radial->cdy -
radial->dr * radial->dr);
image->common.property_changed = radial_gradient_property_changed;
return image;
}