📄 pixman-source.c
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*(buffer) = _gradient_walker_pixel (&walker, t); } buffer += 1; t += inc; } } } } else /* projective transformation */ { pixman_fixed_48_16_t t; if (pict->class == SOURCE_IMAGE_CLASS_VERTICAL) { register uint32_t color; if (v.vector[2] == 0) { t = 0; } else { pixman_fixed_48_16_t x, y; x = ((pixman_fixed_48_16_t) v.vector[0] << 16) / v.vector[2]; y = ((pixman_fixed_48_16_t) v.vector[1] << 16) / v.vector[2]; t = ((a * x + b * y) >> 16) + off; } color = _gradient_walker_pixel( &walker, t ); while (buffer < end) *(buffer++) = color; } else { while (buffer < end) { if (!mask || *mask++ & maskBits) { if (v.vector[2] == 0) { t = 0; } else { pixman_fixed_48_16_t x, y; x = ((pixman_fixed_48_16_t)v.vector[0] << 16) / v.vector[2]; y = ((pixman_fixed_48_16_t)v.vector[1] << 16) / v.vector[2]; t = ((a*x + b*y) >> 16) + off; } *(buffer) = _gradient_walker_pixel (&walker, t); } ++buffer; v.vector[0] += unit.vector[0]; v.vector[1] += unit.vector[1]; v.vector[2] += unit.vector[2]; } } } } 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 */ /* 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; } if (pict->common.type == RADIAL) { 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) = _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) = _gradient_walker_pixel (&walker, t); } ++buffer; rx += cx; ry += cy; rz += cz; } } } else /* SourcePictTypeConical */ { conical_gradient_t *conical = (conical_gradient_t *)pict; double a = conical->angle/(180.*65536); if (affine) { rx -= conical->center.x/65536.; ry -= conical->center.y/65536.; while (buffer < end) { double angle; if (!mask || *mask++ & maskBits) { pixman_fixed_48_16_t t; angle = atan2(ry, rx) + a; t = (pixman_fixed_48_16_t) (angle * (65536. / (2*M_PI))); *(buffer) = _gradient_walker_pixel (&walker, t); } ++buffer; rx += cx; ry += cy; } } else { while (buffer < end) { double x, y; double angle; if (!mask || *mask++ & maskBits) { pixman_fixed_48_16_t t; if (rz != 0) { x = rx/rz; y = ry/rz; } else { x = y = 0.; } x -= conical->center.x/65536.; y -= conical->center.y/65536.; angle = atan2(y, x) + a; t = (pixman_fixed_48_16_t) (angle * (65536. / (2*M_PI))); *(buffer) = _gradient_walker_pixel (&walker, t); } ++buffer; rx += cx; ry += cy; rz += cz; } } } }}
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