我正在尝试编写一个“应用渐变”函数,它采用一堆点和一个线性渐变并计算每个点的颜色。我有这样的东西(给定一个 points
数组和一个由两个点 start
和 end
定义的渐变):
struct Colour {
float red; float green; float blue;
};
struct Point {
int x; int y;
Colour colour;
};
Point points[total_points] = { ... };
Point start = { ... };
Point end = { ... };
for (int i=0; i<total_points; i++) {
// Get normalised position along x and y
float scale_x = end.x-start.x;
float pos_x = (scale_x == 0) ? 0 : (points[i].x-start.x) / scale_x;
float scale_y = end.y-start.y;
float pos_y = (scale_y == 0) ? 0 : (points[i].y-start.y) / scale_y;
// Average the positions
float pos = .5 * (pos_x + pos_y);
// Blend colours
points[i].colour = blend_colour(start.colour, end.colour, pos);
}
我的颜色混合函数是这样简单的:
static Colour blend_colour(Colour start, Colour end, float position) {
Colour blend;
blend.red = (start.red * (1-position)) + (end.red * position);
blend.green = (start.green * (1-position)) + (end.green * position);
blend.blue = (start.blue * (1-position)) + (end.blue * position);
return blend;
}
我在 for 循环中的数学肯定不太正确——我需要使用三角函数计算颜色吗?
最佳答案
代替
// Average the positions
float pos = .5 * (pos_x + pos_y);
尝试
// Get scaled distance to point by calculating hypotenuse
float dist = sqrt(pos_x*pos_x + pos_y*pos_y);
此外,虽然编译器会为您完成,但您应该将比例因子提升到循环之外。 事实上,计算距离的比例因子可能会更好:
Point start = { ... };
Point end = { ... };
float xdelta = end.x - start.x;
float ydelta = end.y - start.y;
float hypot = sqrt(xdelta*xdelta + ydelta*ydelta);
for (int i=0; i<total_points; i++) {
// Get normalised distance to points[i]
xdelta = points[i].x - start.x;
ydelta = points[i].y - start.y;
float dist = sqrt(xdelta*xdelta + ydelta*ydelta);
if (hypot) dist /= hypot;
// Blend colours
points[i].colour = blend_colour(start.colour, end.colour, dist);
}
关于计算沿任意梯度的点的颜色,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17338292/