Question

我正在尝试编写一个'应用渐变'功能，该功能需要一堆点和一个线性渐变，并计算每个点的颜色。我有这样的事情（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);

}

我的色彩混合功能很简单：

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？

Answer 1

而不是

// 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);
}

计算沿任意渐变的点的颜色

1 个答案: