统一优化（-1值）

Question

请考虑以下片段着色器：

uniform PolygonData
{
    int count;
    float points[1024];
} polygon;

out vec4 outColor;

void main()
{
    float j;
    for (int i = 0; i < polygon.count; ++i)
    {
            j = polygon.points[i++];
            j = polygon.points[i++];
            j = polygon.points[i++];
    }

    outColor = vec4(1, 1, j, 1);
}

为什么 polygon.count 和 polygon.points 优化了？

Answer 1

@GuyRT

是。就是这样。从那里我得到了我超过最大统一阵列大小的结论。

这就是我在做的事情：

顶点：

in vec3 inPosition;

void main(void)
{
    gl_Position = vec4(inPosition, 1.0);
}

几何：

layout(lines) in;
layout(triangle_strip) out;
layout(max_vertices = 4) out;

out vec3 worldPos;

uniform mat4 u_proj;

void main()
{

    vec4 pos0 = u_proj * gl_in[0].gl_Position;
    vec4 pos1 = u_proj * gl_in[1].gl_Position;

    //left up
    gl_Position.x = pos0.x;
    gl_Position.y = pos1.y;
    gl_Position.w = pos0.w;
    worldPos = gl_Position.xyz;
    EmitVertex();
    //left down
    gl_Position = pos0;
    worldPos = gl_Position.xyz;
    EmitVertex();
    //right up
    gl_Position = pos1;
    worldPos = gl_Position.xyz;
    EmitVertex();
    //right down
    gl_Position.x = pos1.x;
    gl_Position.y = pos0.y;
    gl_Position.w = pos1.w;
    worldPos = gl_Position.xyz;
    EmitVertex();
    EndPrimitive();
}

片段：

struct PolyData
{
    int count;
    float points[256];
};

uniform PolyData p;

uniform mat4 u_proj;

in vec3 worldPos;

out vec4 outColor;

void main()
{
    float testx = worldPos.x;
    float testy = worldPos.y;

    bool c = false;
    int i;
    int j;
    for (i = 0, j = p.count - 4; i < p.count; j = i, i = i + 3)
    {
        vec4 i_vec = u_proj * vec4(p.points[i], p.points[i + 1], 0, 1.0);
        vec4 j_vec = u_proj * vec4(p.points[j], p.points[j + 1], 0, 1.0);

        if (  (i_vec.y >= testy) != (j_vec.y >= testy) && (testx <= (j_vec.x - i_vec.x) * (testy - i_vec.y) / (j_vec.y - i_vec.y) + i_vec.x))
            c = !c;
}

    if (c == true)
    {
        outColor = vec4(1, 1, 0, 1);
    }
    else
    {
        outColor = vec4(0, 0, 0, 0);
    }
}

感谢您的帮助。