glDepthRange如何工作？

Question

我的顶点线是:

GLfloat vertices[]=
    {
        0.5f,0.5f,0.5f,                                   
        -0.5f,0.5f,0.5f,
        -0.5f,-0.5f,0.5f,
        0.5f,-0.5f,0.5f,//face 1

        0.5f,-0.5f,-0.5f,
        -0.5f,-0.5f,-0.5f,
        -0.5f,0.5f,-0.5f,
        0.5f,0.5f,-0.5f,//face 2

        0.5f,0.5f,0.5f,
        0.5f,-0.5f,0.5f,
        0.5f,-0.5f,-0.5f,
        0.5f,0.5f,-0.5f,//face 3                

        -0.5f,0.5f,0.5f,
        -0.5f,0.5f,-0.5f,
        -0.5f,-0.5f,-0.5f,
        -0.5f,-0.5f,0.5f,//face 4

        0.5f,0.5f,0.5f,
        0.5f,0.5f,-0.5f,
        -0.5f,0.5f,-0.5f,
        -0.5f,0.5f,0.5f,//face 5

        -0.5f,-0.5f,0.5f,
        -0.5f,-0.5f,-0.5f,
        0.5f,-0.5f,-0.5f,
        0.5f,-0.5f,0.5f//face 6     

    };

现在,我正在改变z绳索:

for(int i=0;i<24;i++)
    vertices[i*3+2]*=10
glDepthRange(0,10.0);

现在,我期待由于glDepthRange调用,z线将被映射到-0.5到0.5范围,并且我可以看到一个合适的立方体,但是它给出o/p,就像我在上面用变形几何注释glDepthRange一样.

Answer 1

让我引用一下手册页:

在剪切和除以w之后,深度坐标的范围从-1到1,对应于近剪裁平面和远剪裁平面.glDepthRange指定此范围内的标准化深度坐标与窗口深度坐标的线性映射.无论实际的深度缓冲区实现如何,窗口坐标深度值都被视为从0到1(如颜色分量).因此,glDepthRange接受的值在被接受之前都被钳制到该范围.

突出我的

换句话说,你不能真正使用大于1的深度值; 你必须自己适当地计算它们.它本质上得出结论,以设置适当的near,并far在你的投影矩阵的价值观和翻译中的Z的模型变换矩阵坐标,使所有最终ZS是前述之间.

如果您需要更详细的解释,请发表评论.