OpenGL. 顶点乘以投影矩阵

Question

我正在尝试绘制一个 model，这就是我遇到的情况，下面的代码像 2d 一样工作，尽管应该有一个透视图

Code # 1
mat4 Proj = glFrustum(left, right, bottom, top, zNear, zFar);
mat4 View = gluLookAt(position, direction, up);
mat4 World = mat4(1);
glBegin(GL_TRIANGLES);
for (auto& t : mesh)
{
    norm = Proj * View * World * vec4(t.norm, 0);
    dot1 = Proj * View * World * vec4(t.dot1, 1);
    dot2 = Proj * View * World * vec4(t.dot2, 1);
    dot3 = Proj * View * World * vec4(t.dot3, 1);
 
    glNormal3f(norm);
    glVertex3f(dot1.x, dot1.y, dot1.z);
    glVertex3f(dot2.x, dot2.y, dot3.z);
    glVertex3f(dot3.x, dot3.y, dot3.z);
}
glEnd();

所以透视图出现了：

Code # 2
mat4 Proj = glFrustum(left, right, bottom, top, zNear, zFar);
mat4 View = gluLookAt(position, direction, up);
mat4 World = mat4(1);
glMatrixMode(GL_PROJECTION);
glLoadMatrix(Proj);
glBegin(GL_TRIANGLES);
for (auto& t : mesh)
{
    norm = View * World * vec4(t.norm, 0);
    dot1 = View * World * vec4(t.dot1, 1);
    dot2 = View * World * vec4(t.dot2, 1);
    dot3 = View * World * vec4(t.dot3, 1);
 
    glNormal3f(norm);
    glVertex3f(dot1.x, dot1.y, dot1.z);
    glVertex3f(dot2.x, dot2.y, dot3.z);
    glVertex3f(dot3.x, dot3.y, dot3.z);
}
glEnd();

由此产生的问题是：为什么代码2和代码3得到相同的结果，而代码1和代码2却不同？

Code # 3
mat4 Proj = glFrustum(left, right, bottom, top, zNear, zFar);
mat4 View = gluLookAt(position, direction, up);
mat4 World = mat4(1);
glMatrixMode(GL_PROJECTION);
glLoadMatrix(Proj);
*glMatrixMode(GL_MODELVIEW);
glLoadMatrix(View);*
glBegin(GL_TRIANGLES);
for (auto& t : mesh)
{
    norm = World * vec4(t.norm, 0);
    dot1 = World * vec4(t.dot1, 1);
    dot2 = World * vec4(t.dot2, 1);
    dot3 = World * vec4(t.dot3, 1);
 
    glNormal3f(norm);
    glVertex3f(dot1.x, dot1.y, dot1.z);
    glVertex3f(dot2.x, dot2.y, dot3.z);
    glVertex3f(dot3.x, dot3.y, dot3.z);
}
glEnd();

告诉我我做错了什么？ 如何正确计算顶点乘以投影矩阵？

Answer 1

与投影矩阵的变换生成齐次坐标。 齐次坐标有 4 个分量，但您只需将 3 个分量传递给顶点坐标。 使用所有 4 个组件来指定顶点坐标。 例如：

glVertex3f(dot1.x, dot1.y, dot1.z);

glVertex4f(dot1.x, dot1.y, dot1.z, dot1.w);

或者

glVertex4fv(glm::value_ptr(dot1));