OpenGL ES 2.0：矩阵乘法问题
我正在尝试使用 OpenGL ES 2.0 绘制一个四边形。 我的顶点着色器看起来像这样：
uniform mat4 mvp;
attribute vec4 position;
attribute vec4 color;
varying vec4 colorVarying;
void main()
{
gl_Position = mvp * position;
colorVarying = color;
}
如果我使用这些顶点：
-1.0f, -1.0f, 0.f, 1.f,
1.0f, -1.0f, 0.f, 1.f,
-1.0f, 1.0f, 0.f, 1.f,
1.0f, 1.0f, 0.f, 1.f
和这个矩阵（按列主顺序，与所有以下矩阵一样）：
0.5,0,0,0,
0, 1,0,0,
0, 0,1,0,
0.5,0,0,1
四边形覆盖视口的右半部分。事实上，如果我将这两个矩阵相乘，我会得到顶点
0, -1, 0, 1,
1, -1, 0, 1,
0, 1, 0, 1,
1 , 1, 0, 1,
，如果我使用这些顶点和单位矩阵，我会得到相同的结果。所以看起来乘法是正确完成的。
现在，如果我将这些顶点（A）：
-100.0f, -100.0f, 301.0f, 1.0f,
100.0f, -100.0f, 301.0f, 1.0f,
-100.0f, 100.0f, 301.0f, 1.0f,
100.0f, 100.0f, 301.0f, 1.0f,
乘以这个矩阵（M）：
0.4170 0 0 0
0 0.5560 0 0
0 0 -1.2220 -1.0000
0 0 444.4440 0
我得到这些顶点（B）：
0.2075, 0.2766, -0.9892, 1.0000
-0.2075, 0.2766, -0.9892, 1.0000,
0.2075, -0.2766, -0.9892, 1.0000,
-0.2075, -0.2766, -0.9892, 1.0000,
如果我将B顶点和单位矩阵传递给我的着色器，它会在矩阵的中间渲染一个小矩形视口。但如果我向它传递 A 顶点和 M 矩阵，它不会渲染任何内容。
这是怎么回事？
I'm trying to draw a quadrilateral using OpenGL ES 2.0.
My vertex shader looks like this:
uniform mat4 mvp;
attribute vec4 position;
attribute vec4 color;
varying vec4 colorVarying;
void main()
{
gl_Position = mvp * position;
colorVarying = color;
}
If I use these vertices:
-1.0f, -1.0f, 0.f, 1.f,
1.0f, -1.0f, 0.f, 1.f,
-1.0f, 1.0f, 0.f, 1.f,
1.0f, 1.0f, 0.f, 1.f
and this matrix (in column-major order, as are all following matrices):
0.5,0,0,0,
0, 1,0,0,
0, 0,1,0,
0.5,0,0,1
the quad covers the right half of the viewport. Indeed, if I multiply those two matrices I get the vertices
0, -1, 0, 1,
1, -1, 0, 1,
0, 1, 0, 1,
1 , 1, 0, 1,
and if I use those vertices and an identity matrix, I get the same result. So it seems that the multiplication is done correctly.
Now, if I multiply these vertices (A):
-100.0f, -100.0f, 301.0f, 1.0f,
100.0f, -100.0f, 301.0f, 1.0f,
-100.0f, 100.0f, 301.0f, 1.0f,
100.0f, 100.0f, 301.0f, 1.0f,
by this matrix (M):
0.4170 0 0 0
0 0.5560 0 0
0 0 -1.2220 -1.0000
0 0 444.4440 0
I get these vertices (B):
0.2075, 0.2766, -0.9892, 1.0000
-0.2075, 0.2766, -0.9892, 1.0000,
0.2075, -0.2766, -0.9892, 1.0000,
-0.2075, -0.2766, -0.9892, 1.0000,
and if I pass my shader the B vertices and an identity matrix, it renders a small rectangle in the middle of the viewport. But if I pass it the A vertices and the M matrix, it renders nothing.
What's going on?
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问题是，顶点着色器完成后，gl_Position 会经过裁剪，与范围 [-WW] 进行比较，如果 gl_Position 超出该范围，则进行裁剪（不绘制）。
如果您乘以 A * M 并将结果设置为 gl_Position，则给定您使用的矩阵，W 坐标将为 -301，一个负数，这将反转裁剪比较。
在进行乘法之前尝试将矩阵 M 乘以 -1，这应该产生与矩阵 B 乘以单位矩阵相同的结果。
来自“红皮书”：
The problem is that gl_Position after the vertex shader completes goes through clipping where it is compared against the range [-W W] and clipped ( not drawn ) if it gl_Position falls outside of that range.
IF you multiply A * M and set the result to gl_Position, then given the matrices you are using the W coordinate will be -301, a negative number, which will invert the clipping comparison.
Try multiplying the matrix M by -1 before doing the multiplication, that should yield the same results as multiplying matrix B by the identity matrix.
From the "Red Book":
您使用的数学库可能不使用与 GLSL 相同的矩阵顺序。在将矩阵 M 传递给着色器之前尝试对其进行转置。
The math library you are using might not use the same matrix order as GLSL. Try transposing your matrix M before passing it to the shader.
问题是，当 wz 单元格中的值为 -1 时，三角形最终其法线方向相反，并且由于它们是单面的，因此未绘制它们。我通过将 M 更改为这样解决了问题：
我还更改了 z 列的符号（实际上是矩阵转置后的行），以便在 z 轴上保持 -1 尽可能近，+1 尽可能远。
The problem was that with that -1 in the wz cell, the triangles ended up with their normal in the opposite directions, and since they are one-sided they were not drawn. I solved the problem by changing M to this:
I also changed the sign of the z column (actually row since the matrix is transposed) in order to keep -1 as near and +1 as far on the z axis.