glRotate 除零

Question

我想我明白为什么调用 glRotate(#, 0, 0, 0) 会导致被零除。旋转向量 a 已标准化：a' = a/|a| = a/0

这是 glRotate 可能导致被零除的唯一情况吗？是的，我知道 glRotate 已被弃用。是的，我知道矩阵位于 OpenGL 手册上。不，我对线性代数的了解不足以自信地回答矩阵的问题。是的，我认为这会有帮助。是的，我已经在 #opengl 中问过这个问题了（你能告诉我吗？）。不，我没有得到答案。

Answer 1

我会说是的。我想说，你对标准化步骤的看法也是正确的。 OpenGL 手册中显示的矩阵仅由乘法组成。向量相乘也会得到相同的结果。当然，如果你得到一个 (0,0,0) 向量，它会做奇怪的事情。 OpenGL 在同一个手册中声明 |x， y,z|=1 （或者 OpenGL 将标准化）。

因此，如果它不正常化，您最终会得到一个非常空的矩阵：

0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1

这将以最奇怪的方式使您的对象内爆。所以不要用零向量调用这个函数。如果您愿意，请告诉我原因。

我建议使用像 GLM 这样的库来进行矩阵计算，如果它对于一些简单的<代码>glRotates。

Answer 2

当您可以检查时，为什么它应该除以零那？：

/**
 * Generate a 4x4 transformation matrix from glRotate parameters, and
 * post-multiply the input matrix by it.
 *
 * \author
 * This function was contributed by Erich Boleyn ([email protected]).
 * Optimizations contributed by Rudolf Opalla ([email protected]).
 */
void
_math_matrix_rotate( GLmatrix *mat,
             GLfloat angle, GLfloat x, GLfloat y, GLfloat z )
{
   GLfloat xx, yy, zz, xy, yz, zx, xs, ys, zs, one_c, s, c;
   GLfloat m[16];
   GLboolean optimized;

   s = (GLfloat) sin( angle * DEG2RAD );
   c = (GLfloat) cos( angle * DEG2RAD );

   memcpy(m, Identity, sizeof(GLfloat)*16);
   optimized = GL_FALSE;

#define M(row,col)  m[col*4+row]

   if (x == 0.0F) {
      if (y == 0.0F) {
         if (z != 0.0F) {
            optimized = GL_TRUE;
            /* rotate only around z-axis */
            M(0,0) = c;
            M(1,1) = c;
            if (z < 0.0F) {
               M(0,1) = s;
               M(1,0) = -s;
            }
            else {
               M(0,1) = -s;
               M(1,0) = s;
            }
         }
      }
      else if (z == 0.0F) {
         optimized = GL_TRUE;
         /* rotate only around y-axis */
         M(0,0) = c;
         M(2,2) = c;
         if (y < 0.0F) {
            M(0,2) = -s;
            M(2,0) = s;
         }
         else {
            M(0,2) = s;
            M(2,0) = -s;
         }
      }
   }
   else if (y == 0.0F) {
      if (z == 0.0F) {
         optimized = GL_TRUE;
         /* rotate only around x-axis */
         M(1,1) = c;
         M(2,2) = c;
         if (x < 0.0F) {
            M(1,2) = s;
            M(2,1) = -s;
         }
         else {
            M(1,2) = -s;
            M(2,1) = s;
         }
      }
   }

   if (!optimized) {
      const GLfloat mag = SQRTF(x * x + y * y + z * z);

      if (mag <= 1.0e-4) {
         /* no rotation, leave mat as-is */
         return;
      }

      x /= mag;
      y /= mag;
      z /= mag;


      /*
       *     Arbitrary axis rotation matrix.
       *
       *  This is composed of 5 matrices, Rz, Ry, T, Ry', Rz', multiplied
       *  like so:  Rz * Ry * T * Ry' * Rz'.  T is the final rotation
       *  (which is about the X-axis), and the two composite transforms
       *  Ry' * Rz' and Rz * Ry are (respectively) the rotations necessary
       *  from the arbitrary axis to the X-axis then back.  They are
       *  all elementary rotations.
       *
       *  Rz' is a rotation about the Z-axis, to bring the axis vector
       *  into the x-z plane.  Then Ry' is applied, rotating about the
       *  Y-axis to bring the axis vector parallel with the X-axis.  The
       *  rotation about the X-axis is then performed.  Ry and Rz are
       *  simply the respective inverse transforms to bring the arbitrary
       *  axis back to its original orientation.  The first transforms
       *  Rz' and Ry' are considered inverses, since the data from the
       *  arbitrary axis gives you info on how to get to it, not how
       *  to get away from it, and an inverse must be applied.
       *
       *  The basic calculation used is to recognize that the arbitrary
       *  axis vector (x, y, z), since it is of unit length, actually
       *  represents the sines and cosines of the angles to rotate the
       *  X-axis to the same orientation, with theta being the angle about
       *  Z and phi the angle about Y (in the order described above)
       *  as follows:
       *
       *  cos ( theta ) = x / sqrt ( 1 - z^2 )
       *  sin ( theta ) = y / sqrt ( 1 - z^2 )
       *
       *  cos ( phi ) = sqrt ( 1 - z^2 )
       *  sin ( phi ) = z
       *
       *  Note that cos ( phi ) can further be inserted to the above
       *  formulas:
       *
       *  cos ( theta ) = x / cos ( phi )
       *  sin ( theta ) = y / sin ( phi )
       *
       *  ...etc.  Because of those relations and the standard trigonometric
       *  relations, it is pssible to reduce the transforms down to what
       *  is used below.  It may be that any primary axis chosen will give the
       *  same results (modulo a sign convention) using thie method.
       *
       *  Particularly nice is to notice that all divisions that might
       *  have caused trouble when parallel to certain planes or
       *  axis go away with care paid to reducing the expressions.
       *  After checking, it does perform correctly under all cases, since
       *  in all the cases of division where the denominator would have
       *  been zero, the numerator would have been zero as well, giving
       *  the expected result.
       */

      xx = x * x;
      yy = y * y;
      zz = z * z;
      xy = x * y;
      yz = y * z;
      zx = z * x;
      xs = x * s;
      ys = y * s;
      zs = z * s;
      one_c = 1.0F - c;

      /* We already hold the identity-matrix so we can skip some statements */
      M(0,0) = (one_c * xx) + c;
      M(0,1) = (one_c * xy) - zs;
      M(0,2) = (one_c * zx) + ys;
/*    M(0,3) = 0.0F; */

      M(1,0) = (one_c * xy) + zs;
      M(1,1) = (one_c * yy) + c;
      M(1,2) = (one_c * yz) - xs;
/*    M(1,3) = 0.0F; */

      M(2,0) = (one_c * zx) - ys;
      M(2,1) = (one_c * yz) + xs;
      M(2,2) = (one_c * zz) + c;
/*    M(2,3) = 0.0F; */

/*
      M(3,0) = 0.0F;
      M(3,1) = 0.0F;
      M(3,2) = 0.0F;
      M(3,3) = 1.0F;
*/
   }
#undef M

   matrix_multf( mat, m, MAT_FLAG_ROTATION );
}