void display(void) {
  int sign = 1;
  if (lastFrameTime == 0) {
    /*
     * sets lastFrameTime to be the number of milliseconds since
     * Init() was called;
     */
    lastFrameTime = glutGet(GLUT_ELAPSED_TIME);
  }

  int now = glutGet(GLUT_ELAPSED_TIME);
  int elapsedMilliseconds = now - lastFrameTime;
  float elapsedTime = float(elapsedMilliseconds) / 1000.0f;
  lastFrameTime = now;

  int windowWidth = glutGet(GLUT_WINDOW_WIDTH);

  if (BoxX > windowWidth) {
    BoxX -= windowWidth;
  }
  BoxX += (sign)*256.0f * elapsedTime;

  glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);

  glPushMatrix();
  //creates a new matrix at the top that we can do things to?
  glTranslatef(BoxX,0.0f,0.0f);

  /*
   * draw a "quad" (rectangle)
   */
  glBegin(GL_QUADS);
  glVertex2f(0.0f,0.0f);
  glVertex2f(128.0f,128.0f);
  glVertex2f(0.0f,128.0f);
  glEnd();
  glPopMatrix();
  //pops that matrix off the stack so we can have a "clean" version to do something next time.?

  glutSwapBuffers();
}

现在,我了解glPushMatrix()和glPopMatrix()的方式是,glPushMatrix()在堆栈上放置(或者推送)一个新的矩阵,以便您进行操作,以便在弹出后,您将再次有一个“干净的”.这就是为什么如果我在glEnd()之后忽略glPopMatrix(),我的平方似乎加速而不是以恒定的速度移动.

但是,如何保留在glPushMatrix()和glPopMatrix()中进行的更改？当我使用glPushMatrix()并对顶部矩阵进行更改时,它可以显示变化,但是当我使用glPopMatrix()时,并不是所有这些更改都消失了？当我再次恢复到“干净的”石板时,我的盒子是如何移动到屏幕上的？

如果我在进行更改后再次弹出矩阵,那么该翻译的状态如何记录？