Fewer Colors Than Vertices

Question

In the old deprecated OpenGL, we could do something like this:

glBegin(...);
   glColor3f(r_1,g_1,b_1);
   glVertex3f(x_1, y_1, z_1);
   glVertex3f(x_2, y_2, z_2);
   ...
   glVertex3f(x_n, y_n, z_n);

   glColor3f(r_2, g_2, b_2);
   glVertex3f(x_(n+1), y_(n+1), z_(n+1));
   glVertex3f(x_(n+2), y_(n+2), z_(n+2));
   ...
   glVertex3f(x_2n, y_2n, z_2n);

   ...
glEnd();

That is, I am saying that each n consecutive vertices share the same color. Is the same possible to be done with the new and non-deprecated OpenGL?

For example, if I have a cube, it means that I have 36 vertices. If I want each face to have 1 color, then each consecutive 6 vertices must share that color. Currently I have artificially copied the color data 6 times for each color so that the sizes of vertex array and color array are the same. Is there any other way around this? Hope my question was clear.

Answer 1

Maybe this kind of pseudocode clears things up for you:

GLfloat color_state{r,g,b};
GLfloat normal_state{x,y,z};
...

glColor4f(r,g,b,a):
    color_state.r = r
    color_state.g = g
    color_state.b = b
    color_state.a = a

glNormalf(x,y,z):
    normal_state.x = x
    normal_state.y = y
    normal_state.z = z

glVertex3f(x,y,z):
    __glinternal_submit_vertex(
        position = {x,y,z},
        normal = normal_state,
        color = color_state,
        ... );

This is how OpenGL immediate works internally. Colors were not shared among vertices. A copy was created from the current state for each vertex submitted. Now, with Vertex Arrays and Vertex Buffer Object the burden lies upon you, to do the right data duplication.