Polygon Filling

@DrWatson

Published in 

atari

 · 11 months ago

Third part of The 3D Coding Blackhole tutorial series

• The polygon structure
• Finding the triangles
• Scanning the triangles lines

The polygon structure

How will we store our polygons? First of all, you must know that at this state, our polygons will be flat 2D polygons. And since the original polygons will be 3D, we will only need one temporary 2D polygon, so we can set the maximum number of 2D vertices to a constant number, without wasting an important amount of memory:

typedef struct 
{ 
   _2D Points[20]; 
    int PointsCount; 
   int Texture; 
}Polygon2D_t;

And our 3D structure:

typedef struct 
{ 
    int Count; 
   int * Vertex; 
   int Texture; 

   Vertex_t P,M,N; 
}Polygon_t;

Wondering why the array of vertices consists of integer values? Well... If you think about it, in a cube three polygons use the same vertex. So storing and transforming the same vertex in 3 polygons would be wasting both time and memory. We will rather store them in an object structure, and in the polygons, we will put the indexes of the appropriate vertices. Look at our object structure:

typedef struct 
{ 
   int VertexCount; 
   int PolygonCount; 
   Vertex_t * Vertex; 
   Polygon_t * Polygon; 
   _3D Scaling; 
   _3D Position; 
   _3D Angle; 
   int NeedUpdate; 
}Object_t;

Finding the triangles

As you will realize in the next chapter, drawing a triangle is much simpler than drawing an arbitrary polygon. So we'll start by cutting our polygons into 3-vertices shapes. The method is really simple and straightforward:

void POLY_Draw(Polygon2D_t *Polygon) 
{ 
    _2D P1,P2,P3; 
   int i; 

   P1 = Polygon->Points[0]; 
   for(i=1; i PointsCount-1; i++) 
   { 
       P2=Polygon->Points[i]; 
       P3=Polygon->Points[i+1]; 
       POLY_Triangle(P1,P2,P3,Polygon->Texture); 
   } 
}

Scanning the triangles lines

Now what about the Triangle function? What we have to do is to draw every horizontal scan line, and to do so we have to find the starting and ending x coordinate of every row. We will start by defining two simple useful macros to class two points vertically and two numbers:

#define MIN(a,b) ((a<b)?(a):(b)) 
#define MAX(a,b) ((a>b)?(a):(b)) 
#define MaxPoint(a,b) ((a.y > b.y) ? a : b) 
#define MinPoint(a,b) ((b.y > a.y) ? a : b)

Then we will define three others to class three points:

#define MaxPoint3(a,b,c) MaxPoint(MaxPoint(a,b),MaxPoint(b,c)) 
#define MidPoint3(a,b,c) MaxPoint(MinPoint(a,b),MinPoint(a,c)) 
#define MinPoint3(a,b,c) MinPoint(MinPoint(a,b),MinPoint(b,c))

You will notice that the MidPoint3 macro doesn't always work properly, depending on the order of the 3 points. We will fix this with a if statement. Now our declarations:

void POLY_Triangle(_2D   p1,_2D p2,_2D p3,char c) 
{ 
   _2D p1d,p2d,p3d; 
   int xd1,yd1,xd2,yd2,i; 
   int Lx,Rx;

And we will do a first sorting of our 3 points:

   p1d = MinPoint3(p1,p2,p3); 
   p2d = MidPoint3(p2,p3,p1); 
   p3d = MaxPoint3(p3,p1,p2);

Why is there a rotation of the points when calling these macros? I'll tell you I'm not sure myself, but it probably has something to do with the fact that points are passed counter-clockwise. Try to change the macro and you will see junk on your screen! Now, we're not sure about our MidPoint, so we do a little check, and when we're in this condition, it seems that the MinPoint can be wrong, so we correct that too:

   if(p2.y < p1.y) 
   { 
      p1d=MinPoint3(p2,p1,p3); 
      p2d=MidPoint3(p1,p3,p2); 
   }

I know these orders seem strange, but just try to change them and everything turns to garbage so either try to understand them (and then please mail me your conclusion so I can add it here), or accept them the way they are. Anyway, now we must compute the deltas:

   xd1=p2d.x-p1d.x; 
   yd1=p2d.y-p1d.y; 
   xd2=p3d.x-p1d.x; 
   yd2=p3d.y-p1d.y;

Ok, here is the first side, that we only bother to draw if there is a delta y:

   if(yd1) 
      for(i=p1d.y; i<=p2d.y; i++) 
      {

We compute the x values with the starting x coordinate, adding the delta y between the current point and our starting point, multiplied by the inverse of the slope ( x / y ).

          Lx = p1d.x + ((i - p1d.y) * xd1) / yd1; 
          Rx = p1d.x + ((i - p1d.y) * xd2) / yd2;

If we are not on the same point, draw the horizontal run, passing the two points in order:

          if(Lx!=Rx) 
             VID_HLine(MIN(Lx,Rx),MAX(Lx,Rx),i,c); 
      }

Now we recompute the first deltas and do the second side:

   xd1=p3d.x-p2d.x; 
   yd1=p3d.y-p2d.y; 

   if(yd1) 
      for(i   = p2d.y;   i <= p3d.y;   i++) 
      { 
         Lx =   p1d.x   + ((i   - p1d.y)   * xd2) / yd2; 
         Rx =   p2d.x   + ((i   - p2d.y)   * xd1) / yd1; 
         if(Lx!=Rx) 
            VID_HLine(MIN(Lx,Rx),MAX(Lx,Rx),i,c); 
      } 
}

That's it! You've got your polygon filler! The implementation for a flat filling system is only a simple for:

void VID_HLine(int x1, int x2, int y, char c) 
{ 
   int x; 
   for(x=x1; x<=x2; x++) 
      putpixel(x, y, c); 
}

Copyright © 1996-1998 Jerome St-Louis