OpenGL GL_TRIANGLE_STRIP creating grid with degenerate triangles / for one draw call

Question

I need to create a grid ready for GL_TRIANGLE_STRIP rendering with One drawcall - so i need to degenerate the triangles.

I am almost there but missing last row/column and can't figure out why.

My grid is just a :

Position ( self explanatory )
Size ( the x,y size )
Resolution ( the spacing between each vertex in x,y )

Here is the method used to create verts/indices and return them:

int iCols = vSize.x / vResolution.x;
int iRows = vSize.y / vResolution.y;


// Create Vertices
for(int y = 0; y <= iRows; y ++)
{
    for(int x = 0; x <= iCols; x ++)
    {
        float startu  = (float)x / (float)vSize.x;
        float startv  = (float)y / (float)vSize.y; 


        tControlVertex.Color    = vColor;
        tControlVertex.Position = CVector3(x * vResolution.x,y * vResolution.y,0);
        tControlVertex.TexCoord = CVector2(startu, startv - 1.0 );

        vMeshVertices.push_back(tControlVertex);   
    }
}


// Create Indices
rIndices.clear();

for (int r = 0; r < iRows - 1; r++)
{
    rIndices.push_back(r*iCols);

    for (int c = 0; c < iCols; c++)
    {
        rIndices.push_back(r*iCols+c);
        rIndices.push_back((r+1)*iCols+c);
    }
    rIndices.push_back((r + 1) * iCols + (iCols - 1));
}

And to visualise that, few examples first.

1) Size 512x512 Resolution 64x64, so it should be made of 8 x 8 quads, but i get 7x7 only

2) Size 512x512 Resolution 128x128, so it should be made of 4 x 4 quads, but i get 3x3 only

3) Size 128x128 Resolution 8x8 so it should be made of 16 x 16 quads but i get 15x15 only

So as you can see, i am missing the Last Row and Last Column somewhere. Where am I going wrong?

CiscoIPPhone · Accepted Answer · 2013-01-25 19:57:37Z

2

If iCols is 10, then 11 horizontal vertices are created:

for(int x = 0; x <= iCols; x ++)

When creating the last column of indices you do:

rIndices.push_back((r + 1) * iCols + (iCols - 1));

We can ignore the first part to just concentrate on a single row:

rIndices.push_back(iCols - 1);

(iCols - 1) would be 9, so it's not going to address the last vertex (10). I think to fix the problem with columns you should remove the -1.

To fix the problem with the amount of rows change

(int r = 0; r < iRows - 1; r++)

to

(int r = 0; r < iRows; r++)

Because the difference between y <= iRows; and r < iRows - 1 is two, but there should only be one more vertical vertex than there are rows.

answered Jan 25, 2013 at 19:57

CiscoIPPhone

5,2783 gold badges34 silver badges39 bronze badges

\$\begingroup\$ +1 Rule of thumb: Are you simultaneously dealing with the notion of both tiles and vertices? It's probably an off-by-one error. \$\endgroup\$

michael.bartnett
– michael.bartnett

2013-01-25 20:00:56 +00:00
Commented Jan 25, 2013 at 20:00
\$\begingroup\$ CiscoIPPhone, unfortunately it does not work.. The code with your suggestions: fotoszok.pl/upload/44b05342.png The results: fotoszok.pl/upload/6c54ac19.png \$\endgroup\$

PeeS
– PeeS

2013-01-25 20:23:53 +00:00
Commented Jan 25, 2013 at 20:23
\$\begingroup\$ Try changing the rIndices.push_back(iCols); to rIndices.push_back((r + 1) * iCols + iCols)); I was just ignoring the first part of that line so it was easier to work out the last vertex addressed. \$\endgroup\$

CiscoIPPhone
– CiscoIPPhone

2013-01-25 20:27:35 +00:00
Commented Jan 25, 2013 at 20:27
\$\begingroup\$ @CiscoIPPhone Results: fotoszok.pl/upload/06a4a8a6.png Code: fotoszok.pl/upload/e19e52e8.png Still nothing.. \$\endgroup\$

PeeS
– PeeS

2013-01-25 20:31:38 +00:00
Commented Jan 25, 2013 at 20:31
\$\begingroup\$ @PeeS I think there are still some off by one errors. For example rIndices.push_back(r*iCols); is going to have the same index as rIndices.push_back(r*iCols+c); when c is 0. \$\endgroup\$

CiscoIPPhone
– CiscoIPPhone

2013-01-25 20:52:57 +00:00
Commented Jan 25, 2013 at 20:52

| Show 3 more comments

PeeS · Accepted Answer · 2013-01-25 22:04:48Z

Ok, i have solved this. I think we're all getting old, that's why we couldn't spot it.

Here is the code:

    // Create Vertices
// Notice <= to create exactly as many vertices in X and Y as required
for(int y = 0; y <= iRows ; y++)
{
    for(int x = 0; x <= iCols; x++)
    {
        float u  = (float)x / (float)vSize.x;
        float v  = (float)y / (float)vSize.y;

        tControlVertex.Color    = vColor;
        tControlVertex.Position = CVector3(x * vResolution.x, y * vResolution.y,0);
        tControlVertex.TexCoord = CVector2(u, v - 1.0 );

        vMeshVertices.push_back(tControlVertex);   
    }

}


// Create Indices
rIndices.clear();

// Create Indices with degenerate triangles ( stitching )
// Notice < iRows, <= iCols
for (int y = 0; y < iRows; y++)
{
    // Degenerate first triangle in this column
    rIndices.push_back((unsigned short)((y + 0) * (iCols+1) + 0));

    for (int x = 0; x < iCols + 1; x++)
    {
        // Vertex in actual row
        rIndices.push_back((unsigned short)((y + 0) * (iCols+1) + x));

        // Vertex row below
        rIndices.push_back((unsigned short)((y + 1) * (iCols+1) + x));
    }

    // Degenerate last triangle in this column
    rIndices.push_back((unsigned short)((y + 0) * (iCols+1) + iCols+1));
}

And the results are:

512x512 with Resolution of 64x64 ( called stepping ) produces 8*64 x 8*64 = 512x512.

enter image description here

Thanks CiscoIPPhone for looking into this.

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