DX9 Vbuffer Ibuffer question
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Hi everyone,
I thought I figured out how DrawIndexedPrimitive works until it completely confused me 10 minutes ago.
I was trying to draw an arbitrary polygon. It has 6 sides and 6 vertices. So basically you start with a vertex and play connect the dots pretty much an voila! there is the polygon. In case you want to picture it:
Vertices:
v[0] = VertexPos(-1.0f, 2.0f, 1.0f);
v[1] = VertexPos(2.0f, 1.75f, 1.0f);
v[2] = VertexPos( 1.0f, 0.0f, 1.0f);
v[3] = VertexPos( 3.0f, -0.25f, 1.0f);
v[4] = VertexPos(0.0f, -2.0f, 1.0f);
v[5] = VertexPos(-2.0f, 1.0f, 1.0f);
Indices:
k[0] = 0; k[1] = 1;
k[2] = 1; k[3] = 2;
k[4] = 2; k[5] = 3;
k[6] = 3; k[7] = 4;
k[8] = 4; k[9] = 5;
k[10] = 5; k[11] = 0;
So, I figured, to the "number of vertices" paramter of the function, I needed to pass 6 vertices since there are 6. Also for the "number of primitives", I thought I should pass 6 again, because it has 6 sides. And I am using line strips. However it didn't completely draw my shape. So I decided to play with numbers, completely confused and found out that the least parameters that work are like this:
HR(gd3dDevice->DrawIndexedPrimitive(D3DPT_LINESTRIP, 0, 0, 0, 0, 11));
Can somebody explain to me, why even though I pass 0 vertices, it draws the whole shape and why I need to pass at least 11 primitives?
Thanks
In your situation, an index buffer is barely beneficial. Try the following :
EDIT : As for why it draws even if you specify 0 vertices, you should know that, when drawing indexed primitives, the vertices you specify are only a hint to the graphics driver so it knows which vertices will be indexed by your draw call. If you pass 0 vertices, I bet that the driver consider that any vertex in your vertex buffer could be indexed. It's a simple matter of optimization.
EDIT 2 : The way you defined your indices, by pairs of two points, corresponds to the D3DPT_LINELIST primitive type, not the D3DPT_LINESTRIP primitive type. More on this on msdn
// Indicesk[0] = 0; k[1] = 1; // Connect vertex 0 to vertex 1 tok[2] = 2; k[3] = 3; // vertex 2 to vertex 3 tok[4] = 4; k[5] = 5; // vertex 4 to vertex 5 and back tok[6] = 0; // vertex 0// Draw callgd3dDevice->DrawIndexedPrimitive(D3DPT_LINESTRIP,0, // You'll basically always be using zero here0, 6, // Starting from vertex 0, you're using all of your 6 vertices0, // The first index you're using is index 06); // You're going to draw 6 primitives (6 lines)EDIT : As for why it draws even if you specify 0 vertices, you should know that, when drawing indexed primitives, the vertices you specify are only a hint to the graphics driver so it knows which vertices will be indexed by your draw call. If you pass 0 vertices, I bet that the driver consider that any vertex in your vertex buffer could be indexed. It's a simple matter of optimization.
EDIT 2 : The way you defined your indices, by pairs of two points, corresponds to the D3DPT_LINELIST primitive type, not the D3DPT_LINESTRIP primitive type. More on this on msdn
What you were making was a LINELIST. In a linelist, you can have seperate unconnected lines, and requires the extra indices to point out all the start and end points.
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