Question

From this site: http://www.toymaker.info/Games/html/vertex_shaders.html

We have the following code snippet:

// transformations provided by the app, constant Uniform data
float4x4 matWorldViewProj: WORLDVIEWPROJECTION;

// the format of our vertex data
struct VS_OUTPUT
{
  float4 Pos  : POSITION;
};

// Simple Vertex Shader - carry out transformation
VS_OUTPUT VS(float4 Pos  : POSITION)
{
  VS_OUTPUT Out = (VS_OUTPUT)0;
  Out.Pos = mul(Pos,matWorldViewProj);
  return Out;
}

My question is: why does the struct VS_OUTPUT have a 4 dimensional vector as its position? Isn’t position just x, y and z?

Answer 1

Because you need the w coordinate for perspective calculation. After you output from the vertex shader than DirectX performs a perspective divide by dividing by w.

Essentially if you have 32768, -32768, 32768, 65536 as your output vertex position then after w divide you get 0.5, -0.5, 0.5, 1. At this point the w can be discarded as it is no longer needed. This information is then passed through the viewport matrix which transforms it to usable 2D coordinates.

Edit: If you look at how a matrix multiplication is performed using the projection matrix you can see how the values get placed in the correct places.

Taking the projection matrix specified in D3DXMatrixPerspectiveLH

2*zn/w  0       0              0
0       2*zn/h  0              0
0       0       zf/(zf-zn)     1
0       0       zn*zf/(zn-zf)  0

And applying it to a random x, y, z, 1 (Note for a vertex position w will always be 1) vertex input value you get the following

x' = ((2*zn/w) * x) + (0 * y) + (0 * z) + (0 * w)
y' = (0 * x) + ((2*zn/h) * y) + (0 * z) + (0 * w)
z' = (0 * x) + (0 * y) + ((zf/(zf-zn)) * z) + ((zn*zf/(zn-zf)) * w)
w' = (0 * x) + (0 * y) + (1 * z) + (0 * w)

Instantly you can see that w and z are different. The w coord now just contains the z coordinate passed to the projection matrix. z contains something far more complicated.

So .. assume we have an input position of (2, 1, 5, 1) we have a zn (Z-Near) of 1 and a zf (Z-Far of 10) and a w (width) of 1 and a h (height) of 1.

Passing these values through we get

x' = (((2 * 1)/1) * 2
y' = (((2 * 1)/1) * 1
z' = ((10/(10-1)  * 5 + ((10 * 1/(1-10)) * 1)
w' = 5

expanding that we then get

x' = 4
y' = 2
z' = 4.4
w' = 5

We then perform final perspective divide and we get

x'' = 0.8
y'' = 0.4
z'' = 0.88
w'' = 1

And now we have our final coordinate position. This assumes that x and y ranges from -1 to 1 and z ranges from 0 to 1. As you can see the vertex is on-screen.

As a bizarre bonus you can see that if |x’| or |y’| or |z’| is larger than |w’| or z’ is less than 0 that the vertex is offscreen. This info is used for clipping the triangle to the screen.

Anyway I think thats a pretty comprehensive answer 😀

Edit2: Be warned i am using ROW major matrices. Column major matrices are transposed.