Display Quad in top left corner.

Graphics and GPU Programming Programming DX11

Started by Yann_A June 28, 2013 11:26 PM

3 comments, last by Yann_A 10 years, 9 months ago

Yann_A

258

Author

June 28, 2013 11:26 PM

Hi there,

I am very new to rendering, and i have a problem displaying a quad on the top left corner of my screen.

I already have a small 3D engine, and i wanted to use this code http://www.rastertek.com/dx11tut22.html to display my quad.

The issue is that the quad shows, but way off the -x axis and up the y axis. So i actually have to move my camera down the -z axis and tilt it to the left to see the quad.

Camera position and aiming when starting :

After going way back and turning a on the left

I just want it to stick in front of my camera in the top left corner of my screen.

My guts tell me there is something wrong with my view matrix, or at least, it is not setup the way the rastertek tutorial expected it, but i can't find exactly what it is.

Vertex Shader


cbuffer PerFrameBuffer
{
float4x4 gWorldViewProj;
};

PixelInputType VS(VertexInputType input)
{
    PixelInputType output;    

// Change the position vector to be 4 units for proper matrix calculations.
    input.position.w = 1.0f;

// Calculate the position of the vertex against the world, view, and projection matrices.
    output.position = mul(float4(input.position.xyz, 1.0f), gWorldViewProj);
    
// Store the texture coordinates for the pixel shader.
output.tex = input.tex;
    
    return output;
}

Quad code sending the matrix


XMMATRIX worldViewProj = world * camera->View() * camera->Proj(); 
mTextureEffect->mfxWorldViewProj->SetMatrix(reinterpret_cast<float*>(&(worldViewProj)));

Camera code (from http://www.d3dcoder.net/d3d11.htm)


XMMATRIX Camera::View()const
{
return XMLoadFloat4x4(&mView);
}

void Camera::UpdateViewMatrix()
{
XMVECTOR R = XMLoadFloat3(&mRight);
XMVECTOR U = XMLoadFloat3(&mUp);
XMVECTOR L = XMLoadFloat3(&mLook);
XMVECTOR P = XMLoadFloat3(&mPosition);

// Keep camera's axes orthogonal to each other and of unit length.
L = XMVector3Normalize(L);
U = XMVector3Normalize(XMVector3Cross(L, R));

// U, L already ortho-normal, so no need to normalize cross product.
R = XMVector3Cross(U, L); 

// Fill in the view matrix entries.
float x = -XMVectorGetX(XMVector3Dot(P, R));
float y = -XMVectorGetX(XMVector3Dot(P, U));
float z = -XMVectorGetX(XMVector3Dot(P, L));

XMStoreFloat3(&mRight, R);
XMStoreFloat3(&mUp, U);
XMStoreFloat3(&mLook, L);

mView(0,0) = mRight.x; 
mView(1,0) = mRight.y; 
mView(2,0) = mRight.z; 
mView(3,0) = x;   

mView(0,1) = mUp.x;
mView(1,1) = mUp.y;
mView(2,1) = mUp.z;
mView(3,1) = y;  

mView(0,2) = mLook.x; 
mView(1,2) = mLook.y; 
mView(2,2) = mLook.z; 
mView(3,2) = z;   

mView(0,3) = 0.0f;
mView(1,3) = 0.0f;
mView(2,3) = 0.0f;
mView(3,3) = 1.0f;
}

The shader output vertices with very big negative values, when my camera is almost centered at the world origin, so there is no way it can see those.

So i am a bit confused about how i am supposed to use the view transformation (if the problem indeed lies there) so i can always see my quad no matter where i look.

Any help or suggestion would be greatly appreciated :-)

Thanks !

Zaoshi Kaba

8,470

June 29, 2013 03:13 AM

You don't really need view matrix to render 2D image on screen.

Try to take a look at this tutorial, which is specifically made to render 2D things on the screen, not 3D space.

It should explain how to position things on the screen, and you should be able to easily see where's your mistake.

Yann_A

258

Author

June 29, 2013 02:41 PM

You don't really need view matrix to render 2D image on screen.

Try to take a look at this tutorial, which is specifically made to render 2D things on the screen, not 3D space.

It should explain how to position things on the screen, and you should be able to easily see where's your mistake.

Thanks a lot for the reply !

I read that tutorial, and i have to admit i am still confused because i think this is extacly what i am doing.

But i kept diging into http://www.d3dcoder.net/d3d11.htm and i found a similar code in the shadowmap sample which works for me :

I just don't know why yet.

Contrary to rastertek code, the d3dcoder code creates vertices directly in NDC space


	// Load the vertex array with data.
	vertices[0].position = XMFLOAT3(-1.0f, -1.0f, 0.0f);
	vertices[0].texture = XMFLOAT2(0.0f, 0.0f);

	vertices[1].position = XMFLOAT3(-1.0f, +1.0f, 0.0f);
	vertices[1].texture = XMFLOAT2(1.0f, 1.0f);

	vertices[2].position = XMFLOAT3(+1.0f, +1.0f, 0.0f);
	vertices[2].texture = XMFLOAT2(0.0f, 1.0f);

	vertices[3].position = XMFLOAT3(+1.0f, -1.0f, 0.0f);
	vertices[3].texture = XMFLOAT2(0.0f, 0.0f);

and only sends this world transform matrix to the shader :


XMMATRIX world(
		0.5f, 0.0f, 0.0f, 0.0f,
		0.0f, 0.5f, 0.0f, 0.0f,
		0.0f, 0.0f, 1.0f, 0.0f,
		0.5f, -0.5f, 0.0f, 1.0f);

Nothing else.

No projection or view matrix sent at all. And no matter where i look, the quad remains visible.

Its like the quad is not part of the rest of the "world" and "bypassed" the whole 3d part.

How is this possible ?

Juliean

7,343

June 29, 2013 02:58 PM

Its like the quad is not part of the rest of the "world" and "bypassed" the whole 3d part.

How is this possible ?

Its just how graphics work. Your GPU does not know nor care what a view, projection matrix is, or whats drawn 3d and what not. It just sees vertices, and performs transformations on those in the vertex shader. You could even display a quad from (-1, -1, 0) to (1, 1, 0) without any transformation applied. What that matrix you supplied here does is scale the quad to 50% of its size, and move it 0.5 to the right, and 0.5 down. You don't need anything else, try to replace that matrix with identity and see what happens.

Yann_A

258

Author

July 06, 2013 12:04 AM

Thanks for the reply.
I think your answer helped me understand something very fundamental !
Every vertices in the frustrum ends up being projected on the screen, and my screen surface IS the NDC space, ranging from -1 to 1.

So the whole point of world * view * proj operation is to convert my vertex from its 3D position to a 2D position in the NDC space.
And nothing stops me from sending vertices with NDC coordinates directly to the GPU.
I feel very stupid for realizing this only now... The whole thing makes sense now !

Thank you very much :-)

Note: Please do not hesitate to correct me if i came to a wrong conclusion, i would much rather know it now ;-)

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