What's the best way to draw a fullscreen quad

问题:

I'm doing ray casting in the fragment shader. I can think of a couple ways to draw a fullscreen quad for this purpose. Either draw a quad in clip space with the projection matrix set to the identity matrix, or use the geometry shader to turn a point into a triangle strip. The former uses immediate mode, deprecated in OpenGL 3.2. The latter I use out of novelty, but it still uses immediate mode to draw a point.

回答1:

You can send two triangles creating a quad, with their vertex attributes set to -1/1 respectively.

You do not need to multiply them with any matrix in the vertex/fragment shader.

Here are some code samples, simple as it is :)

Vertex Shader:

const vec2 madd=vec2(0.5,0.5);
attribute vec2 vertexIn;
varying vec2 textureCoord;
void main() {
   textureCoord = vertexIn.xy*madd+madd; // scale vertex attribute to [0-1] range
   gl_Position = vec4(vertexIn.xy,0.0,1.0);
}

Fragment Shader :

varying vec2 textureCoord;
void main() {
   vec4 color1 = texture2D(t,textureCoord);
   gl_FragColor = color1;
}

回答2:

To output a fullscreen quad geometry shader can be used:

#version 330 core

layout(points) in;
layout(triangle_strip, max_vertices = 4) out;

out vec2 texcoord;

void main() 
{
    gl_Position = vec4( 1.0, 1.0, 0.5, 1.0 );
    texcoord = vec2( 1.0, 1.0 );
    EmitVertex();

    gl_Position = vec4(-1.0, 1.0, 0.5, 1.0 );
    texcoord = vec2( 0.0, 1.0 ); 
    EmitVertex();

    gl_Position = vec4( 1.0,-1.0, 0.5, 1.0 );
    texcoord = vec2( 1.0, 0.0 ); 
    EmitVertex();

    gl_Position = vec4(-1.0,-1.0, 0.5, 1.0 );
    texcoord = vec2( 0.0, 0.0 ); 
    EmitVertex();

    EndPrimitive(); 
}

Vertex shader is just empty:

#version 330 core

void main()
{
}

To use this shader you can use dummy draw command with empty VBO:

glDrawArrays(GL_POINTS, 0, 1);

回答3:

A hint by Chistophe Riccio:

A large triangle is more efficient for reason I have illustrated in videos:

http://www.youtube.com/watch?v=WFx6StqpRdY

http://www.youtube.com/watch?v=WnUS8kzA3dI&feature=related

回答4:

The following comes from the draw function of the class that draws fbo textures to a screen aligned quad.

Gl.glUseProgram(shad);      

Gl.glBindBuffer(Gl.GL_ARRAY_BUFFER, vbo);           
Gl.glEnableVertexAttribArray(0);
Gl.glEnableVertexAttribArray(1);
Gl.glVertexAttribPointer(0, 3, Gl.GL_FLOAT, Gl.GL_FALSE, 0, voff);
Gl.glVertexAttribPointer(1, 2, Gl.GL_FLOAT, Gl.GL_FALSE, 0, coff);  

Gl.glActiveTexture(Gl.GL_TEXTURE0);
Gl.glBindTexture(Gl.GL_TEXTURE_2D, fboc);
Gl.glUniform1i(tileLoc, 0);

Gl.glDrawArrays(Gl.GL_QUADS, 0, 4);

Gl.glBindTexture(Gl.GL_TEXTURE_2D, 0);
Gl.glBindBuffer(Gl.GL_ARRAY_BUFFER, 0); 

Gl.glUseProgram(0); 

The actual quad itself and the coords are got from:

private float[] v=new float[]{  -1.0f, -1.0f, 0.0f,
                                1.0f, -1.0f, 0.0f,
                                1.0f, 1.0f, 0.0f,
                                -1.0f, 1.0f, 0.0f,

                                0.0f, 0.0f,
                                1.0f, 0.0f,
                                1.0f, 1.0f,
                                0.0f, 1.0f
};

The binding and set up of the vbo's I leave to you.

The vertex shader:

#version 330

layout(location = 0) in vec3 pos;
layout(location = 1) in vec2 coord;

out vec2 coords;

void main() {
    coords=coord.st;
    gl_Position=vec4(pos, 1.0);
}

Because the position is raw, that is, not multiplied by any matrix the -1, -1::1, 1 of the quad fit into the viewport. Look for Alfonse's tutorial linked off any of his posts on openGL.org.

回答5:

No need to use a geometry shader, a VBO or any memory at all.

A vertex shader can generate the quad.

layout(location = 0) out vec2 uv;

void main() 
{
    float x = float(((uint(gl_VertexID) + 2u) / 3u)%2u); 
    float y = float(((uint(gl_VertexID) + 1u) / 3u)%2u); 

    gl_Position = vec4(-1.0f + x*2.0f, -1.0f+y*2.0f, 0.0f, 1.0f);
    uv = vec2(x, y);
}

Bind an empty VAO. Send a draw call for 6 vertices.