Display Part of Texture in GLSL - opengl

I'm using GLSL to draw sprites from a sprite-sheet. I'm using jME 3, yet there are only small differences, and only with regards to deprecated functions.
The most important part of drawing a sprite from a sprite sheet is to draw only a subset/range of pixels, for example the range from (100, 0) to (200, 100). In the following test case sprite-sheet, and using the previous bounds, only the green part of the sprite-sheet would be drawn.
.
This is what I have so far:
Definition:
MaterialDef Solid Color {
//This is the list of user-defined variables to be used in the shader
MaterialParameters {
Vector4 Color
Texture2D ColorMap
}
Technique {
VertexShader GLSL100: Shaders/tc_s1.vert
FragmentShader GLSL100: Shaders/tc_s1.frag
WorldParameters {
WorldViewProjectionMatrix
}
}
}
.vert file:
uniform mat4 g_WorldViewProjectionMatrix;
attribute vec3 inPosition;
attribute vec4 inTexCoord;
varying vec4 texture_coordinate;
void main(){
gl_Position = g_WorldViewProjectionMatrix * vec4(inPosition, 1.0);
texture_coordinate = vec4(inTexCoord);
}
.frag:
uniform vec4 m_Color;
uniform sampler2D m_ColorMap;
varying vec4 texture_coordinate;
void main(){
vec4 color = vec4(m_Color);
vec4 tex = texture2D(m_ColorMap, texture_coordinate);
color *= tex;
gl_FragColor = color;
}
In jME 3, inTexCoord refers to gl_MultiTexCoord0, and inPosition refers to gl_Vertex.
As you can see, I tried to give the texture_coordinate a vec4 type, rather than a vec2, so as to be able to reference its p and q values (texture_coordinate.p and texture_coordinate.q). Modifying them only resulted in different hues.
m_Color refers to the color, inputted by the user, and serves the purpose of altering the hue. In this case, it should be disregarded.
So far, the shader works as expected and the texture displays correctly.
I've been using resources and tutorials from NeHe (http://nehe.gamedev.net/article/glsl_an_introduction/25007/) and Lighthouse3D (http://www.lighthouse3d.com/tutorials/glsl-tutorial/simple-texture/).
Which functions/values I should alter to get the desired effect of displaying only part of the texture?

Generally, if you want to only display part of a texture, then you change the texture coordinates associated with each vertex. Since you don't show your code for how you're telling OpenGL about your vertices, I'm not sure what to suggest. But in general, if you're using older deprecated functions, instead of doing this:
// Lower Left of triangle
glTexCoord2f(0,0);
glVertex3f(x0,y0,z0);
// Lower Right of triangle
glTexCoord2f(1,0);
glVertex3f(x1,y1,z1);
// Upper Right of triangle
glTexCoord2f(1,1);
glVertex3f(x2,y2,z2);
You could do this:
// Lower Left of triangle
glTexCoord2f(1.0 / 3.0, 0.0);
glVertex3f(x0,y0,z0);
// Lower Right of triangle
glTexCoord2f(2.0 / 3.0, 0.0);
glVertex3f(x1,y1,z1);
// Upper Right of triangle
glTexCoord2f(2.0 / 3.0, 1.0);
glVertex3f(x2,y2,z2);
If you're using VBOs, then you need to modify your array of texture coordinates to access the appropriate section of your texture in a similar manner.
For the sampler2D the texture coordinates are normalized so that the leftmost and bottom-most coordinates are 0, and the rightmost and topmost are 1. So for your example of a 300-pixel-wide texture, the green section would be between 1/3rd and 2/3rds the width of the texture.

Related

sampling GL_DEPTH_COMPONENTs of type GL_UNSIGNED_SHORT in GLSL shader

I have access to a depth camera's output. I want to visualise this in opengl using a compute shader.
The depth feed is given as a frame and i know the width and height ahead of time. How do I sample the texture and retrieve the depth value in the shader? Is this possible? I've read through the OpenGl types here and can't find anything on unsigned shorts so am starting to worry. Are there any workarounds?
My current compute shader
#version 430
layout(local_size_x = 1, local_size_y = 1) in;
layout(rgba32f, binding = 0) uniform image2D img_output;
uniform float width;
uniform float height;
uniform sampler2D depth_feed;
void main() {
// get index in global work group i.e x,y position
vec2 sample_coords = ivec2(gl_GlobalInvocationID.xy) / vec2(width, height);
float visibility = texture(depth_feed, sample_coords).r;
vec4 pixel = vec4(1.0, 1.0, 0.0, visibility);
// output to a specific pixel in the image
imageStore(img_output, ivec2(gl_GlobalInvocationID.xy), pixel);
}
The depth texture definition is as follows:
glTexImage2D(GL_TEXTURE_2D, 0, GL_DEPTH_COMPONENT16, width, height, 0,GL_DEPTH_COMPONENT, GL_UNSIGNED_SHORT, nullptr);
Currently my code produces a plain yellow screen.
If you use perspective projection, then the depth value is not linear. See LearnOpenGL - Depth testing.
If all the depth values are near 0.0, and you use the following expression:
vec4 pixel = vec4(vec3(visibility), 1.0);
then all the pixels appear almost black. Actually the pixels are not completely black, but the difference is barely noticeable.
This happens, when the far plane is "too" far away. To verify that you can compute the power of 1.0 - visibility, to make the different depth values ​​recognizable. For instance:
float exponent = 5.0;
vec4 pixel = vec4(vec3(pow(1.0-visibility, exponent)), 1.0);
If you want a more sophisticated solution, you can linearize the depth values as explained in the answer to How to render depth linearly in modern OpenGL with gl_FragCoord.z in fragment shader?.
Please note that for a satisfactory visualization you should use the entire range of the depth buffer ([0.0, 1.0]). The geometry must be between the near and far planes, but try to move the near and far planes as close to the geometry as possible.

How can I add different color overlays to each iteration of the texture in this glsl shader?

So I'm working on this shader right now, and the goal is to have a series of camera based tiles, each overlayed with a different color (think Andy Warhol). I've got the tiles working (side issue - the tiles on the ends are currently being cut off ~50%) but I want to add a color filter to each iteration. I'm looking for the cleanest possible way of doing this. Any ideas?
frag shader:
#define N 3.0 // number of columns
#define M 3. // number of rows
#import "GPUImageWarholFilter.h"
NSString *const kGPUImageWarholFragmentShaderString = SHADER_STRING
(
precision highp float;
varying highp vec2 textureCoordinate;
uniform sampler2D inputImageTexture;
void main()
{
vec4 color = texture2D(inputImageTexture, vec2(fract(textureCoordinate.x * N)/(M/ N), fract(textureCoordinate.y * M) / (M/N)));
gl_FragColor = color;
}
);
Let's say you're texture is getting applied to a quad and your vertices are:
float quad[] = {
0.0, 0.0,
1.0, 0.0,
1.0, 0.0, // First triangle
1.0, 0.0,
1.0, 0.0,
1.0, 1.0 // Second triangle
};
You can specify texture coordinates which corresponds to each of these vertices and describe the relationship in your vertex array object.
Now, if you quadruple the number of vertices, you'll have access to mid points between the extents of the quad (create four sub-quads) and you can associate (0,0) -> (1,1) texture coordinates with each of these subquads. The effect would be a rendering of the full texture in each subquad.
You could then do math on your vertices in your vertex shader to calculate a color component to assign to each subquad. The color component would be passed to your fragment shader. Use uniforms to specify number of rows and columns then define the color component based on which cell you've calculated the current vertex to be in.
You could also try to compute everything in the fragment shader, but I think it could get expensive pretty fast. Just a hunch though.
You could also try the GL_REPEAT texture parameter, but you'll have less control of the outcome: https://open.gl/textures

OpenGL Missing Triangles Diffuse Shader

I'm using C++ and when I implemented a diffuse shader, it causes every other triangle to disappear.
I can post my render code if need be, but I believe the issue is with my normal matrix (which I wrote it to be the transpose inverse of the model view matrix). Here is the shader code which is sort of similar to that of a tutorial on lighthouse tutorials.
VERTEX SHADER
#version 330
layout(location=0) in vec3 position;
layout(location=1) in vec3 normal;
uniform mat4 transform_matrix;
uniform mat4 view_model_matrix;
uniform mat4 normal_matrix;
uniform vec3 light_pos;
out vec3 light_intensity;
void main()
{
vec3 tnorm = normalize(normal_matrix * vec4(normal, 1.0)).xyz;
vec4 eye_coords = transform_matrix * vec4(position, 1.0);
vec3 s = normalize(vec3(light_pos - eye_coords.xyz)).xyz;
vec3 light_set_intensity = vec3(1.0, 1.0, 1.0);
vec3 diffuse_color = vec3(0.5, 0.5, 0.5);
light_intensity = light_set_intensity * diffuse_color * max(dot(s, tnorm), 0.0);
gl_Position = transform_matrix * vec4(position, 1.0);
}
My fragment shader just outputs the "light_intensity" in the form of a color. My model is straight from Blender and I have tried different exporting options like keeping vertex order, but nothing has worked.
This is not related to you shader.
It appears to be depth test related. Here, the order of triangles in the depth relative to you viewport is messed up, because you do not make sure that only the nearest pixel to your camera gets drawn.
Enable depth testing and make sure you have a z buffer bound to your render target.
Read more about this here: http://www.opengl.org/wiki/Depth_Test
Only the triangles highlighted in red should be visible to the viewer. Due to the lack of a valid depth test, there is no chance to guarantee, what triangle is painted top most. Thus blue triangles of the faces that should not be visible will cover parts of previously drawn red triangles.
The depth test would omit this, by comparing the depth in the z buffer with the depth of the pixel to be drawn at the current moment. Only the color information of a pixel that is closer to the viewer, i.e. has a smaller z value than the z value in the buffer, shall be written to the framebuffer in order to achieve a correct result.
(Backface culling, would be nice too and, if the model is correctly exported, also allow to show it correctly. But it would only hide the main problem, not solve it.)

GLSL Fragment Position

In my .cpp code I create a list of quads, a few of them have a flag, in the pixel shader I check if this flag is set or not, if the flag is not set, the quad gets colored in red for example, if the flag is set, I want to decide the color of every single pixel, so if I need to colour half of the flagged quad in red and the other half in blue I can simply do something like :
if coordinate in quad < something color = red
else colour = blue;
In this way I can get half of the quad colored in blue and another half colored in red, or I can decide where to put the red color or where to put the blue one.
Imagine I've got a quad 50x50 pixels
[frag]
if(quad.flag == 1)
{
if(Pixel_coordinate.x<25 ) gl_fragColor = vec4(1.0, 0.0, 0.0, 1.0);
else gl_fragColor = vec4(0.0, 1.0, 0.0, 1.0);
}
else
gl_FragColor = vec4(1.0, 0.0, 0.0, 1.0);
In this case I would expect that a quad with the flag set will get two colors per face.
I hope I have been more specific now.
thanks.
Just to add something I can't use any texture.
Ok i do this now :
Every quad has 4 textures coordinated (0,0), (0,1), (1,1), (1,0);
I enable the texture coordinates using :
glTexCoordPointer(2, GL_SHORT, sizeof(Vertex), BUFFER_OFFSET(sizeof(float) * 7));
[vert]
varying vec2 texCoord;
main()
{
texCoord = gl_MultiTexCoord0.xy;
}
[frag]
varying vec2 texCoord;
main()
{
float x1 = texCoord.s;
float x2 = texCoord.t;
gl_FragColor = vec4(x1, x2, 0.0, 1.0);
}
I get always the yellow color so x1 =1 and x2 = 1 almost always and some quad is yellow/green.
I would expect that the texture coordinates change in the fragment shader and so I should get a gradient, am I wrong?
If you want to know the coordinate within the quad, you need to calculate it yourself. In order to that, you'll need to create a new interpolant (call it something like vec2 quadCoord), and set it appropriately for each vertex, which means you'll likely also need to add it as an attribute and pass it through your vertex shader. eg:
// in the vertex shader
attribute vec2 quadCoordIn;
varying vec2 quadCoord;
main() {
quadCoord = quadCoordIn;
:
You'll need to feed in this attribute in your drawing code when drawing your quads. For each quad, the vertexes will have likely have quadCoordIn values of (0,0), (0,1), (1,1) and (1,0) -- you could use some other coordinate system if you prefer, but this is the easiest.
Then, in your fragment program, you can access quadCoord.xy to determine where in the quad you are.
In addition to Chris Dodd's answer, you can also access the screen-space coordinate (in pixels, though actually pixel centers and thus ?.5) of the currently processed fragment through the special fragment shader variable gl_FragCoord:
gl_FragColor = (gl_FragCoord.x<25.0) ? vec4(1.0, 0.0, 0.0, 1.0) : vec4(0.0, 1.0, 0.0, 1.0);
But this gives you the position of the fragment in screen-space and thus relative to the lower left corner of you viewport. If you actually need to know the position inside the individual quad (which makes more sense if you want to actually color each quad half-by-half, since the "half-cut" would otherwise vary with the quad's position), then Chris Dodd's answer is the correct approach.

What's the best way to draw a fullscreen quad in OpenGL 3.2?

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.
I'm going to argue that the most efficient approach will be in drawing a single "full-screen" triangle. For a triangle to cover the full screen, it needs to be bigger than the actual viewport. In NDC (and also clip space, if we set w=1), the viewport will always be the [-1,1] square. For a triangle to cover this area just completely, we need to have two sides to be twice as long as the viewport rectangle, so that the third side will cross the edge of the viewport, hence we can for example use the following coordiates (in counter-clockwise order): (-1,-1), (3,-1), (-1,3).
We also do not need to worry about the texcoords. To get the usual normalized [0,1] range across the visible viewport, we just need to make the corresponding texcoords for the vertices tiwce as big, and the barycentric interpolation will yield exactly the same results for any viewport pixel as when using a quad.
This approach can of course be combined with attribute-less rendering as suggested in demanze's answer:
out vec2 texcoords; // texcoords are in the normalized [0,1] range for the viewport-filling quad part of the triangle
void main() {
vec2 vertices[3]=vec2[3](vec2(-1,-1), vec2(3,-1), vec2(-1, 3));
gl_Position = vec4(vertices[gl_VertexID],0,1);
texcoords = 0.5 * gl_Position.xy + vec2(0.5);
}
Why will a single triangle be more efficient?
This is not about the one saved vertex shader invocation, and the one less triangle to handle at the front-end. The most significant effect of using a single triangle will be that there are less fragment shader invocations
Real GPUs always invoke the fragment shader for 2x2 pixel sized blocks ("quads") as soon as a single pixel of the primitive falls into such a block. This is necessary for calculating the window-space derivative functions (those are also implicitly needed for texture sampling, see this question).
If the primitive does not cover all 4 pixels in that block, the remaining fragment shader invocations will do no useful work (apart from providing the data for the derivative calculations) and will be so-called helper invocations (which can even be queried via the gl_HelperInvocation GLSL function). See also Fabian "ryg" Giesen's blog article for more details.
If you render a quad with two triangles, both will have one edge going diagonally across the viewport, and on both triangles, you will generate a lot of useless helper invocations at the diagonal edge. The effect will be worst for a perfectly square viewport (aspect ratio 1). If you draw a single triangle, there will be no such diagonal edge (it lies outside of the viewport and won't concern the rasterizer at all), so there will be no additional helper invocations.
Wait a minute, if the triangle extends across the viewport boundaries, won't it get clipped and actually put more work on the GPU?
If you read the textbook materials about graphics pipelines (or even the GL spec), you might get that impression. But real-world GPUs use some different approaches like Guard-band clipping. I won't go into detail here (that would be a topic on it's own, have a look at Fabian "ryg" Giesen's fine blog article for details), but the general idea is that the rasterizer will produce fragments only for pixels inside the viewport (or scissor rect) anyway, no matter if the primitive lies completely inside it or not, so we can simply throw bigger triangles at it if both of the following are true:
a) the triangle does only extend the 2D top/bottom/left/right clipping planes (as opposed to the z-Dimension near/far ones, which are more tricky to handle, especially because vertices may also lie behind the camera)
b) the actual vertex coordinates (and all intermediate calculation results the rasterizer might be doing on them) are representable in the internal data formats the GPU's hardware rasterizer uses. The rasterizer will use fixed-point data types of implementation-specific width, while vertex coords are 32Bit single precision floats. (That is basically what defines the size of the Guard-band)
Our triangle is only factor 3 bigger than the viewport, so we can be very sure that there is no need to clip it at all.
But is it worth it?
Well, the savings on fragment shader invocations are real (especially when you have a complex fragment shader), but the overall effect might be barely measurable in a real-world scenario. On the other hand, the approach is not more complicated than using a full-screen quad, and uses less data, so even if might not make a huge difference, it won't hurt, so why not using it?
Could this approach be used for all sorts of axis-aligned rectangles, not just fullscreen ones?
In theory, you can combine this with the scissor test to draw some arbitrary axis-aligned rectangle (and the scissor test will be very efficient, as it just limits which fragments are produced in the first place, it isn't a real "test" in HW which discards fragments). However, this requires you to change the scissor parameters for each rectangle you want to draw, which implies a lot of state changes and limits you to a single rectangle per draw call, so doing so won't be a good idea in most scenarios.
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;
}
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.
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);
This is similar to the answer by demanze, but I would argue it's easier to understand. Also this is only drawn with 4 vertices by using TRIANGLE_STRIP.
#version 300 es
out vec2 textureCoords;
void main() {
const vec2 positions[4] = vec2[](
vec2(-1, -1),
vec2(+1, -1),
vec2(-1, +1),
vec2(+1, +1)
);
const vec2 coords[4] = vec2[](
vec2(0, 0),
vec2(1, 0),
vec2(0, 1),
vec2(1, 1)
);
textureCoords = coords[gl_VertexID];
gl_Position = vec4(positions[gl_VertexID], 0.0, 1.0);
}
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.