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Is it possible to have a dynamic array in a GLSL shader? For instance, what if I have something like this in my GLSL Shader:

uniform int size;
uniform SceneLights lights[size];

void main()
{
    for(int i = 0; i < size; i++) {
        /* Do Some Calculation */
    }
}

And this would be my C++ file:

for (GLuint i = 0; i < pointLights.size(); i++) {
    glUniform3f(glGetUniformLocation(shader, ("pointLights[" + std::to_string(i) + "].position").c_str()), lights[i].someValue);
}

I would like to do this because I have a file with the positions of all the light sources in my scene (this may vary for each scene) so I would like to have a dynamic array which will allow me to send different numbers of lights to my shader.

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  • 1
    $\begingroup$ There are also Shader Storage Buffer Objects that you can read about here: khronos.org/opengl/wiki/Shader_Storage_Buffer_Object And in vulkan there are "unbounded arrays" that are declared in GLSL like this: layout(binding=10) uniform sampler2D TextureUBA[];" (sampler2D is just an example) I'm not sure how that is set up in OpenGL though. The last version has very large limits but both of these have a performance penalty which varies depending on how the memory is accessed. $\endgroup$ – pmw1234 Dec 23 '20 at 23:24
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I don't think uniform arrays can be dynamically sized. In your case you should define the array as the maximum number of lights you will process and then use a uniform to control the number of iterations you do on this array. On the CPU side you can set a subset of the lights[] array according to the 'size' variable.

e.g.

#define MAX_LIGHTS 128

uniform int size;
uniform SceneLights lights[MAX_LIGHTS];

void main()
{
    for(int i = 0; i < size; i++) {
        /* Do Some Calculation */
    }
}

I think (speculate) the likely reason for this is that it would be impossible to determine the location of a uniform if there are variable sized arrays in your uniform list which depend on the value of another uniform.. e.g.

uniform int arraySize;               // offset 0
uniform int myArray[arraySize];      // offset 4
uniform int anotherVar;              // offset ???? 

GLSL would not know were to place anotherVar, because it would need arraySize to be set with a value before it can compute the offset of 'anotherVar'. I suppose the compiler could be clever and re-arrange the order of uniforms to get around this, but this will fail again if you have 2 or more variable sized arrays..

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You could add the variable as code when you upload the shader code.

char header[64];
sprintf (header, "const int size = %u;\n", pointLights.size());

Then add the header (assuming your code is in char code[]):

GLchar  *source[2] = {(GLchar  *)header, (GLchar  *)code};
glShaderSource(shader, 2, (const GLchar **)source, NULL);
..

That way you can vary the size and add it to the shader on the fly. Bit of a hack, but it's the way I did it.

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I am suprised that it was only mentioned in comments but yes you can use variable length arrays through SSBO. Here is the excerpt from wiki:

On the host side:

int data[SOME_SIZE];
...
GLuint ssbo;
glGenBuffers(1, &ssbo);
glBindBuffer(GL_SHADER_STORAGE_BUFFER, ssbo);
glBufferData(GL_SHADER_STORAGE_BUFFER, sizeof(data), data​, GLenum usage); //sizeof(data) only works for statically sized C/C++ arrays.
glBindBufferBase(GL_SHADER_STORAGE_BUFFER, 3, ssbo);
glBindBuffer(GL_SHADER_STORAGE_BUFFER, 0); // unbind

On the glsl side:

layout(std430, binding = 3) buffer layoutName
{
    int data_SSBO[];
};

Note that it has to be the last element in an SSBO:

layout(std430, binding = 2) buffer anotherLayoutName
{
    int some_int;
    float fixed_array[42];
    float variable_array[];
};

The only limitation is that you can use one variable array per SSBO. You can use multiple SSBO's in a shader though.

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  • $\begingroup$ Please remove the last statement of the answer. It is obviously wrong. Maybe you did mean that you can have only one dynamic array per ssbo? $\endgroup$ – user14063792468 Jan 9 at 23:12
  • $\begingroup$ @yvw yes that was my intention $\endgroup$ – Kaan E. Jan 10 at 10:10
  • $\begingroup$ removed the last statement as requested $\endgroup$ – Kaan E. Jan 10 at 10:11

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