OpenGL: passing dynamically allocated matrices to a uniform

Question

The title says it all.

For some reason this code is working:

float m[3][3] = {
    {1.0f, 0.0f, 0.0f},
    {0.0f, 1.0f, 0.0f},
    {0.0f, 0.0f, 1.0f}
};
glUniformMatrix3fv(basicShader->uniformAt(5), 1, GL_TRUE, &m[0][0]);

But this one isn't:

float **m = new float*[3];
for(int i = 0; i < 3; i++) {
    m[i] = new float[3];
    for(int j = 0; j < 3; j++) m[i][j] = 0;
}
m[0][0] = 1.0f;
m[1][1] = 1.0f;
m[2][2] = 1.0f;
glUniformMatrix3fv(basicShader->uniformAt(5), 1, GL_TRUE, &m[0][0]);

float[3][3] is a 2 dimensional array, in a coherent memory block. float ** is an array of pointers and each pointer points to a new 1 dimensional array. — Rabbid76
– Rabbid76, Commented Oct 7, 2018 at 14:00
Ohhh, that makes sense. So do I have to manually convert float ** to float[3][3] in order to send it to opengl? — Roy Varon
– Roy Varon, Commented Oct 7, 2018 at 14:02

birdfreeyahoo · Accepted Answer · 2018-10-07 14:09:48Z

1

float* m = new float[9] {1.0f, 0.0f, 0.0f, 
                         0.0f, 1.0f, 0.0f,
                         0.0f, 0.0f, 1.0f };



glUniformMatrix3fv(basicShader->uniformAt(5), 1, GL_TRUE, m);

2D arrays are actually laid out continuously.

EDIT: Just saying, in your first example you don't need &m[0][0]. Just &m is sufficient because the address points automatically to the first element.

edited Oct 7, 2018 at 14:09

answered Oct 7, 2018 at 14:05

birdfreeyahoo

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2 Comments

Roy Varon Over a year ago

I was hoping I could keep float **m somehow but I guess I can't. But your solution will work. Thanks!

birdfreeyahoo Over a year ago

It doesn't work just the way the OpenGL function parses the data. It parses 9 consecutive floats from the dereferenced pointer and doesn't dereference the single rows.

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