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uglybdavis

glm matrix layout

5 posts in this topic

Does anyone know the matrix layout for glm? I know it's laid out linearly in memory, i'm wondering where the basis vectors are. 

 

To my knowledge it looks like this

Xx, Xy, Xz, 0

Yx, Yy, Yz, 0

Zx, Zy, Zz, 0

Tx, Ty, Tz, 1

 

So, assuming a linear array of memory

flat* matrix = glm::value_ptr(mat);

 

The right vector would be row 1

glm::vec3 right = glm::vec3(matrix[0], matrix[1], matrix[2]);

 

The up vector would be row 2

glm::vec3 up = glm::vec3(matrix[4], matrix[5], matrix[6]);

 

The forward vector would be row 3

glm::vec3 forward = glm::vec3(matrix[8], matrix[9], matrix[10]);

 

and the position would be row 4

glm::vec3 position = glm::vec3(matrix[12], matrix[13], matrix[14]);

 

 

Also, how would one go from a 4x4 matrix to a quaternion?

Right now my assumption is

glm::quat a = glm::quat_cast(mat);

 

 

Can anyone please shed some light on weather or not i'm doing this right?

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Yeah, you are doing it right, but I'd flip that drawing of the matrix layout because the matrices are accessed as columns. You can access the columns with [] operator. For example matrix[3] will give the last column in mat4, which should be the translation part. So you can say glm::vec3 position = glm::vec3(mat[3]);

 

Imagine that the mat4 contains vec4 columns[4], and the value_ptr of that would be &columns[0].x.

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Wait, i tought the matrix in OpenGL was laid out linearly, so it indexes:

00 01 02 03

04 05 06 07

08 09 10 11

12 13 14 15

 

glm::value_ptr converts a mat4 into a float* for linear access.

 

But, the overloaded [] operator of glm accessess these elements column first, 

that is element 12 would be accessed as mat[3][0]

 

I think.......

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Yes, the matrix is laid out linearly as columns, you're just printing them as rows. It's still the same if you replace glm::value_ptr(mat) with &mat[0][0]. mat[3][0] == (&mat[0][0])[12].

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In the manual of glm (v0.9.3, section 4.11) it is mentioned that glm matrices use column-major order. This term means that the indices of consecutive memory places a mapped onto a matrix like so

| 00 04 08 12 |
| 01 05 09 13 |
| 02 06 10 14 |
| 03 07 11 15 |

[ 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 ]

This obviously allows for a simple access to columns, because those are given by 4 consecutive memory places.

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