CODEWITHSUNDEEP
opengl
Mattias Cibien
Mattias Cibien

Reputation: 296

OpenGL stride glVertexPointer

I am working on a OpenGL 2.0 renderer (https://bitbucket.org/mattiascibien/anima-render) and I actually came across smething I cannot understand at all which is the stride for the glVertexPointer() instruction. Actually my code looks like this

glBindBuffer(GL_ARRAY_BUFFER, data.vertex_buffer);
glEnableClientState(GL_VERTEX_ARRAY);
glVertexPointer(
    3,
    GL_FLOAT,
    sizeof(GLfloat)*3,
    (void*)0
    );

but I actually found that by reading OpenGL documentation the stride should actually be 0. Changing it does not change how the model is rendered.

What value is correct? Why does changing this parameter makes no difference?

EDIT: the data is stored in a 

std::vector<GLFloat>

same happens with glTexCoordPointer using sizeof(GLfloat)*2 or 0

Upvotes: 0

Views: 264

Answers (2)

cfrick
cfrick

Reputation: 37033

https://www.opengl.org/sdk/docs/man2/xhtml/glInterleavedArrays.xml gives more infos about it.

If stride is 0, the aggregate elements are stored consecutively. Otherwise, stride bytes occur between the beginning of one aggregate array element and the beginning of the next aggregate array element.

while you are providing a stride, that is 0 or your "struct" size (3*sizeof(GLFloat)) this is the same. Yet I'd go with 0 in this case.

Upvotes: 2

Lev
Lev

Reputation: 434

Stride is the distance from the beginning of one entity, to the beginning of the following entity. 0 is special: it means the distance from the end of one entity to the beginning of the next entity is 0. So basically 0 is the same as sizeof(entity).

Basically they packed 2 different mechanisms in one parameter, which I find suboptimal.

Upvotes: 1

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