Jointts - 1 month ago 13

C++ Question

Hello I am having trouble figuring out why does passing the first element of an array of GLfloat positions (xyz) suddenly give the buffer all the elements of that array.

`glBufferData(GL_ARRAY_BUFFER, sizeof(Vertex) * vertices.size(), &vertices[0], GL_STATIC_DRAW);`

Here the argument for data is

The sane thing to do would be to pass the whole vector

Perhaps I am having a misunderstanding how OpenGL buffers work?

I have been following the tutorials from here: http://learnopengl.com/#!Model-Loading/Mesh

To sum up:

Answer

I'm assuming `vertices`

is a `std::vector<Vertex>`

. Since it is, that means that `&vertices`

is of type `std::vector<Vertex>*`

, or a pointer to the vector.

However, OpenGL doesn't want a vector. OpenGL was written for C, which has no notion of vectors. Therefore, it expects a C-style array. (If you're not sure what that is, see here.)

According to the C++ standard, a `std::vector`

must store all of its elements in continuous memory. That means that `vertices[1]`

must come right after `vertices[0]`

, `vertices[2]`

after `vertices[1]`

, and so forth. This is essentially exactly what C-style array is: a pointer to a continuous set of items.

If `vertices[0]`

is the first element of the vector, then `&vertices[0]`

is a pointer to the first element. Like I said above, all elements are stored continuously, so `&vertices[0]`

is really just a pointer to a C-style array.

In C++11, you can instead use `vertices.data()`

, which is essentially the same thing but cleaner.