How does OpenGL fill buffers and read them back?

问题

I was using an OpenGL buffer with a bunch of GLfloats as a vertex buffer and all was well. The format of the GLfloats being [x1, y1, z1, x2, y2, z2, ...].

But then, while following this tutorial, it tells me to use glm::vec3 instead:

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

glBufferData(GL_ARRAY_BUFFER, vertices.size() * sizeof(glm::vec3), &vertices[0], GL_STATIC_DRAW);

Now this code is valid, and I wonder how would OpenGL know how to fill in the buffer with glm::vec3 instead of GLfloats. Then I wonder, when I read the data back from the buffer, using:

std::vector<glm::vec3> data;
glGetBufferSubData(mTarget, offset, vertexCount * sizeof(glm::vec3), &data[0]);`

Will this make a bunch of glm::vec3? So the question is, how does OpenGL fill buffers with glm::vec3, and does (and if so, how does) it read it back?

回答1:

According to OpenGL's documentation, glBufferData() needs a pointer to the data (i.e. an array, i.e. the coordinates of the vertices).

Let's first have a look at glm::vec3's implementation.

If you check out glm's Github repo, you'll see that, depending on your compilation flags, glm::vec3 is a typedef of highp_vec3 which is a typedef of tvec3<float, highp>.

tvec3 is declared in type_vec3.hpp (included by vec3.hpp) and the class (template) methods are defined in type_vec3.inl.

In particular, operator[]'s definition is:

template <typename T, precision P>
GLM_FUNC_QUALIFIER T & tvec3<T, P>::operator[](typename tvec3<T, P>::length_type i)
{
    assert(i >= 0 && static_cast<detail::component_count_t>(i) < detail::component_count(*this));
    return (&x)[i];
}

Given that piece of code, one would assume that x is the first element of the "array" containing the coordinates of glm::vec3. However, when we go back to type_vec3.h, we find:

union { T x, r, s; };
union { T y, g, t; };
union { T z, b, p; };

So x, y and z are separate attributes. But thanks to how class/struct members are laid out, they can be viewed as a single array starting from &x.

We know now, that glm::vec3 (actually tvec3) stores the coordinates in a contiguous manner. But does it also store other attributes ?

Well, we can continue to dive into the code, or use a simple program to give us the answer:

#include <iostream>
#include <ios>

#include <glm/vec3.hpp>

int main()
{
    const glm::vec3 v;

    const size_t sizeof_v   = sizeof(v);
    const size_t sizeof_xyz = sizeof(v.x) + sizeof(v.y) + sizeof(v.z);

    std::cout << "sizeof(v)  : " << sizeof_v   << std::endl;
    std::cout << "sizeof(xyz): " << sizeof_xyz << std::endl;

    std::cout << "sizeof(v) == sizeof(xyz) : " << std::boolalpha << (sizeof_v == sizeof_xyz) << std::endl;
}

Which prints, on my machine:

sizeof(v)  : 12
sizeof(xyz): 12
sizeof(v) == sizeof(xyz) : true

Therefore, glm::vec3 stores only the (x, y, z) coordinates.

Now, if we create a std::vector<glm::vec3> vertices;, it is safe to say that the layout of the data pointed by &vertices[0] (which, in C++11 is vertices.data()) is:

vertices == [vertice1 vertice2 ...]
         == [vertice1.x vertice1.y vertice1.z vertice2.x vertice2.y vertice2.z ...]

Going back to the original issue -- glBufferData()'s requirements: when you pass &vertices[0] you are actually passing the address (i.e. pointer) of the data, just as expected by glBufferData(). The same logic applies to glGetBufferSubData().

回答2:

glm::vec3 is just three floats in a struct. So passing the adress of a glm::vec3 to gl function effectively does the same thing as passing the adress to the first element of a float array. GLfloat is just a typedef of float btw.

The same principle apply when reading data from gl. An array of x elements of glm::vec3 in memory is equivalent to an array of GLfloat (float) with 3x elements.

来源：https://stackoverflow.com/questions/33058927/how-does-opengl-fill-buffers-and-read-them-back

标签

c++

opengl

buffer

glm-math