带有 std::vector 的 VBO

问题描述

我用 C++ 和 OpenGL 编写了一个模型加载器.我已经使用 std::vectors 来存储我的顶点数据，但现在我想将它传递给 glBufferData()，但是数据类型却大不相同.我想知道是否有办法在 std::vector 和 glBufferData() 的文档const GLvoid * 之间进行转换.

I've written a model loader in C++ an OpenGL. I've used std::vectors to store my vertex data, but now I want to pass it to glBufferData(), however the data types are wildly different. I want to know if there's a way to convert between std::vector to the documented const GLvoid * for glBufferData().

typedef struct
{
    float x, y, z;
    float nx, ny, nz;
    float u, v;
}
Vertex;

vector<Vertex> vertices;

glBufferData() 调用

glBufferData(GL_ARRAY_BUFFER, vertices.size() * 3 * sizeof(float), vertices, GL_STATIC_DRAW);

我收到以下(预期的)错误:

I get the following (expected) error:

error: cannot convert ‘std::vector<Vertex>’ to ‘const GLvoid*’ in argument passing

如何将向量转换为与 glBufferData() 兼容的类型?

How can I convert the vector to a type compatible with glBufferData()?

注意.我现在不在乎正确的内存分配；vertices.size() * 3 * sizeof(float) 很可能会出现段错误，但我想先解决类型错误.

NB. I don't care about correct memory allocation at the moment; vertices.size() * 3 * sizeof(float) will most likely segfault, but I want to solve the type error first.

推荐答案

如果你有一个 std::vectorv，你可以获得一个 T* 指向连续数据的开始(这是 OpenGL 所追求的)，表达式 &v[0].

If you have a std::vector<T> v, you may obtain a T* pointing to the start of the contiguous data (which is what OpenGL is after) with the expression &v[0].

就您而言，这意味着将 Vertex* 传递给 glBufferData:

In your case, this means passing a Vertex* to glBufferData:

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

或者像这样，这是一样的:

Or like this, which is the same:

glBufferData(
   GL_ARRAY_BUFFER,
   vertices.size() * sizeof(Vertex),
   &vertices.front(),
   GL_STATIC_DRAW
);

<小时>

您可以在此处依赖从 Vertex* 到 void const* 的隐式转换；这应该不会造成问题.

You can rely on implicit conversion from Vertex* to void const* here; that should not pose a problem.

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