当修改基础OpenGL状态时，我应该声明方法const吗？

Question

The following class encapsulates the OpenGL name of a buffer and provides a few methods for changing the state of the buffer:

class BufferObject {
    public:
        explicit BufferObject( GLenum type );
        virtual ~BufferObject();

        // some methods omitted

        void    dataStore( GLsizeiptr size, const GLvoid* data, int usage );
        void*   mapBufferRange( GLintptr offset, GLsizeiptr length, int accessFlag );
        void    unmapBuffer() const;
    private:
        GLuint object_;
};

None of these methods change the state of the BufferObject object, so they could all be declared with const . However, dataStore and mapBufferRange both call OpenGL methods which change the state of the object on the GPU ( glBufferData and glMapBufferRange , respectively). I would like to declare them without const to indicate that they are modifying the state on the GPU.

What is the best practise to follow in this situation?

Answer 1

You're right that since they don't modify the actual state of the object itself, you can choose.

While there is no hard-and-fast rule, "use const wherever possible" is definitely not the universal way to go. Refer to functions like std::vector::operator[] —that doesn't change the members of the vector object, but stills provide a non- const version (and a different const version).

One good way of looking at this is: assume you have a BufferObject , and you pass it to a function which takes a const BufferObject& . Will it mess up your expectations (invariants you expect to hold) if that function calls dataStore() ? If so, do not mark dataStore() as const .

To address your particular case, I think you're correct and should leave these functions non- const . While they don't modify the physical contents of the C++ object, they do modify the logical state of the entity represented by that C++ object.