Cuda and OpenGL Interop

I've been reading through the CUDA documentation and it seems to me, that every buffer that needs to interface with OpenGL needs to be created in the glBuffer.

According to the nvidia programming guide, this has to be done like this:

GLuint positionsVBO;
struct cudaGraphicsResource* positionsVBO_CUDA;

int main() {

    // Explicitly set device
    cudaGLSetGLDevice(0);
    // Initialize OpenGL and GLUT
    ...
    glutDisplayFunc(display);
    // Create buffer object and register it with CUDA
    glGenBuffers(1, positionsVBO);
    glBindBuffer(GL_ARRAY_BUFFER, &vbo);
    unsigned int size = width * height * 4 * sizeof(float);
    glBufferData(GL_ARRAY_BUFFER, size, 0, GL_DYNAMIC_DRAW);
    glBindBuffer(GL_ARRAY_BUFFER, 0);
    cudaGraphicsGLRegisterBuffer(&positionsVBO_CUDA, positionsVBO, cudaGraphicsMapFlagsWriteDiscard);

    // Launch rendering loop
    glutMainLoop();
}
void display() {
    // Map buffer object for writing from CUDA
    float4* positions;
    cudaGraphicsMapResources(1, &positionsVBO_CUDA, 0);
    size_t num_bytes;
    cudaGraphicsResourceGetMappedPointer((void**)&positions, &num_bytes, positionsVBO_CUDA));
    // Execute kernel
    dim3 dimBlock(16, 16, 1);
    dim3 dimGrid(width / dimBlock.x, height / dimBlock.y, 1);
    createVertices<<<dimGrid, dimBlock>>>(positions, time, width, height);
    // Unmap buffer object
    cudaGraphicsUnmapResources(1, &positionsVBO_CUDA, 0);
    // Render from buffer object
    glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
    glBindBuffer(GL_ARRAY_BUFFER, positionsVBO);
    glVertexPointer(4, GL_FLOAT, 0, 0);
    glEnableClientState(GL_VERTEX_ARRAY);
    glDrawArrays(GL_POINTS, 0, width * height);
    glDisableClientState(GL_VERTEX_ARRAY);
    // Swap buffers
    glutSwapBuffers();
    glutPostRedisplay();
}
void deleteVBO() {
    cudaGraphicsUnregisterResource(positionsVBO_CUDA);
    glDeleteBuffers(1, &positionsVBO);
}

__global