Cuda and OpenGL Interop

Question

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.

Acc

Answer 1

For an example of how to use CUDA-GL interop without having to re-map every frame, please refer to this example:

https://github.com/nvpro-samples/gl_cuda_interop_pingpong_st

Answer 2

As of CUDA 4.0, OpenGL interop is one-way. That means to do what you want (run a CUDA kernel that writes data to a GL buffer or texture image), you have to map the buffer to a device pointer, and pass that pointer to your kernel, as shown in your example.

As for your side note: cudaGraphicsResourceGetMappedPointer is called every time display() is called because cudaGraphicsMapResource is called every frame. Any time you re-map a resource you should re-get the mapped pointer, because it may have changed. Why re-map every frame? Well, OpenGL sometimes moves buffer objects around in memory, for performance reasons (especially in memory-intensive GL applications). If you leave the resource mapped all the time, it can't do this, and performance may suffer. I believe GL's ability and need to virtualize memory objects is also one of the reasons the current GL interop API is one-way (the GL is not allowed to move CUDA allocations around, and therefore you can't map a CUDA-allocated device pointer into a GL buffer object).