为什么cudaGLSetGLDevice失败，即使它在main函数的第一行中调用 [英] Why cudaGLSetGLDevice failed, even though it's called in the first line of main function

问题描述

我想使用OpenGL和CUDA之间的互操作性。我知道，正如一些教程说的，第一步是选择设备。但是，当我在主函数的第一行调用cudaGLSetGLDevice（0）时，程序退出时带有信息cudaSafeCall（）Runtime API错误36：在此过程中设备处于活动状态时无法设置。

I want to use interoperability between OpenGL and CUDA. I know, as some tutorials said, the first step is to choose device. However, when I called the cudaGLSetGLDevice(0) in the first line of the main function, the program exited with the information "cudaSafeCall() Runtime API error 36: cannot set while device is active in this process."

实际上，即使我在调用cudaGLSetDevice之前使用cudaDeviceProp和cudaChooseDevice，错误仍然存​​在。

Actually, even though I use cudaDeviceProp and cudaChooseDevice before calling cudaGLSetDevice, the error still exists.

相信我，我的电脑只有一个GPU，9800GT。我知道，cudaGLSetGLDevice的调用应该在任何其他CUDA函数之前，这就是为什么我把它放在main函数的第一行。

Believe me, my computer just has one GPU, 9800GT. And I do know that the calling of cudaGLSetGLDevice should be prior to any other CUDA function so that's why I put it in the first line of main function.

我使用Windows SDK来渲染OpenGL而不是glut，这是有问题的吗？

And I use the Windows SDK to render OpenGL instead of glut, is problem about this?

推荐答案

您需要在调用 cudaGLSetDevice 之前初始化opengl上下文。在glut glutInit（argc，argv）; 被调用的情况下。在使用任何cuda< - opengl互操作性函数调用之前初始化opengl上下文。

You need in opengl context initialized before call cudaGLSetDevice. In case of glut glutInit(argc, argv); is called. Initialize opengl context before use any cuda<->opengl interoperability functions call.

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