cudaDeviceReset用于多个gpu

Question

我目前正在开发一台拥有4台Tesla T10 gpu的gpu服务器.虽然我一直在测试内核并且必须经常使用ctrl-C终止进程,但我在简单设备查询代码的末尾添加了几行代码.代码如下:

#include <stdio.h>

 // Print device properties
 void printDevProp(cudaDeviceProp devProp)
{
    printf("Major revision number:         %d\n",  devProp.major);
    printf("Minor revision number:         %d\n",  devProp.minor);
    printf("Name:                          %s\n",  devProp.name);
    printf("Total global memory:           %u\n",  devProp.totalGlobalMem);
    printf("Total shared memory per block: %u\n",  devProp.sharedMemPerBlock);
    printf("Total registers per block:     %d\n",  devProp.regsPerBlock);
    printf("Warp size:                     %d\n",  devProp.warpSize);
    printf("Maximum memory pitch:          %u\n",  devProp.memPitch);
    printf("Maximum threads per block:     %d\n",  devProp.maxThreadsPerBlock);
    for (int i = 0; i < 3; ++i)
    printf("Maximum dimension %d of block:  %d\n", i, devProp.maxThreadsDim[i]);
    for (int i = 0; i < 3; ++i)
    printf("Maximum dimension %d of grid:   %d\n", i, devProp.maxGridSize[i]);
    printf("Clock rate:                    %d\n",  devProp.clockRate);
    printf("Total constant memory:         %u\n",  devProp.totalConstMem);
    printf("Texture alignment:             %u\n",  devProp.textureAlignment);
    printf("Concurrent copy and execution: %s\n",  (devProp.deviceOverlap ? "Yes" : "No"));
    printf("Number of multiprocessors:     %d\n",  devProp.multiProcessorCount);
    printf("Kernel execution timeout:      %s\n",  (devProp.kernelExecTimeoutEnabled ? "Yes" : "No"));
    return;
}

 int main()
{
    // Number of CUDA devices
    int devCount;
    cudaGetDeviceCount(&devCount);
    printf("CUDA Device Query...\n");
    printf("There are %d CUDA devices.\n", devCount);

    // Iterate through devices
    for (int i = 0; i < devCount; ++i)
    {
        // Get device properties
        printf("\nCUDA Device #%d\n", i);
        cudaDeviceProp devProp;
        cudaGetDeviceProperties(&devProp, i);
        printDevProp(devProp);
    }

    printf("\nPress any key to exit...");
    char c;
    scanf("%c", &c);

    **for (int i = 0; i < devCount; i++) {
        cudaSetDevice(i);
        cudaDeviceReset();
    }**

    return 0;
}

我的查询与main()结束之前的for循环有关,我在其中逐个设置每个设备,然后使用cudaResetDevice命令.我有一种奇怪的感觉,这段代码,虽然没有产生任何错误,但我无法重置所有设备.相反,程序每次仅重置默认设备,即设备0.谁能告诉我怎么做才能重置4个设备中的每一个.

谢谢

Answer 1

看起来您可以向GPU程序添加一个函数来捕获ctrl + c信号(SIGINT)并为程序使用的每个设备调用cudaDeviceReset()函数.

可以在此处找到捕获SIGINT时调用函数的示例代码:

/sf/answers/33790781/

对于你编写的每个GPU程序都包含这样的代码似乎是一种很好的做法,我也会这样做:-)

我没有时间写完整的详细答案,所以请阅读其他答案及其评论.

Answer 2

这可能已经太晚了，但是如果您编写一个信号处理函数，您可以消除内存泄漏并以某种确定的方式重置设备：

// State variables for 
extern int no_sigint;
int no_sigint = 1;
extern int interrupts;
int interrupts = 0;

/* Catches signal interrupts from Ctrl+c.
   If 1 signal is detected the simulation finishes the current frame and
   exits in a clean state. If Ctrl+c is pressed again it terminates the
   application without completing writes to files or calculations but
   deallocates all memory anyway. */
void
sigint_handler (int sig)
{
  if (sig == SIGINT)
    {
      interrupts += 1;
      std::cout << std::endl
                << "Aborting loop.. finishing frame."
                << std::endl;

      no_sigint = 0;

      if (interrupts >= 2)
        {
          std::cerr << std::endl
                    << "Multiple Interrupts issued: "
                    << "Clearing memory and Forcing immediate shutdown!"
                    << std::endl;

          // write a function to free dynamycally allocated memory
          free_mem ();

          int devCount;
          cudaGetDeviceCount (&devCount);

          for (int i = 0; i < devCount; ++i)
            {
              cudaSetDevice (i);
              cudaDeviceReset ();
            }
          exit (9);
        }
    }
}

....

int main(){ 
.....
for (int simulation_step=1 ; simulation_step < SIM_STEPS && no_sigint; ++simulation_step)
{
   .... simulation code
}
free_mem();
... cuda device resets
return 0;
}

如果您使用此代码（您甚至可以在外部标头中包含第一个代码段，它就可以工作。您可以对 ctrl+c 进行 2 级控制：第一次按下会停止模拟并正常退出，但应用程序会完成渲染该步骤很高兴优雅地停止并获得正确的结果，如果您再次按 ctrl+c，它将关闭应用程序并释放所有内存。