cudaDeviceReset为多个gpu的 [英] cudaDeviceReset for multiple gpu's

问题描述

我目前正在使用一个gpu服务器，它有4 Tesla T10 gpu的。虽然我不断测试内核，并经常使用ctrl-C杀死进程，我在一个简单的设备查询代码的末尾添加了几行。代码如下：

I am currently working on a gpu server which has 4 Tesla T10 gpu's. While I keep testing the kernels and have to frequently kill the processes using ctrl-C, I added a few lines to the end of a simple device query code. The code is given below :

#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我设置每个设备一个一个，然后使用cudaResetDevice命令。我得到一个奇怪的感觉，这个代码，虽然不产生任何错误，但我不能重置所有的设备。相反，程序每次仅重置默认设备，即设备0。任何人都可以告诉我应该如何重置4个设备。

My query is related to the for loop just before the main() ends in which I set each device one by one and then use cudaResetDevice command. I get a strange feeling that this code, although doesnt produce any error but I am not able to reset all the devices. Instead, the program is resetting only the default device i.e device 0 each time. Can anyone tell me what should I do to reset each of the 4 devices.

感谢

推荐答案

这可能太晚了，但如果你写一个信号处理函数，你可以摆脱内存泄漏，并以一种确定的方式重置设备：

This is probably too late but if you write a signal-handler function you can get rid of the memory leaks and reset the device in a sure way:

// 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);
        }
    }
}

p>

....

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（您甚至可以将第一个代码段包含在外部你可以有ctrl + c的两个级别的控制：第一次按停止你的模拟并正常退出，但应用程序完成渲染的步骤是伟大的停止优雅和正确的结果，如果你按ctrl + c它再次关闭应用程序释放所有内存。

If you use this code (you can even include the first snippet in an external header, it works. You can have 2 levels of control of ctrl+c: the first press stops your simulation and exits normally but the application finishes rendering the step which is great to stop gracefully and have correct results, if you press ctrl+c again it closes the application freeing all memory.

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