Question

Is there a way to invoke kernel failure using cuda-gdb? I've tried stepping through the kernel code and setting invalid index positions, odd values to variables, but I'm unable to trigger a "kernel Execution Failed" after continuing from an erroneous setting.

Does anyone know of a proper way to do this through cuda-gdb? I've read through the cuda-gdb documentation twice but might have missed some clues on how to achieve this if it is at all possible. If anyone knows of any tools/techniques that would be most appreciated, thanks.

I'm on CentOS 7 and my device's compute capability is 2.1. See below for the output of the uname -a command.

Linux john 3.10.0-327.10.1.el7.x86_64 #1 SMP Tue Feb 16 17:03:50 UTC 2016 x86_64 x86_64 x86_64 GNU/Linux

Answer 1

有没有办法使用cuda-gdb调用内核失败？

是的，这是可能的。这是一个完整的例子：

$ cat t678.cu
#include <stdio.h>
__global__ void kernel(int *data){

  int idx = 0;  // line 4
  idx += data[0];
  int tval = data[idx];
  data[1] =  tval;
}

int main(){

  int *d_data;
  cudaMalloc(&d_data, 32*sizeof(int));
  cudaMemset(d_data, 0, 32*sizeof(int));
  kernel<<<1,1>>>(d_data);
  cudaDeviceSynchronize();
  cudaError_t err = cudaGetLastError();
  if (err != cudaSuccess) printf("kernel fail %s\n", cudaGetErrorString(err));
}
$ nvcc -g -G -o t678 t678.cu
$ cuda-gdb ./t678
NVIDIA (R) CUDA Debugger
7.5 release
Portions Copyright (C) 2007-2015 NVIDIA Corporation
GNU gdb (GDB) 7.6.2
Copyright (C) 2013 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.  Type "show copying"
and "show warranty" for details.
This GDB was configured as "x86_64-unknown-linux-gnu".
For bug reporting instructions, please see:
<http://www.gnu.org/software/gdb/bugs/>...
Reading symbols from /home/user2/misc/t678...done.
(cuda-gdb) break t678.cu:4
Breakpoint 1 at 0x4026d5: file t678.cu, line 4.
(cuda-gdb) run
Starting program: /home/user2/misc/./t678
[Thread debugging using libthread_db enabled]
Using host libthread_db library "/lib64/libthread_db.so.1".
[New Thread 0x7ffff700a700 (LWP 8693)]
[Switching focus to CUDA kernel 0, grid 2, block (0,0,0), thread (0,0,0), device 0, sm 14, warp 2, lane 0]

Breakpoint 1, kernel<<<(1,1,1),(1,1,1)>>> (data=0x13047a0000) at t678.cu:4
4         int idx = 0;  // line 4
(cuda-gdb) step
5         idx += data[0];
(cuda-gdb) print idx
$1 = 0
(cuda-gdb) set idx=1000000
(cuda-gdb) step
6         int tval = data[idx];
(cuda-gdb) print idx
$2 = 1000000
(cuda-gdb) step

CUDA Exception: Device Illegal Address
The exception was triggered in device 0.

Program received signal CUDA_EXCEPTION_10, Device Illegal Address.
kernel<<<(1,1,1),(1,1,1)>>> (data=0x13047a0000) at t678.cu:7
7         data[1] =  tval;
(cuda-gdb)

在上面的cuda-gdb输出中，您可以看到在将idx变量设置为较大值后，执行以下行时会导致索引越界（非法地址）错误在调试器中：

  int tval = data[idx];

Invoke kernel failure through cuda-gdb?

1 个答案: