如何在CUDA中使用指针

Question

I have a problem with using of pointer to pointer in cuda. Code snippet is below.

char** d_ppcPtr, *d_pcPtr, *h_pcPtr;
cudaMalloc(&d_ppcPtr, sizeof(char*) * 10);

h_pcPtr = (char*)malloc(sizeof(char) * 100);
for(int i = 0; i < 10; i ++)
{
      cudaMalloc(&d_pcPtr, sizeof(char) * 100);
      cudaMemset(d_pcPtr, 1, sizeof(char) * 100);
      cudaMemcpy(&d_ppcPtr[i], &d_pcPtr, sizeof(char*), cudaMemcpyHostToDevice);
      cudaMemcpy(h_pcPtr, d_ppcPtr[i], sizeof(char) * 100, cudaMemcpyDeviceToHost); //crash here
      cudaFree(d_ppcPtr[i]); //crash also here
}
cudaFree(d_ppcPtr);

how can i fix above two crashes? Thanks in advance.

Answer 1

The following modification will "fix" your code (fully worked example, including host and device verification):

$ cat t583.cu
#include <stdio.h>

__global__ void testkernel(char **data, unsigned n){
  for (int i = 0; i < 100; i++) if (data[n][i] != 1) printf("kernel error\n");
}

int main(){
  char** d_ppcPtr, *d_pcPtr, *h_pcPtr;
  cudaMalloc(&d_ppcPtr, sizeof(char*) * 10);

  h_pcPtr = (char*)malloc(sizeof(char) * 100);
  for(int i = 0; i < 10; i ++)
  {
      cudaMalloc(&d_pcPtr, sizeof(char) * 100);
      cudaMemset(d_pcPtr, 1, sizeof(char) * 100);
      cudaMemcpy(&d_ppcPtr[i], &d_pcPtr, sizeof(char*), cudaMemcpyHostToDevice);
      memset(h_pcPtr, 0, sizeof(char)*100);
      testkernel<<<1,1>>>(d_ppcPtr, i);
      cudaMemcpy(h_pcPtr, d_pcPtr, sizeof(char) * 100, cudaMemcpyDeviceToHost);
      cudaFree(d_pcPtr);
      for (int i = 0; i < 100; i++) if (h_pcPtr[i] != 1) printf("Error!");
  }
  cudaFree(d_ppcPtr);
}
$ nvcc -arch=sm_20 -o t583 t583.cu
$ cuda-memcheck ./t583
========= CUDA-MEMCHECK
========= ERROR SUMMARY: 0 errors

Note that conceptually, there is no difference between my code and yours, because the pointer that you are attempting to use in location d_ppcPtr[i] , (and is crashing, because it is located on the device,) is already contained in d_pcPtr , which is on the host.