一个结构的 cudaMalloc 和一个相同结构的元素

A proper code would be something like this:

$ cat t363.cu
#include <stdio.h>

typedef struct {
  int *a;
  int foo;
} StructA;

__global__ void kernel(StructA *data){

  printf("The value is %d
", *(data->a + 2));
}

int main()
{
  int  numS = 1; // defined at runtime

  //allocate host memory for the structure storage
  StructA *h_A = (StructA*)malloc(numS * sizeof(StructA));
  //allocate host memory for the storage pointed to by the embedded pointer
  h_A->a = (int *)malloc(10*sizeof(int));
  // initialize data pointed to by the embedded pointer
  for (int i = 0; i <10; i++) *(h_A->a+i) = i;
  StructA *d_A;  // pointer for device structure storage
  //allocate device memory for the structure storage
  cudaMalloc( (void**)&(d_A), numS * sizeof(StructA) );
  // create a pointer for cudaMalloc to use for embedded pointer device storage
  int *temp;
  //allocate device storage for the embedded pointer storage
  cudaMalloc((void **)&temp, 10*sizeof(int));
  //copy this newly created *pointer* to it's proper location in the device copy of the structure
  cudaMemcpy(&(d_A->a), &temp, sizeof(int *), cudaMemcpyHostToDevice);
  //copy the data pointed to by the embedded pointer from the host to the device
  cudaMemcpy(temp, h_A->a, 10*sizeof(int), cudaMemcpyHostToDevice);

  kernel<<<1, 1>>>(d_A); // Passing pointer to StructA in device
  cudaDeviceSynchronize();
}
$ nvcc -arch=sm_20 -o t363 t363.cu
$ cuda-memcheck ./t363
========= CUDA-MEMCHECK
The value is 2
========= ERROR SUMMARY: 0 errors
$

您会注意到，我还没有解决您正在处理 StructA 数组(即 numS > 1)的情况，这将需要环形.我将留给您处理我在这里和我的 previous linked回答 看看您是否可以计算出该循环的详细信息.此外，为了清楚/简洁，我已经放弃了通常的 cuda 错误检查 但请在您的代码中使用它.最后，如果您还没有得出结论，这个过程(有时称为深拷贝操作")在普通 CUDA 中有些乏味.以前的建议是扁平化"此类结构(以便它们不包含指针)，但您也可以探索 cudaMallocManaged 即 CUDA 6 中的统一内存.