结构中指向数组的 cudaFree 指针上的 CUDA 分段错误

Question

我在 CUDA 设备上得到了一个结构，它包含一个指向数组的指针。计算、访问元素和一切正常，但是当我尝试成为一个好孩子并打电话时

cudaFree(my_struct->pointer_to_array)

我遇到了分段错误。 cudaFree(my_struct) 但是工作得很好。有什么我遗漏的吗？

请找到以下最小示例：

#include <stdio.h>

#include <cuda.h>
#include <cuda_runtime.h>
#include <cassert>

typedef struct {
  int n;
  float *arr;
} DummyStruct;

__global__ void check(DummyStruct *d) {
  printf("EL %f", d->arr[0]);
}

int main() {
  cudaError_t status;

  // create host pointer to dummy struct
  DummyStruct *dummy;
  dummy = (DummyStruct *)malloc(sizeof(DummyStruct));

  int arr_size = 32;

  dummy->n = 0;
  float *arr = (float *) malloc(sizeof(float) * arr_size);

  for (int i=0; i < 32; i++) {
    arr[i] = i;
  }

  // allocate device array
  float *d_arr;
  status = cudaMalloc(&d_arr, arr_size * sizeof(float));
  assert( status == cudaSuccess );

  status = cudaMemcpy(d_arr, arr, arr_size * sizeof(float), cudaMemcpyHostToDevice);
  assert( status == cudaSuccess );

  free(arr);

  // for some reason this should happen here and not d_sp->coeff = d_coeff ...
  dummy->arr = d_arr;

  // allocate and ship struct to device
  DummyStruct* d_dummy;
  status = cudaMalloc(&d_dummy, sizeof(DummyStruct));
  assert( status == cudaSuccess );

  status = cudaMemcpy(d_dummy, dummy, sizeof(DummyStruct), cudaMemcpyHostToDevice);
  assert( status == cudaSuccess );

  // free host struct
  free(dummy);


  // check whether array access works
  check<<<1, 1>>>(d_dummy);


  // THIS causes Segmentation fault (core dumped)
  status = cudaFree(d_dummy->arr);
  assert( status == cudaSuccess );

  status = cudaFree(d_dummy);
  assert( status == cudaSuccess );
}

Answer 1

此声明：

status = cudaFree(d_dummy->arr);

需要在 host 代码中取消引用 device 指针（d_dummy - 使用设备分配器分配，即 cudaMalloc）。这在 CUDA 中是非法的。

既然您已经知道 (d_dummy->arr) == d_arr，那么释放嵌入指针的一种可能方法是：

status = cudaFree(d_arr);

类似的概念（取消引用主机代码中的设备指针）是此处注释的基础：

// for some reason this should happen here and not d_sp->coeff = d_coeff ...