Question

[Q1]在执行结构成员cudaMalloc时，是否需要转换为（void **）？示例（请参阅代码问题）：

结构：

typedef struct {
  int a;
  int *b;
} Matrix;

分配和复制到设备的主要功能：

int main void() 
{
  int rows, cols, numMat = 2;

  //[Q2] What would be the problem of no allocating (numMat * sizeof()) here?
  //for example, allocating just for sizeof(Matrix)?
  Matrix *data = (Matrix*)malloc(numMat * sizeof(Matrix));

  // ... Successfully read from file into "data" ...
  //[Q3] Do we really need to copy "data" to host?
  //[A3] No necessary
  Matrix *h_data = (Matrix*)malloc(numMat * sizeof(Matrix));
  memcpy(h_data, data, numMat * sizeof(Matrix));

  // ... Copy matrix data is now on the gpu ...
  //[Q4] Do we need to cast (void**)&(h_data->a)? 'a' not a pointer.
  //[A4] An int cannot be copied in this fashion
  // cudaMalloc(&(h_data->a), rows*cols*sizeof(int));
  // cudaMemcpy(h_data->a, data->a, rows*cols*sizeof(int), cudaMemcpyHostToDevice);

  //[Q5] Do we need to cast (void**)&(h_data->b)? 'b' is a pointer
  cudaMalloc(&(h_data->b), rows*cols*sizeof(int));
  cudaMemcpy(h_data->b, data->b, rows*cols*sizeof(int), cudaMemcpyHostToDevice);

  // ... Copy the "meta" data to gpu ...
  //[Q6] Can we just copy h_data instead? Why creating another pointer "d_data"?
  //[A6] Yes
  Matrix *d_data;

  //[Q7] Wouldn't we need to cast (void**)&d_data?
  cudaMalloc(&d_data, numMat*sizeof(Matrix));

  //[Q8] h_data is in host and device. Can we just copy "data" to device?
  cudaMemcpy(d_data, h_data, numMat*sizeof(Matrix));
  // ... Do other things ...
}

最终，我们只想将Matrix作为指针传递：

// Kernel call
doThings<<<dimGrid, dimBlock>>>(d_data);

内核定义：

__global__ doThings(Matrix *matrices)
{
  matrices->a = ...;
  matrices->b = ...;
}

提前感谢您的时间和工作，帮助我解决疑惑！

cudaMalloc cast（void **）和struct member allocation / copy和sizeof

0 个答案: