在CUDA设备中获取原始矩阵索引

Question

I am passing a vectorized representation of a 2D square matrix to a CUDA device. I have found online how to perform matrix multiplication with two matrices on this format on a CUDA device.

However, I now need to obtain the original indices of my matrix before the device.

This is my code to pass to my cuda_kernel

#define MATRIX_SIZE 20
#define BLOCK_SIZE 2
#define TILE_SIZE  2

void cuda_stuff(int sz, double **A)
{
  double* A1d = matrix_to_vector(sz, A);
  double* d_A
  size_t sizeA = sz * sz * sizeof(double);
  cudaMalloc(&d_A, sizeA);
  cudaMemcpy(d_A, A1d, sizeA, cudaMemcpyHostToDevice);
  dim3 threads(BLOCK_SIZE, BLOCK_SIZE);
  dim3 grid(MATRIX_SIZE / threads.x, MATRIX_SIZE / threads.y);
  cudakernel<<<grid, threads>>>(sz, d_A);
}

This is my cudakernel

__global__ void cudakernel(int sz, double* A_d);
{
  int tx = blockIdx.x * TILE_SIZE + threadIdx.x;
  int ty = blockIdx.y * TILE_SIZE + threadIdx.y;

  /* Need to get original i, j from my matrix double* A */
}

How can I get the original indices [i][j] of my matrix double* A ?

Answer 1

Your code will only work properly if MATRIX_SIZE is evenly divisible by BLOCK_SIZE (and BLOCK_SIZE must be the same as TILE_SIZE ). This code appears to be set up to handle square matrices only, so I am assuming your original A matrix is of size ( MATRIX_SIZE , MATRIX_SIZE ).

Given that proviso, the following should retrieve the original element A corresponding to a given thread:

double my_A_element  = A_d[ty*MATRIX_SIZE+tx];

if you prefer, (again, given the above proviso) you can use the built-in variables:

double my_A_element  = A_d[ty*(blockDim.x*gridDim.x)+tx];

or, equivalently:

double my_A_element  = A_d[ty*sz+tx];

Regarding the indices, the tx variable is properly defined to give you the original column index into A , and the ty variable is properly defined to give you the original row index into A , for the above defined my_A_element variables.

Therefore the original element of A (corresponding to my_A_element ) is just A[ty][tx]