Dimension of Grid in CUDAFunctionLoad for Matrix Multiplication

Question

I want to do nonsquare matrix multiplication using CUDA The input is two matrices: a and b

a = RandomReal[{0, 5}, {100, 4}];
b = RandomReal[{0, 5}, {4, 6}];

m = Dimensions[a][[1]];
nn = Dimensions[a][[2]];
k = Dimensions[b][[2]];

cc = ConstantArray[0, {m, k}]; (*output*)

for this case, the output matrix will have dimensions, 100 x 6

the CUDA code:

    code = "
  __global__ void squarematrixmult(double *a, double *b, double *c, \
mint m, mint nn, mint k)
  { 
      int row = blockIdx.y * blockDim.y + threadIdx.y; 
      int col = blockIdx.x * blockDim.x + threadIdx.x;
      float sum = 0;
      if( col < k && row < m) 
      {
          for(int i = 0; i < nn; i++) 
          {
              sum += a[row * nn + i] * b[i * k + col];
          }
          c[row * k + col] = sum;
      }
  } ";

the computation:

cudaFun = 
  CUDAFunctionLoad[code, 
   "squarematrixmult", {{_Real, _, "Input"}, {_Real, _, 
     "Input"}, {_Real, _, 
     "Output"}, _Integer, _Integer, _Integer}, {32, 32}];
res = cudaFun[a, b, cc, m, nn, k];

However, since the defined block size is {32,32}, the given answer is only for 32 elements. elements after 32 are all zeros. for this case, the output is 100 x 6, but only 32 x 6 have values (correct values), but else are zeros.

So how can i fix the way to define threads/blocks/grids in CUDAFunctionLoad? thank you so much

Answer 1

it was long ago, maybe you already solved this issue, but you have to exchange the definitions of "col" and "row" and your code should be fine for any block size.

You probably took this formulas from

https://www.tutorialspoint.com/cuda/cuda_matrix_multiplication.htm

If you notice, they show this definition twice, you probably took the reversed one.