Cuda program for matrix batch multiplication - c

Question

I am a novice in the field of CUDA program and I am trying to repeat the function of cublasSgemmBatched, which means that I want to perform the matrix-matrix multiplication of a batch of matrices. I try to implement my idea as the following code.
#include <stdio.h>
__global__ void BatchMulCUDA(float* array1, float* array2, int narray1, int dim, float* result)
{
int tx = blockIdx.x * blockDim.x + threadIdx.x;
if (tx < narray1 * dim)
{
float temp = 0;
int index = tx / dim;
#pragma
for (int i = 0; i < dim; i++)
{
temp += array1[tx * dim + i] * array2[index * dim + i];
}
result[tx] = temp;
}
}
void BatchMulGPU(float* array1, float* array2, int narray1, int dim, float* result)
{
dim3 threads(1024, 1);
dim3 grid(narray1 / 1024 + 1, 1);
int threadsPerBlock = threads.x * threads.y;
int blocksPerGrid = grid.x * grid.y;
printf("CUDA kernel launch with %d blocks of %d threads\n", blocksPerGrid, threadsPerBlock);
BatchMulCUDA<<<grid, threads>>>(array1, array2, narray1, dim, result);
}
However, strangely, I found that I can get the right output before the index 19730. After the element of 19730, the output of GPU is always 0. I do not know what the problem is. The CPU version of my code and test function are as the following. Is there any hardware limitation that I do not realize?
#include "kernel.h"
#include <cuda_runtime.h>
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <sys/time.h>
#include <math.h>
double cpuSecond()
{
struct timeval tp;
gettimeofday(&tp, NULL);
return ((double) tp.tv_sec + (double)tp.tv_usec*1e-6);
}
void BatchMulCPU(float* array1, float* array2, int narray1, int dim, float* result)
{
for (int i = 0; i < narray1 * dim; i++)
{
float temp = 0;
int index = i / dim;
for (int j = 0; j < dim; j++)
{
temp += array1[i * dim + j] * array2[index * dim + j];
}
result[i] = temp;
}
}
int main(int argc, char** argv)
{
int narray1 = 6980;
int dim = 4;
float* array1 = new float[narray1 * dim * dim];
float* array2 = new float[narray1 * dim];
float* resultGPU = new float[narray1 * dim];
float* resultCPU = new float[narray1 * dim];
float* d_array1;
float* d_array2;
float* d_result;
for (int i = 0; i < narray1 * dim * dim; i++)
{
array1[i] = static_cast<float> (rand() / (static_cast<float> (RAND_MAX / 10)));
}
for (int i = 0; i < narray1 * dim; i++)
{
array2[i] = static_cast<float> (rand() / (static_cast<float> (RAND_MAX / 10)));
}
cudaError_t err;
double iStart = cpuSecond();
err = cudaMalloc((void**)&d_array1, narray1 * dim * dim * sizeof(float));
err = cudaMalloc((void**)&d_array2, narray1 * dim * sizeof(float));
err = cudaMalloc((void**)&d_result, narray1 * dim * sizeof(float));
err = cudaMemcpy(d_array1, array1, narray1 * dim * dim * sizeof(float), cudaMemcpyHostToDevice);
err = cudaMemcpy(d_array2, array2, narray1 * dim * sizeof(float), cudaMemcpyHostToDevice);
BatchMulGPU(d_array1, d_array2, narray1, dim, d_result);
err = cudaMemcpy(resultGPU, d_result, narray1 * dim * sizeof(float), cudaMemcpyDeviceToHost);
double iElaps = cpuSecond() - iStart;
printf("Total GPU computation time is %lf \n" , iElaps);
iStart = cpuSecond();
BatchMulCPU(array1, array2, narray1, dim, resultCPU);
iElaps = cpuSecond() - iStart;
printf("Total CPU computation time is %lf \n" , iElaps);
float error = 0;
float temp = 0;
for (long i = 0; i < narray1 * dim; i++)
{
// temp = abs(resultCPU[i] - resultGPU[i]);
// if (temp > 0.5)
// {
// std::cout << i << std::endl;
// }
error += abs(resultCPU[i] - resultGPU[i]);
}
printf("Error is %f \n", error);
// for (int i = 19730; i < 19750; i++)
// {
// std::cout << "GPU " << resultGPU[i] << std::endl;
// std::cout << "CPU " << resultCPU[i] << std::endl;
// }
cudaFree(d_array1);
cudaFree(d_array2);
cudaFree(d_result);
return 0;
}

Answer 1

Apart from the possibility of a WDDM TDR timeout as discussed in the comments, the code has an error.
Its evident that the kernel design expects that a total grid size (total number of threads) will be launched that is equal to or greater than the number of arrays times the side dimension:
int tx = blockIdx.x * blockDim.x + threadIdx.x;
if (tx < narray1 * dim)
i.e. narray1*dim are the needed number of threads
However the number being launched is only narray1:
dim3 threads(1024, 1);
dim3 grid(narray1 / 1024 + 1, 1);
If we change the last line above to:
dim3 grid((narray1*dim) / 1024 + 1, 1);
this code design error will be addressed.
The reason the code works correctly for small number of matrices (anything up to 256) is because of the rounding-up effect in the grid sizing to a minimum of 1024 threads, which is 256*4 (narray1 * dim).
As an aside, this code is not functionally similar to cublasSgemmBatched from what I can see. I don't recognize this code as being any matrix multiplication (matrix dot product) that I am familiar with.