How to use 2D Arrays in CUDA? - cuda

Question

How to allocate a 2D array of size MXN? And how to traverse that array in CUDA?
__global__ void test(int A[BLOCK_SIZE][BLOCK_SIZE], int B[BLOCK_SIZE][BLOCK_SIZE],int C[BLOCK_SIZE][BLOCK_SIZE])
{
int i = blockIdx.y * blockDim.y + threadIdx.y;
int j = blockIdx.x * blockDim.x + threadIdx.x;
if (i < BLOCK_SIZE && j < BLOCK_SIZE)
C[i][j] = A[i][j] + B[i][j];
}
int main()
{
int d_A[BLOCK_SIZE][BLOCK_SIZE];
int d_B[BLOCK_SIZE][BLOCK_SIZE];
int d_C[BLOCK_SIZE][BLOCK_SIZE];
int C[BLOCK_SIZE][BLOCK_SIZE];
for(int i=0;i<BLOCK_SIZE;i++)
for(int j=0;j<BLOCK_SIZE;j++)
{
d_A[i][j]=i+j;
d_B[i][j]=i+j;
}
dim3 dimBlock(BLOCK_SIZE, BLOCK_SIZE);
dim3 dimGrid(GRID_SIZE, GRID_SIZE);
test<<<dimGrid, dimBlock>>>(d_A,d_B,d_C);
cudaMemcpy(C,d_C,BLOCK_SIZE*BLOCK_SIZE , cudaMemcpyDeviceToHost);
for(int i=0;i<BLOCK_SIZE;i++)
for(int j=0;j<BLOCK_SIZE;j++)
{
printf("%d\n",C[i][j]);
}
}

Answer 1

How to allocate 2D array:
int main(){
#define BLOCK_SIZE 16
#define GRID_SIZE 1
int d_A[BLOCK_SIZE][BLOCK_SIZE];
int d_B[BLOCK_SIZE][BLOCK_SIZE];
/* d_A initialization */
dim3 dimBlock(BLOCK_SIZE, BLOCK_SIZE); // so your threads are BLOCK_SIZE*BLOCK_SIZE, 256 in this case
dim3 dimGrid(GRID_SIZE, GRID_SIZE); // 1*1 blocks in a grid
YourKernel<<<dimGrid, dimBlock>>>(d_A,d_B); //Kernel invocation
}
How to traverse that array:
__global__ void YourKernel(int d_A[BLOCK_SIZE][BLOCK_SIZE], int d_B[BLOCK_SIZE][BLOCK_SIZE]){
int row = blockIdx.y * blockDim.y + threadIdx.y;
int col = blockIdx.x * blockDim.x + threadIdx.x;
if (row >= h || col >= w)return;
/* whatever you wanna do with d_A[][] and d_B[][] */
}
i hope this is helpful
and also you can refer to CUDA Programming Guide page 22 about Matrix Multiplication

Answer 2

The best way would be storing a two-dimensional array A in its vector form.
For example you have a matrix A size nxm, and it's (i,j) element in pointer to pointer representation will be
A[i][j] (with i=0..n-1 and j=0..m-1).
In a vector form you can write
A[i*n+j] (with i=0..n-1 and j=0..m-1).
Using one-dimensional array in this case will simplify the copy process, which would be simple:
double *A,*dev_A; //A-hous pointer, dev_A - device pointer;
A=(double*)malloc(n*m*sizeof(double));
cudaMalloc((void**)&dev_A,n*m*sizeof(double));
cudaMemcpy(&dev_A,&A,n*m*sizeof(double),cudaMemcpyHostToDevice); //In case if A is double