How to use 2D Arrays in CUDA?

I am new to CUDA. How to allocate 2D array of size MXN ?. How to traverse that array in CUDA?. Give me a sample code. ............................................................................................

Hi..Thanks for reply. I used your code in following program. But I am not getting correct results.

__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]);

      }
}

Please help me.

Solution 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

Solution 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