如何在 CUDA 中使用 2D 数组？

如何分配大小为 MXN 的二维数组？那么如何在 CUDA 中遍历该数组呢？

__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]);
    
      }
}

如何分配二维数组：

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
}

如何遍历该数组：

__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[][] */
}

希望对大家有帮助，也可以参考一下CUDA 编程指南关于矩阵乘法