Question

我是CUDA的新手。如何分配大小为MXN的2D数组？如何在CUDA中遍历该数组？给我一个示例代码。 .................................................. ..........................................

嗨..谢谢你的回复。我在下面的程序中使用了你的代码。但我得不到正确的结果。

__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]);

      }
}

请帮帮我。

Answer 1

如何分配2D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[][] */
}

我希望这很有用

您也可以参考有关Matrix Multiplication的CUDA Programming Guide第22页

Answer 2

最好的方法是以矢量形式存储二维数组A. 例如，你有一个矩阵A大小为nxm，它指向指针表示的那个（i，j）元素将是

A[i][j] (with i=0..n-1 and j=0..m-1).

在矢量形式中，您可以编写

A[i*n+j] (with i=0..n-1 and j=0..m-1).

在这种情况下使用一维数组将简化复制过程，这很简单：

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

如何在CUDA中使用2D数组？

2 个答案: