Question

我想编写一个矩阵乘法算法，基于CUDA的共享内存示例，即同时执行计算和数据加载。我的代码看起来像这样：

float As[BLOCK_SIZE][BLOCK_SIZE];
float Bs[BLOCK_SIZE][BLOCK_SIZE];
As[ty][tx] = A[aBegin + wA * ty + tx];
Bs[ty][tx] = B[bBegin + wB * ty + tx];
for (int a = aBegin, b = bBegin; a <= aEnd; a += aStep, b += bStep)
{
    __shared__ float A2s[BLOCK_SIZE][BLOCK_SIZE];
    __shared__ float B2s[BLOCK_SIZE][BLOCK_SIZE];
    A2s[ty][tx] = As[ty][tx];
    B2s[ty][tx] = Bs[ty][tx];
    __syncthreads();
    if (a+1 <= aEnd)
    {
        As[ty][tx] = A[a+1 + wA * ty + tx];
        Bs[ty][tx] = B[b+1 + wB * ty + tx]; 
    }
#pragma unroll
    for (int k = 0; k < BLOCK_SIZE; ++k)
    {
         Csub += A2s[ty][k] * B2s[k][tx];
    }   
    __syncthreads();
}

但它的工作速度比原始解决方案慢，因为第二次数据加载是按计算顺序执行的。我怎样才能并行？

Answer 1

您应避免将数据A和B移至本地数组As和Bs，即

As[ty][tx] = A[aBegin + wA * ty + tx];
Bs[ty][tx] = B[bBegin + wB * ty + tx];

您可以直接将它们移动到共享内存A2s和B2s，即

A2s[ty][tx] = A[aBegin + wA * ty + tx];
B2s[ty][tx] = B[bBegin + wB * ty + tx];

此外，数据加载

As[ty][tx] = A[a+1 + wA * ty + tx];
Bs[ty][tx] = B[b+1 + wB * ty + tx];

似乎未被开发。

最后，您应该将共享内存数组的声明移到for循环之外，并且还缺少对输出矩阵的最终赋值。

尝试类似：

__global__ void TiledMatrixMultiplicationKernel(float* A, float* B, float* C, int Width)
{
    __shared__float As[BLOCK_SIZE][BLOCK_SIZE];
    __shared__float Bs[BLOCK_SIZE][BLOCK_SIZE];
    int bx = blockIdx.x; int by = blockIdx.y;
    int tx = threadIdx.x; int ty = threadIdx.y;
    int Row = by * BLOCK_SIZE + ty;
    int Col = bx * BLOCK_SIZE + tx;
    float Csub = 0;
    for (int m = 0; m < Width/BLOCK_SIZE; ++m) {
    As[ty][tx] = A[Row*Width + (m*BLOCK_SIZE + tx)];
    Bs[ty][tx] = B[Col + (m*BLOCK_SIZE + ty)*Width];
    __syncthreads();
    for (int k = 0; k < BLOCK_SIZE; ++k) {
       Csub += As[ty][k] * Bs[k][tx];
       __syncthreads();
    }
    C[Row*Width+Col] = Csub;
}

共享内存的同时计算和数据加载：平铺矩阵 - 矩阵乘法的情况

1 个答案: