Question

您好我正在CUDA内核中处理矩阵矢量产品。我想通过平铺和共享内存来提高性能。问题是，使用此代码，M矩阵或N向量不能正确加载。

您是否知道如何将M和N中的切片加载到共享内存数组中？

M是矩阵，N是向量，P是矩阵向量积的结果

    __global__ void matrixMul( float* P, float* M, float* N, int Mw, int Nw)
{
int bx = blockIdx.x;     int by = blockIdx.y;
int tx = threadIdx.x;    int ty = threadIdx.y;
__shared__ float Ms[BLOCK_SIZE][BLOCK_SIZE];
__shared__ float Ns[BLOCK_SIZE];

// ===================================================================
// Code segment 1
// Determine the update values for the tile indices in the loop
// ===================================================================

int mBegin = Mw * BLOCK_SIZE * by;
int mEnd   = mBegin + Mw - 1;
int mStep  = BLOCK_SIZE;
int nBegin = BLOCK_SIZE * bx;
//int nStep  = BLOCK_SIZE*Nw;
int nStep = 1;
float Psub = 0.0f;

// ===================================================================
// Code segment 2
// Do matrix-matrix multiplication inside a tile
// ===================================================================

for (int m = mBegin, n = nBegin; m <= mEnd; m += mStep, n += nStep) {

    // Load a tile from M and N into the shared memory arrays
    Ms[ty][tx] = M[bx*mStep*Mw+m];
    Ns[ty] = N[by*nStep*Nw+n];

    // Synchronize the threads
    __syncthreads();

    // Multiply the two tiles together, each thread accumulating
    // the partial sum of a single dot product.
    for (int i = 0; i < BLOCK_SIZE; i++) {
        Psub += Ms[i][tx] * Ns[i];
    }

    // Synchronize again.
    __syncthreads();
}

// ===================================================================
// Code segment 3
// Store the data back to global memory
// ===================================================================

int p = Nw * BLOCK_SIZE * by + BLOCK_SIZE * bx;
P[p + nStep] = Psub;
}

Answer 1

我发现了一个类似的例子（处理相同大小的方形矩阵，请注意），它还将矩阵的一部分加载到共享内存中。看来你的声明是正确的，它可能只取决于你用来确定哪些元素去哪里的代数。

__global__ void MatrixMulKernel(float* Md, float* Nd, float* Pd, int Width){

    __shared__float Mds[TILE_WIDTH][TILE_WIDTH];  // Shared memory
    __shared__float Nds[TILE_WIDTH][TILE_WIDTH];  //   declarations

   int bx = blockIdx.x; int by = blockIdx.y; // ID thread
   int tx = threadIdx.x; int ty = threadIdx.y;

   // Identify the row and column of the Pd element to work on

   int Row = by * TILE_WIDTH + ty;
   int Col = bx * TILE_WIDTH + tx;

   float Pvalue = 0; // REGISTER!

   // Loop over the Md and Nd tiles required to compute the Pd element
   for (int m = 0; m < Width/TILE_WIDTH; ++m) { 
       // Collaborative loading of Md and Nd tiles into shared memory
        Mds[ty][tx] = Md[Row*Width + (m*TILE_WIDTH + tx)];
        Nds[ty][tx] = Nd[Col + (m*TILE_WIDTH + ty)*Width];

        __syncthreads();

        for (int k = 0; k < TILE_WIDTH; ++k)
           Pvalue +=  Mds[ty][k] * Nds[k][tx];

        __syncthreads();
   }
Pd[Row*Width+Col] = Pvalue;
}

矩阵向量产品CUDA通过平铺和共享内存提高性能

1 个答案: