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我正在尝试让自己熟悉 CUDA 编程，并且玩得很开心。我目前正在查看这个处理矩阵乘法的 pdf，无论是否使用共享内存。两个版本的完整代码可以在这里找到。此代码与 CUDA 矩阵乘法示例中的代码几乎完全相同。尽管非共享内存版本能够以任何矩阵大小运行，无论块大小如何，共享内存版本必须使用块大小的倍数的矩阵（我设置为 4，默认值最初为 16） .

pdf末尾建议的问题之一是对其进行更改，以便共享内存版本也可以与块大小的非倍数一起使用。我认为这将是一个简单的索引检查，就像在非共享版本中一样：

int row = blockIdx.y * blockDim.y + threadIdx.y;
int col = blockIdx.x * blockDim.x + threadIdx.x;
if(row > A.height || col > B.width) return;

但这不起作用。这是完整的代码，减去 main 方法（有点乱，抱歉），我对它进行了一些修改：

void MatMul(const Matrix A, const Matrix B, Matrix C) { 
  // Load A and B to device memory 
  Matrix d_A; 
  d_A.width = d_A.stride = A.width; 
  d_A.height = A.height; 
  size_t size = A.width * A.height * sizeof(float); 
  cudaError_t err = cudaMalloc(&d_A.elements, size); 
  printf("CUDA malloc A: %s\n",cudaGetErrorString(err)); 
  err = cudaMemcpy(d_A.elements, A.elements, size, cudaMemcpyHostToDevice); 
  printf("Copy A to device: %s\n",cudaGetErrorString(err)); 

  Matrix d_B; 
  d_B.width = d_B.stride = B.width; 
  d_B.height = B.height; 
  size = B.width * B.height * sizeof(float); 
  err = cudaMalloc(&d_B.elements, size); 
  printf("CUDA malloc B: %s\n",cudaGetErrorString(err));
  err = cudaMemcpy(d_B.elements, B.elements, size, cudaMemcpyHostToDevice);
  printf("Copy B to device: %s\n",cudaGetErrorString(err)); 

  Matrix d_C; 
  d_C.width = d_C.stride = C.width; 
  d_C.height = C.height; 
  size = C.width * C.height * sizeof(float); 
  err = cudaMalloc(&d_C.elements, size); 
  printf("CUDA malloc C: %s\n",cudaGetErrorString(err));

  dim3 dimBlock(BLOCK_SIZE, BLOCK_SIZE); 
    dim3 dimGrid((B.width + dimBlock.x - 1) / dimBlock.x, (A.height + dimBlock.y-1) / dimBlock.y);
    MatMulKernel<<<dimGrid, dimBlock>>>(d_A, d_B, d_C); 
    err = cudaThreadSynchronize();
    printf("Run kernel: %s\n", cudaGetErrorString(err));

  // Read C from device memory 
  err = cudaMemcpy(C.elements, d_C.elements, size, cudaMemcpyDeviceToHost); 
  printf("Copy C off of device: %s\n",cudaGetErrorString(err));

  // Free device memory
  cudaFree(d_A.elements); 
  cudaFree(d_B.elements); 
  cudaFree(d_C.elements); 
} 

// Get a matrix element
__device__ float GetElement(const Matrix A, int row, int col) { 
  return A.elements[row * A.stride + col]; 
} 

// Set a matrix element 
__device__ void SetElement(Matrix A, int row, int col, float value) { 
  A.elements[row * A.stride + col] = value; 
} 

// Get the BLOCK_SIZExBLOCK_SIZE sub-matrix Asub of A that is 
// located col sub-matrices to the right and row sub-matrices down 
// from the upper-left corner of A 
__device__ Matrix GetSubMatrix(Matrix A, int row, int col) { 
  Matrix Asub; 
  Asub.width = BLOCK_SIZE; 
  Asub.height = BLOCK_SIZE; 
  Asub.stride = A.stride; 
  Asub.elements = &A.elements[A.stride * BLOCK_SIZE * row + BLOCK_SIZE * col]; 
  return Asub; 
}


// Matrix multiplication kernel called by MatMul() 
__global__ void MatMulKernel(Matrix A, Matrix B, Matrix C) { 
  // Block row and column 
  int blockRow = blockIdx.y; 
  int blockCol = blockIdx.x; 

  int rowTest = blockIdx.y * blockDim.y + threadIdx.y;
  int colTest = blockIdx.x * blockDim.x + threadIdx.x;
  if (rowTest>A.height || colTest>B.width)
    return;
  // Each thread block computes one sub-matrix Csub of C
  Matrix Csub = GetSubMatrix(C, blockRow, blockCol); 

  // Each thread computes one element of Csub 
  // by accumulating results into Cvalue 
  float Cvalue = 0.0; 
  // Thread row and column within Csub 
  int row = threadIdx.y; 
  int col = threadIdx.x; 
  // Loop over all the sub-matrices of A and B that are 
  // required to compute Csub 
  // Multiply each pair of sub-matrices together 
  // and accumulate the results 
  for (int m = 0; m < (BLOCK_SIZE + A.width - 1)/BLOCK_SIZE; ++m) {
    // Get sub-matrix Asub of A 
    Matrix Asub = GetSubMatrix(A, blockRow, m); 

    // Get sub-matrix Bsub of B 
    Matrix Bsub = GetSubMatrix(B, m, blockCol); 

    // Shared memory used to store Asub and Bsub respectively 
    __shared__ float As[BLOCK_SIZE][BLOCK_SIZE]; 
    __shared__ float Bs[BLOCK_SIZE][BLOCK_SIZE]; 

    // Load Asub and Bsub from device memory to shared memory 
    // Each thread loads one element of each sub-matrix 
    As[row][col] = GetElement(Asub, row, col); 
    Bs[row][col] = GetElement(Bsub, row, col); 

    // Synchronize to make sure the sub-matrices are loaded 
    // before starting the computation 
    __syncthreads(); 

    // Multiply Asub and Bsub together 
    for (int e = 0; e < BLOCK_SIZE; ++e) 
    {
      Cvalue += As[row][e] * Bs[e][col];
    }
    // Synchronize to make sure that the preceding 
    // computation is done before loading two new 
    // sub-matrices of A and B in the next iteration 
    __syncthreads();  
  }
  // Write Csub to device memory 
  // Each thread writes one element 
  SetElement(Csub, row, col, Cvalue); 
}

我改变的值得注意的事情：我在 MatMulKernel 中添加了一个检查，检查我们当前的线程是否正在尝试在不存在的 C 中工作。这似乎不起作用。虽然它确实改变了结果，但这些变化似乎没有任何模式，除了后来（更高的 x 或 y 值）条目似乎受到更大的影响（而且我得到了更多的非整数结果）。我还更改了给定的 dimGrid 计算方法和 MatMulKernel 中 m 的循环条件（之前它只是宽度或高度除以块大小，这似乎是错误的）。

甚至我为本指南找到的解决方案指南似乎也暗示它应该只是一个简单的索引检查，所以我认为我错过了一些真正基本的东西。

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1 回答 1

28

当矩阵维度不是瓦片维度的倍数时，可能会发生一些瓦片仅部分覆盖矩阵的情况。落在不完全重叠的瓦片之外的瓦片元素应正确归零。因此，将您的代码扩展到任意大小的矩阵很容易，但并不意味着简单的索引检查。下面，我正在复制并粘贴我的平铺矩阵-矩阵乘法内核版本，其中包含任意大小的矩阵

__global__ void MatMul(float* A, float* B, float* C, int ARows, int ACols, int BRows,
    int BCols, int CRows, int CCols)
{
    float CValue = 0;

    int Row = blockIdx.y*TILE_DIM + threadIdx.y;
    int Col = blockIdx.x*TILE_DIM + threadIdx.x;

    __shared__ float As[TILE_DIM][TILE_DIM];
    __shared__ float Bs[TILE_DIM][TILE_DIM];

    for (int k = 0; k < (TILE_DIM + ACols - 1)/TILE_DIM; k++) {

         if (k*TILE_DIM + threadIdx.x < ACols && Row < ARows)
             As[threadIdx.y][threadIdx.x] = A[Row*ACols + k*TILE_DIM + threadIdx.x];
         else
             As[threadIdx.y][threadIdx.x] = 0.0;

         if (k*TILE_DIM + threadIdx.y < BRows && Col < BCols)
             Bs[threadIdx.y][threadIdx.x] = B[(k*TILE_DIM + threadIdx.y)*BCols + Col];
         else
             Bs[threadIdx.y][threadIdx.x] = 0.0;

         __syncthreads();

         for (int n = 0; n < TILE_DIM; ++n)
             CValue += As[threadIdx.y][n] * Bs[n][threadIdx.x];

         __syncthreads();
    }

    if (Row < CRows && Col < CCols)
        C[((blockIdx.y * blockDim.y + threadIdx.y)*CCols) +
           (blockIdx.x * blockDim.x)+ threadIdx.x] = CValue;
}
于 2013-09-17T17:21:02.667 回答